Concise Mathematics Selina Solutions Class 10 ICSE Chapter 25 Probability

This chapter provides the complete ICSE Class 10 Selina Maths Probability Solutions, helping you master the principles of chance and likelihood. In your Class 10 Concise Mathematics Selina textbook, the chapter on Probability introduces you to the fundamental concepts of predicting outcomes in random experiments. You will learn how to define a sample space, identify favourable outcomes, and calculate the probability of an event occurring. We will work through classic examples involving coins, dice, and playing cards, which are very common in board exams. Understanding these core ideas is crucial for solving complex problems and for building a strong foundation in statistics.

If you are stuck on a specific question about complementary events or calculating the odds from a deck of cards, you have come to the right place. This page is designed to help you verify your methods and understand the logic behind each answer. We provide clear, step-by-step solutions for all 76 questions found in Exercise 25(A), Exercise 25(B), and the Test Yourself section. Each solution is crafted to follow the exact method prescribed by the ICSE board, ensuring you learn the correct way to present your answers in examinations. Here you will find reliable and easy-to-follow solutions for every problem in the chapter.

Exercise 25(A)

Question 1(a)

If A and B are two complementary events then the relation between P(A) and P(B) is :

• (a) P(A) = P(B)
• (b) P(A) + P(B) = 0
• (c) P(A) + P(B) = 1
• (d) none of these

Consider that for any two complementary events, the sum of their probabilities is always equal to 1. This is a fundamental principle in probability theory.

∴ P(A) + P(B) = 1

Hence, Option 3 is the correct option.

Question 1(b)

Out of the vowels of English alphabet, one letter is selected at random. The probability of selecting the letter ‘O’ is :

• (a) 1
• (b) \dfrac{4}{5}
• (c) \dfrac{1}{26}
• (d) \dfrac{1}{5}

The vowels in the English alphabet are: a, e, i, o, u.

Here, the letter ‘O’ is just one among these vowels.

∴ The number of favourable outcomes = 1

The probability of selecting ‘O’ is calculated as:

P(selecting the letter ‘O’) = \dfrac{\text{No. of favourable outcomes}}{\text{Total no. of possible outcomes}} = \dfrac{1}{5}.

Hence, Option 4 is the correct option.

Question 1(c)

When a die is thrown, the probability of getting an even number greater than 4 is :

• (a) \dfrac{1}{3}
• (b) \dfrac{2}{3}
• (c) \dfrac{1}{6}
• (d) \dfrac{11}{2}

When you roll a die, the outcomes you can get are 1, 2, 3, 4, 5, and 6. We are interested in finding an even number that is greater than 4. Here, the only number that fits this criterion is 6.

∴ The number of favourable outcomes is 1.

Now, calculate the probability of this event:

P(getting an even number greater than 4) = \dfrac{\text{No. of favourable outcomes}}{\text{Total no. of possible outcomes}} = \dfrac{1}{6}.

Hence, Option 3 is the correct option.

Question 1(d)

If a letter is drawn from the letters of English alphabet, then the probability, that it is a letter of the word ‘DELHI’ is :

• (a) \dfrac{1}{5}
• (b) \dfrac{5}{26}
• (c) \dfrac{1}{26}
• (d) \dfrac{21}{26}

Consider the total number of letters in the English alphabet, which is 26.

The letters that make up the word ‘DELHI’ are ‘D’, ‘E’, ‘L’, ‘H’, and ‘I’.

∴ The number of favourable outcomes, which are the letters from ‘DELHI’, is 5.

Now, the probability of selecting one of these letters when drawing from the entire alphabet is given by:

Hence, Option 2 is the correct option.

Question 1(e)

A card is selected at random from a well-shuffled deck of 52 cards. The probability of it being a face card is :

• (a) \dfrac{3}{13}
• (b) \dfrac{4}{13}
• (c) \dfrac{6}{13}
• (d) \dfrac{8}{13}

In a standard deck of 52 cards, there are 12 face cards (which are the Jacks, Queens, and Kings from each of the four suits).

The probability of drawing a face card from the deck is determined by dividing the number of face cards by the total number of cards. Thus, the probability is given by:

Hence, Option 1 is the correct option.

Question 2

A coin is tossed once. Find the probability of :

(i) getting a tail

(ii) not getting a tail

When you toss a coin once, there are 2 potential outcomes: Head (H) and Tail (T).

(i) For the event of ‘getting a tail’, the favourable outcome is one.

∴ The number of favourable outcomes = 1.

The probability of getting a tail is given by:

P(getting a tail) = \dfrac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \dfrac{1}{2}.

Hence, the probability of getting a tail = \dfrac{1}{2}.

(ii) For the event of ‘not getting a tail’, which is essentially ‘getting a head’, the favourable outcome is also one.

∴ The number of favourable outcomes = 1.

The probability of not getting a tail is given by:

P(not getting a tail) = \dfrac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \dfrac{1}{2}.

Hence, the probability of not getting a tail = \dfrac{1}{2}.

Question 3

A bag contains 3 white, 5 black and 2 red balls, all of the same shape and size. A ball is drawn from the bag without looking into it, find the probability that the ball drawn is :

(i) a black ball.

(ii) a red ball.

(iii) a white ball.

(iv) not a red ball.

(v) not a black ball.

The bag contains a total of 3 white, 5 black, and 2 red balls, adding up to 10 balls altogether. Thus, the total number of possible outcomes is 10.

(i) The bag holds 5 black balls.

∴ The favourable outcomes for drawing a black ball are 5.

The probability of drawing a black ball is given by:

Hence, probability of getting a black ball = \dfrac{1}{2}.

(ii) There are 2 red balls in the bag.

∴ The favourable outcomes for drawing a red ball are 2.

The probability of drawing a red ball is:

Hence, probability of getting a red ball = \dfrac{1}{5}.

(iii) The bag contains 3 white balls.

∴ The favourable outcomes for drawing a white ball are 3.

The probability of drawing a white ball is:

Hence, probability of getting a white ball = \dfrac{3}{10}.

(iv) Since there are 2 red balls, there are 8 balls that are not red (10 – 2).

∴ The favourable outcomes for not drawing a red ball are 8.

The probability of not drawing a red ball is:

Hence, the probability of not getting a red ball = \dfrac{4}{5}.

(v) The bag has 3 white and 2 red balls, totalling 5 balls that are not black.

∴ The favourable outcomes for not drawing a black ball are 5.

The probability of not drawing a black ball is:

Hence, the probability of not getting a black ball = \dfrac{1}{2}.

Question 4

In a single throw of a dice, find the probability of getting a number :

(i) greater than 4.

(ii) less than or equal to 4.

(iii) not greater than 4.

Consider the sample space for a single throw of a die: {1, 2, 3, 4, 5, 6}.

This means there are 6 possible outcomes.

(i) For numbers greater than 4, we have the outcomes {5, 6}.

∴ The number of favourable outcomes is 2.

Therefore, the probability of rolling a number greater than 4 is given by:
P(\text{number > 4}) = \dfrac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \dfrac{2}{6} = \dfrac{1}{3}.

Thus, the probability of getting a number greater than 4 is \dfrac{1}{3}.

(ii) For numbers less than or equal to 4, the possible outcomes are {1, 2, 3, 4}.

∴ The number of favourable outcomes is 4.

So, the probability of rolling a number less than or equal to 4 is:
P(\text{number ≤ 4}) = \dfrac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \dfrac{4}{6} = \dfrac{2}{3}.

Thus, the probability of getting a number less than or equal to 4 is \dfrac{2}{3}.

(iii) For numbers not greater than 4, again the outcomes are {1, 2, 3, 4}.

∴ The number of favourable outcomes is 4.

Thus, the probability of rolling a number not greater than 4 is:
P(\text{number not > 4}) = \dfrac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \dfrac{4}{6} = \dfrac{2}{3}.

Therefore, the probability of getting a number not greater than 4 is \dfrac{2}{3}.

Question 5

From a well shuffled deck of 52 cards, one card is drawn. Find the probability that the card drawn will :

(i) be a black card.

(ii) not be a red card.

(iii) be a red card.

(iv) be a face card.

(v) be a face card of red colour.

To solve this, let’s first recall that a standard deck has 52 cards.

For each part, we’ll determine the probability by dividing the number of favorable outcomes by the total number of outcomes, which is 52.

(i) In a deck, there are 26 black cards (13 spades and 13 clubs).

∴ Number of favorable outcomes = 26

Thus, the probability of drawing a black card is given by:

So, the probability of drawing a black card is \dfrac{1}{2}.

(ii) Since there are 26 red cards in the deck, the non-red (black) cards also number 26.

∴ Number of favorable outcomes = 26

The probability of not drawing a red card is:

Thus, the probability of not drawing a red card is \dfrac{1}{2}.

(iii) With 26 red cards in the deck, the number of favorable outcomes for drawing a red card is 26.

∴ Number of favorable outcomes = 26

The probability of drawing a red card is:

Therefore, the probability of drawing a red card is \dfrac{1}{2}.

(iv) There are 12 face cards in a deck (4 kings, 4 queens, and 4 jacks).

∴ Number of favorable outcomes = 12

The probability of drawing a face card is:

Hence, the probability of drawing a face card is \dfrac{3}{13}.

(v) Out of the 26 red cards, 6 are face cards (2 kings, 2 queens, and 2 jacks).

∴ Number of favorable outcomes = 6

The probability of drawing a red face card is:

Thus, the probability of drawing a red face card is \dfrac{3}{26}.

Question 6(i)

If A and B are two complementary events then what is the relation between P(A) and P(B)?

Complementary events are those that together cover all possible outcomes of an experiment. This means that if one event happens, the other cannot, and vice versa. The sum of the probabilities of all possible outcomes in any experiment is always 1.

∴ P(A) + P(B) = 1

Hence, P(A) + P(B) = 1.

Question 6(ii)

If the probability of happening an event A is 0.46. What will be the probability of not happening of the event A?

Given that the probability of event A occurring, denoted as P(A), is 0.46.

Let P(A') represent the probability of event A not occurring.

According to the fundamental principle of probability, the sum of the probabilities of an event happening and not happening is always 1.

∴ P(A) + P(A') = 1

Rearranging the equation, we find:

⇒ P(A') = 1 - P(A)

Substituting the given value:

⇒ P(A') = 1 - 0.46 = 0.54

Hence, the probability of not happening of event A is 0.54

Question 7

In a T.T. match between Geeta and Ritu, the probability of the winning of Ritu is 0.73. Find the probability of:

(i) winning of Geeta

(ii) not winning of Ritu.

(i) Consider that Ritu’s victory and Geeta’s victory are complementary events.

∴ P(Ritu wins) + P(Geeta wins) = 1

⇒ P(Geeta wins) = 1 – P(Ritu wins)

⇒ P(Geeta wins) = 1 – 0.73

⇒ P(Geeta wins) = 0.27

Therefore, the probability that Geeta wins is 0.27.

(ii) The probability that Ritu does not win is the same as the probability that Geeta wins, which is 0.27.

Thus, the probability that Ritu does not win is 0.27.

Question 8

In a race between Mahesh and John; the probability that John will loose the race is 0.54. Find the probability of :

(i) winning of Mahesh

(ii) winning of John.

(i) Since the competition is solely between Mahesh and John, we know that if John loses, Mahesh wins. Thus, the probability that Mahesh wins is the same as the probability of John losing the race, which is 0.54.

Hence, the probability of winning of Mahesh = 0.54

(ii) Winning and losing are opposite outcomes for John, meaning they are complementary. Therefore, the sum of the probabilities of these two events must equal 1.

∴ P(winning of John) + P(loosing of John) = 1

⇒ P(winning of John) + 0.54 = 1

⇒ P(winning of John) = 1 – 0.54 = 0.46

Hence, the probability of winning of John = 0.46

Question 9

(i) Write the probability of sure event.

(ii) Write the probability of an even which is impossible.

(iii) For an event E, write a relation representing the range of values of P(E).

(i) The probability of a certain event occurring is 1.

(ii) An event that cannot happen at all is termed impossible, and its probability is 0.

(iii) The probability of any event, denoted as P(E), must fall within the range of 0 and 1. Hence, we can express this as 0 \leq P(E) \leq 1.

Question 10

In a single throw of a dice, find the probability of getting :

(i) 5

(ii) 8

(iii) a number less than 8

(iv) a prime number.

When a die is thrown once, the outcomes that can appear are {1, 2, 3, 4, 5, 6}.

∴ Total possible outcomes = 6.

(i) To get the number 5, there is only one favourable outcome, which is 5 itself.

Thus, the probability of rolling a 5 is given by:

P(getting a 5) = \dfrac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \dfrac{1}{6}.

Hence, the probability of getting a 5 = \dfrac{1}{6}.

(ii) Since a die has no face numbered 8, there are no favourable outcomes for 8.

∴ No. of favourable outcomes = 0.

The probability of rolling an 8 is:

P(getting a number 8) = \dfrac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \dfrac{0}{6} = 0.

Hence, the probability of getting 8 = 0.

(iii) All numbers on the die (1, 2, 3, 4, 5, 6) are less than 8.

∴ No. of favourable outcomes = 6.

The probability of rolling a number less than 8 is:

P(getting a number less than 8) = \dfrac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \dfrac{6}{6} = 1.

Hence, the probability of getting a number less than 8 = 1.

(iv) Among the numbers on a die, the primes are 2, 3, and 5.

∴ No. of favourable outcomes = 3.

The probability of rolling a prime number is:

P(getting a prime number) = \dfrac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \dfrac{3}{6} = \dfrac{1}{2}.

Hence, the probability of getting a prime number = \dfrac{1}{2}.

Question 11

A dice is thrown once. Find the probability of getting :

(i) an even number

(ii) a number between 3 and 8

(iii) an even number or a multiple of 3.

When rolling a single die, the outcomes we might see are {1, 2, 3, 4, 5, 6}.

∴ Total number of possible outcomes = 6.

(i) The even numbers among 1, 2, 3, 4, 5, 6 are 2, 4, and 6.

∴ Number of favourable outcomes = 3.

The probability of rolling an even number is given by:

Hence, the probability of getting an even number = \dfrac{1}{2}.

(ii) The numbers between 3 and 8, from the set {1, 2, 3, 4, 5, 6}, are 4, 5, and 6.

∴ Number of favourable outcomes = 3.

The probability of getting a number between 3 and 8 is:

Hence, the probability of getting a number between 3 and 8 = \dfrac{1}{2}.

(iii) The numbers that are either even or multiples of 3 within {1, 2, 3, 4, 5, 6} are 2, 3, 4, and 6.

∴ Number of favourable outcomes = 4.

The probability of getting either an even number or a multiple of 3 is:

Hence, the probability of getting either an even number or a multiple of 3 = \dfrac{2}{3}.

Question 12

Which of the following cannot be the probability of an event ?

(i) \dfrac{3}{5}

(ii) 2.7

(iii) 43%

(iv) -0.6

(v) -3.2

(vi) 0.35

Recall that the probability of any event must satisfy the condition: 0 ≤ P(Event) ≤ 1. This means that a probability cannot exceed 1 or be less than 0.

The value 2.7 is greater than 1, so it cannot represent a probability.

Similarly, probabilities like -0.6 and -3.2 are not valid since they are negative.

Hence, (ii), (iv) and (v) cannot be probability of an event.

Question 13

A bag contains six identical black balls. A child withdraws one ball from the bag without looking into it. What is the probability that he takes out:

(i) a white ball ?

(ii) a black ball ?

Inside the bag, we have six identical black balls.

∴ Total number of possible outcomes = 6.

(i) Since there are no white balls present in the bag, the number of favourable outcomes for drawing a white ball is 0.

Thus, the probability of picking a white ball is given by:

P(getting a white ball) = \dfrac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \dfrac{0}{6} = 0.

Hence, the probability of getting a white ball = 0.

(ii) As all the balls are black, the number of favourable outcomes for drawing a black ball is 6.

Therefore, the probability of selecting a black ball is:

P(getting a black ball) = \dfrac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \dfrac{6}{6} = 1.

Hence, the probability of getting a black ball = 1.

Question 14

A single letter is selected at random from the word ‘Probability’. Find the probability that it is a vowel.

The word ‘Probability’ consists of a total of 11 letters.

∴ The total number of possible outcomes is 11.

The vowels present in the word ‘Probability’ are ‘O’, ‘A’, ‘I’, and another ‘I’.

Thus, the number of favourable outcomes, which is selecting a vowel, is 4.

Therefore, the probability of selecting a vowel is given by:

P(getting a vowel) = \dfrac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \dfrac{4}{11}.

Hence, the probability that letter drawn is a vowel = \dfrac{4}{11}.

Question 15

Ramesh chooses a date at random in January for a party (see the following figure).

Find the probability that he chooses :

(i) a Wednesday

(ii) a Friday

(iii) a Tuesday or a Saturday.

In January, we have a total of 31 days.

∴ Total possible outcomes = 31.

(i) Wednesdays occur on the 1st, 8th, 15th, 22nd, and 29th of January.

∴ Number of favourable outcomes = 5

The probability of selecting a Wednesday is given by:

P(getting a Wednesday) = \dfrac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \dfrac{5}{31}.

Hence, the probability of getting a Wednesday = \dfrac{5}{31}.

(ii) Fridays fall on the 3rd, 10th, 17th, 24th, and 31st of January.

∴ Number of favourable outcomes = 5

The probability of picking a Friday is:

P(getting a Friday) = \dfrac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \dfrac{5}{31}.

Hence, the probability of getting a Friday = \dfrac{5}{31}.

(iii) Tuesdays are on the 7th, 14th, 21st, and 28th, while Saturdays are on the 4th, 11th, 18th, and 25th of January.

∴ Number of favourable outcomes for either Tuesday or Saturday = 8.

The probability of choosing a Tuesday or Saturday is:

P(getting a Tuesday or Saturday) = \dfrac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \dfrac{8}{31}.

Hence, the probability of getting a Tuesday or Saturday = \dfrac{8}{31}.

Exercise 25(B)

Question 1(a)

A card is drawn from a well-shuffled pack of 52 cards. The probability of the card drawn to be king and jack is :

• (a) 1
• (b) 0
• (c) \dfrac{1}{2}
• (d) \dfrac{3}{13}

A single card cannot simultaneously be both a king and a jack.

The probability of drawing a card that is both a king and a jack is calculated as follows:

P(drawing a king and jack) = \dfrac{\text{Number of favourable outcomes}}{\text{Total number of possible outcomes}} = \dfrac{0}{52} = 0.

Hence, Option 2 is the correct option.

Question 1(b)

The probability that a non-leap year has 53 Sundays is :

• (a) 1
• (b) \dfrac{1}{7}
• (c) \dfrac{1}{365}
• (d) \dfrac{7}{365}

In a non-leap year, we have a total of 365 days.

The number of complete weeks in this year can be calculated as \dfrac{365}{7}, which results in 52 weeks with 1 extra day remaining.

Each week includes exactly one Sunday. To have 53 Sundays in a year, the additional day must also be a Sunday.

The probability that this extra day is a Sunday is given by \dfrac{\text{No. of favourable outcomes}}{\text{Total no. of possible outcomes}} = \dfrac{1}{7}.

Hence, Option 2 is the correct option.

Question 1(c)

Out of 800 identical articles some are selected and remaining are rejected. If the probability of an article to be rejected is 0.425; the number of rejected articles is :

• (a) 340
• (b) 800
• (c) 430
• (d) 560

The probability of rejecting an article can be expressed as:

⇒ P(\text{rejecting an item}) = \dfrac{\text{No. of favourable outcomes}}{\text{Total no. of possible outcomes}}

Given that the probability is 0.425, we have:

⇒ 0.425 = \dfrac{\text{No. of favourable outcomes}}{800}

To find the number of articles rejected, calculate:

⇒ No. of favourable outcomes = 0.425 \times 800 = 340.

Hence, Option 1 is the correct option.

Question 1(d)

A bag contains 24 balls of which 4 are red, 8 are white and remaining are blue. A ball is drawn at random, the probability of it to be blue is :

• (a) 0.5
• (b) 1
• (c) \dfrac{1}{3}
• (d) 2

The total number of blue balls can be calculated by subtracting the red and white balls from the total: 24 – 4 – 8 = 12.

The probability of picking a blue ball is given by the formula:

Hence, Option 1 is the correct option.

Question 1(e)

An integer is chosen from integers between 0 and 51, the probability of it to be divisible by 6 is :

• (a) \dfrac{3}{25}
• (b) \dfrac{4}{25}
• (c) \dfrac{1}{50}
• (d) \dfrac{1}{25}

The integers between 0 and 51 are the numbers from 1 to 50. These are the numbers we are selecting from, giving us a total of 50 possible outcomes.

Next, identify the integers within this range that are divisible by 6. These numbers are: 6, 12, 18, 24, 30, 36, 42, and 48. Thus, we have 8 numbers that meet the divisibility condition.

The probability that a randomly chosen integer from this set is divisible by 6 is calculated as:

∴ The probability is \dfrac{4}{25}. Hence, Option 2 is the correct option.

Question 2

Nine cards (identical in all respects) are numbered 2 to 10. A card is selected from them at random. Find the probability that the card selected will be:

(i) an even number

(ii) a multiple of 3

(iii) an even number and a multiple of 3

(iv) an even number or a multiple of 3

There are a total of 9 cards, each numbered from 2 to 10. When a card is drawn at random, the total number of possible outcomes is 9.

(i) Considering the numbers from 2 to 10, the even numbers are 2, 4, 6, 8, and 10. This gives us 5 favorable outcomes.

∴ The probability of drawing a card with an even number is given by:

Thus, the probability that the card selected will be an even number is \dfrac{5}{9}.

(ii) Among the numbers 2 to 10, the multiples of 3 are 3, 6, and 9, resulting in 3 favorable outcomes.

∴ The probability of drawing a card that is a multiple of 3 is:

Hence, the probability that the card selected will be a multiple of 3 is \dfrac{3}{9} = \dfrac{1}{3}.

(iii) From the numbers 2 to 10, only the number 6 is both even and a multiple of 3, giving us 1 favorable outcome.

∴ The probability of drawing a card that is both an even number and a multiple of 3 is:

Therefore, the probability of selecting a card that is both an even number and a multiple of 3 is \dfrac{1}{9}.

(iv) Considering the numbers 2 to 10, the numbers that are either even or multiples of 3 are 2, 3, 4, 6, 8, 9, and 10. This results in 7 favorable outcomes.

∴ The probability of drawing a card that is either an even number or a multiple of 3 is:

Hence, the probability of selecting a card that is either an even number or a multiple of 3 is \dfrac{7}{9}.

Question 3

Hundred identical cards are numbered from 1 to 100. The cards are well shuffled and then a card is drawn. Find the probability that the number on the card drawn is:

(i) a multiple of 5

(ii) a multiple of 6

(iii) between 40 and 60

(iv) greater than 85

(v) less than 48

Consider a scenario where you have 100 cards, each uniquely numbered from 1 to 100. When you draw one card, there are 100 possible outcomes.

(i) Let’s determine the probability of drawing a card that shows a multiple of 5. The multiples of 5 between 1 and 100 are: {5, 10, 15, 20, 25, 30, 35, 40, 45, 50, 55, 60, 65, 70, 75, 80, 85, 90, 95, 100}. This gives us 20 favourable outcomes.

∴ Probability of drawing a multiple of 5 = \dfrac{\text{Number of favourable outcomes}}{\text{Number of possible outcomes}} = \dfrac{20}{100} = \dfrac{1}{5}.

Hence, the probability of selecting a card with a multiple of 5 = \dfrac{1}{5}.

(ii) Now, consider the multiples of 6: {6, 12, 18, 24, 30, 36, 42, 48, 54, 60, 66, 72, 78, 84, 90, 96}. There are 16 such numbers.

∴ Probability of drawing a multiple of 6 = \dfrac{\text{Number of favourable outcomes}}{\text{Number of possible outcomes}} = \dfrac{16}{100} = \dfrac{4}{25}.

Hence, the probability of selecting a card with a multiple of 6 = \dfrac{4}{25}.

(iii) Next, let’s find the probability for a number between 40 and 60. The numbers in this range are: {41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59}, totaling 19 numbers.

∴ Probability of drawing a number between 40 and 60 = \dfrac{\text{Number of favourable outcomes}}{\text{Number of possible outcomes}} = \dfrac{19}{100}.

Hence, the probability of selecting a card between 40 and 60 = \dfrac{19}{100}.

(iv) For numbers greater than 85, we have: {86, 87, 88, 89, 90, 91, 92, 93, 94, 95, 96, 97, 98, 99, 100}, which counts to 15 numbers.

∴ Probability of drawing a number greater than 85 = \dfrac{\text{Number of favourable outcomes}}{\text{Number of possible outcomes}} = \dfrac{15}{100} = \dfrac{3}{20}.

Hence, the probability of selecting a card with number greater than 85 = \dfrac{3}{20}.

(v) Lastly, for numbers less than 48, we consider: {1, 2, …, 46, 47}, which gives us 47 numbers.

∴ Probability of drawing a number less than 48 = \dfrac{\text{Number of favourable outcomes}}{\text{Number of possible outcomes}} = \dfrac{47}{100}.

Hence, the probability of selecting a card with number less than 48 = \dfrac{47}{100}.

Question 4

From 25 identical cards, numbered 1, 2, 3, 4, 5, ….. , 24, 25; one card is drawn at random. Find the probability that the number on the card drawn is a multiple of :

(i) 3

(ii) 5

(iii) 3 and 5

(iv) 3 or 5

We have a total of 25 cards, each uniquely numbered from 1 to 25. A single card is drawn at random, and we need to determine the probability based on different conditions.

(i) First, identify the numbers that are multiples of 3 within this range: these numbers are {3, 6, 9, 12, 15, 18, 21, 24}. This results in 8 favourable outcomes.

∴ Probability of drawing a card with a multiple of 3 = \dfrac{\text{Number of favourable outcomes}}{\text{Total possible outcomes}} = \dfrac{8}{25}.

Thus, the probability of selecting a card with a multiple of 3 is \dfrac{8}{25}.

(ii) Next, consider the numbers that are multiples of 5: {5, 10, 15, 20, 25}. Here, we have 5 favourable outcomes.

∴ Probability of drawing a card with a multiple of 5 = \dfrac{\text{Number of favourable outcomes}}{\text{Total possible outcomes}} = \dfrac{5}{25} = \dfrac{1}{5}.

Therefore, the probability of selecting a card with a multiple of 5 is \dfrac{1}{5}.

(iii) Now, look for numbers that are multiples of both 3 and 5. The only number in this category is 15, giving us 1 favourable outcome.

∴ Probability of drawing a card with a multiple of both 3 and 5 = \dfrac{\text{Number of favourable outcomes}}{\text{Total possible outcomes}} = \dfrac{1}{25}.

Hence, the probability of selecting a card with a multiple of 3 and 5 is \dfrac{1}{25}.

(iv) Finally, for multiples of 3 or 5, the numbers are {3, 5, 6, 9, 10, 12, 15, 18, 20, 21, 24, 25}. This gives us 12 favourable outcomes.

∴ Probability of drawing a card with a multiple of 3 or 5 = \dfrac{\text{Number of favourable outcomes}}{\text{Total possible outcomes}} = \dfrac{12}{25}.

Thus, the probability of selecting a card with a multiple of 3 or 5 is \dfrac{12}{25}.

Question 5

A dice is thrown once. Find the probability of getting a number :

(i) less than 3

(ii) greater than or equal to 4

(iii) less than 8

(iv) greater than 6

When a die is rolled, the possible outcomes are the numbers 1 through 6. Thus, the total number of possible outcomes is 6.

(i) The numbers on the die that are less than 3 are 1 and 2.

∴ The number of favourable outcomes is 2.

The probability of rolling a number less than 3 is given by:

Therefore, the probability of getting a number less than 3 is \dfrac{1}{3}.

(ii) The numbers that are greater than or equal to 4 are 4, 5, and 6.

∴ The number of favourable outcomes is 3.

The probability of rolling a number greater than or equal to 4 is:

Thus, the probability of getting a number greater than or equal to 4 is \dfrac{1}{2}.

(iii) Every number on the die is less than 8, which includes 1, 2, 3, 4, 5, and 6.

∴ The number of favourable outcomes is 6.

The probability of rolling a number less than 8 is:

Hence, the probability of getting a number less than 8 is 1.

(iv) There are no numbers on a die greater than 6.

∴ The number of favourable outcomes is 0.

The probability of rolling a number greater than 6 is:

Thus, the probability of getting a number greater than 6 is 0.

Question 6

A book contains 85 pages. A page is chosen at random. What is the probability that the sum of the digits on the page is 8?

The book has a total of 85 pages.

∴ The total number of possible outcomes is 85.

The pages where the sum of the digits equals 8 are: {08, 17, 26, 35, 44, 53, 62, 71, 80}.

∴ The number of favourable outcomes is 9.

The probability of selecting a page where the sum of the digits is 8 is given by:

Hence, the probability of getting a page with sum of digits equal to 8 is \dfrac{9}{85}.

Question 7

A pair of dice is thrown. Find the probability of getting a sum of 10 or more, if 5 appears on the first dice.

When you roll two dice together, the total number of possible outcomes is calculated by multiplying the number of faces on each die: 6 \times 6 = 36.

Now, let’s focus on the condition where the first die shows a 5. We need the sum of the numbers on both dice to be 10 or more.

The combinations that satisfy this condition are:

• (5, 5), which gives a sum of 10
• (5, 6), which results in a sum of 11

Thus, there are 2 favourable outcomes.

The probability of achieving a sum of 10 or more under these conditions is given by the formula:

P(getting a sum of 10 or more) = \dfrac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \dfrac{2}{36} = \dfrac{1}{18}.

Hence, the probability of getting a sum of 10 or more, if 5 appears on the first dice = \dfrac{1}{18}.

Question 8

If two coins are tossed once, what is the probability of getting :

(i) 2 heads ?

(ii) at least one head ?

(iii) both heads or both tails ?

When tossing two coins, the potential results are: {HH, HT, TH, TT}.

Total possible outcomes = 4.

(i) To find the probability of getting 2 heads, we look at the favourable outcome, which is {HH}.

The number of favourable outcomes = 1.

The probability of getting 2 heads is calculated as:

P(\text{getting 2 heads}) = \dfrac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \dfrac{1}{4}.

Thus, the probability of getting 2 heads is \dfrac{1}{4}.

(ii) For at least one head, the favourable outcomes are {HT, TH, HH}.

Number of favourable outcomes = 3.

The probability of getting at least one head is:

P(\text{getting at least one head}) = \dfrac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \dfrac{3}{4}.

Thus, the probability of getting at least one head is \dfrac{3}{4}.

(iii) For both heads or both tails, the favourable outcomes are {HH, TT}.

Number of favourable outcomes = 2.

The probability of getting both heads or both tails is:

P(\text{getting both heads or both tails}) = \dfrac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \dfrac{2}{4} = \dfrac{1}{2}.

Thus, the probability of getting both heads or both tails is \dfrac{1}{2}.

Question 9

Two dice are rolled together. Find the probability of getting :

(i) a total of at least 10.

(ii) a multiple of 2 on one die and an odd number on the other die.

When rolling two dice together, the total number of possible outcomes is calculated as 6 multiplied by 6, which equals 36.

(i) To find the probability of achieving a total of at least 10, we must identify the favourable outcomes. These are:

(4, 6), (5, 5), (5, 6), (6, 4), (6, 5), and (6, 6).

∴ The number of favourable outcomes = 6.

The probability of getting a total of at least 10 is given by:

Hence, the probability of getting a total of at least 10 = \dfrac{1}{6}.

(ii) To find the probability of having a multiple of 2 on one die and an odd number on the other, we list the favourable outcomes:

(1, 2), (1, 4), (1, 6), (2, 1), (2, 3), (2, 5), (3, 2), (3, 4), (3, 6), (4, 1), (4, 3), (4, 5), (5, 2), (5, 4), (5, 6), (6, 1), (6, 3), (6, 5).

∴ The number of favourable outcomes = 18.

The probability of getting a multiple of 2 on one die and an odd number on the other is:

Hence, the probability of getting a multiple of 2 on one die and an odd number on the other die = \dfrac{1}{2}.

Question 10

A card is drawn from a well-shuffled pack of 52 cards. Find the probability that the card drawn is :

(i) a spade.

(ii) a red card.

(iii) a face card.

(iv) 5 of heart or diamond.

(v) Jack or queen.

(vi) ace and king.

(vii) a red and a king.

(viii) a red or a king.

A standard pack of playing cards consists of 52 cards, divided into 4 suits with 13 cards each.

∴ Total possible outcomes = 52.

(i) Within the deck, there are 13 spades.

∴ Favourable outcomes = 13.

P(drawing a spade) = \dfrac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \dfrac{13}{52} = \dfrac{1}{4}.

Hence, the probability of drawing a spade = \dfrac{1}{4}.

(ii) The deck contains 26 red cards, which include 13 hearts and 13 diamonds.

∴ Favourable outcomes = 26.

P(drawing a red card) = \dfrac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \dfrac{26}{52} = \dfrac{1}{2}.

Hence, the probability of drawing a red card = \dfrac{1}{2}.

(iii) Among the cards, there are 12 face cards, consisting of 4 kings, 4 queens, and 4 jacks.

∴ Favourable outcomes = 12.

P(drawing a face card) = \dfrac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \dfrac{12}{52} = \dfrac{3}{13}.

Hence, the probability of drawing a face card = \dfrac{3}{13}.

(iv) The deck includes two cards that are 5s, one each from hearts and diamonds.

∴ Favourable outcomes = 2.

P(drawing a 5 of heart or diamond) = \dfrac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \dfrac{2}{52} = \dfrac{1}{26}.

Hence, the probability of drawing a 5 of heart or diamond = \dfrac{1}{26}.

(v) There are 4 jacks and 4 queens in the deck.

∴ Favourable outcomes = 8.

P(drawing a jack or queen) = \dfrac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \dfrac{8}{52} = \dfrac{2}{13}.

Hence, the probability of drawing a jack or queen = \dfrac{2}{13}.

(vi) Since a single card cannot simultaneously be an ace and a king, this event is impossible.

∴ Favourable outcomes = 0.

P(drawing an ace and a king) = \dfrac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = 0.

Hence, the probability of drawing an ace and a king = 0.

(vii) The red kings in the deck are the king of diamonds and the king of hearts.

∴ Favourable outcomes = 2.

P(drawing a red and a king) = \dfrac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \dfrac{2}{52} = \dfrac{1}{26}.

Hence, the probability of drawing a red and a king = \dfrac{1}{26}.

(viii) There are 26 red cards, and additionally, the deck has a king of clubs and a king of spades.

∴ Favourable outcomes = 26 + 1 + 1 = 28.

P(drawing a red or a king) = \dfrac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \dfrac{28}{52} = \dfrac{7}{13}.

Hence, the probability of drawing a red or a king = \dfrac{7}{13}.

Question 11

A bag contains 16 coloured balls. Six are green, 7 are red and 3 are white. A ball is chosen, without looking into the bag. Find the probability that the ball chosen is :

(i) red

(ii) not red

(iii) white

(iv) not white

(v) green or red

(vi) white or green

(vii) green or red or white

The total number of coloured balls in the bag is 16.

∴ The total number of possible outcomes = 16.

(i) The bag contains 7 red balls.

∴ The number of favourable outcomes = 7.

The probability of selecting a red ball is given by:

P(\text{drawing a red ball}) = \dfrac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \dfrac{7}{16}.

Hence, the probability of drawing a red ball = \dfrac{7}{16}.

(ii) There are 9 balls that are not red (6 green + 3 white).

∴ The number of favourable outcomes = 9.

The probability of not selecting a red ball is:

P(\text{not drawing a red ball}) = \dfrac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \dfrac{9}{16}.

Hence, the probability of not drawing a red ball = \dfrac{9}{16}.

(iii) There are 3 white balls in the bag.

∴ The number of favourable outcomes = 3.

The probability of selecting a white ball is:

P(\text{drawing a white ball}) = \dfrac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \dfrac{3}{16}.

Hence, the probability of drawing a white ball = \dfrac{3}{16}.

(iv) There are 13 balls that are not white (6 green + 7 red).

∴ The number of favourable outcomes = 13.

The probability of not selecting a white ball is:

P(\text{not drawing a white ball}) = \dfrac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \dfrac{13}{16}.

Hence, the probability of not drawing a white ball = \dfrac{13}{16}.

(v) Considering the only colours are red, green, and white, the probability of drawing a green or red ball is the same as not drawing a white ball:

P(\text{drawing a green or red ball}) = P(\text{not drawing a white ball}) = \dfrac{13}{16}.

Hence, the probability of drawing a green or red ball = \dfrac{13}{16}.

(vi) Similarly, since the colours are red, green, and white, the probability of drawing a white or green ball is the same as not drawing a red ball:

From part (ii),

P(\text{not drawing a red ball}) = \dfrac{9}{16}.

∴ P(\text{drawing a white or green ball}) = \dfrac{9}{16}.

Hence, the probability of drawing a white or green ball = \dfrac{9}{16}.

(vii) Since all balls are either green, red, or white, the probability of drawing any of these colours is:

∴ The number of favourable outcomes = 16.

P(\text{drawing a green or red or white ball}) = \dfrac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \dfrac{16}{16} = 1.

Hence, the probability of drawing a green or red or white ball = 1.

Question 12

A ball is drawn at random from a box containing 12 white, 16 red and 20 green balls. Determine the probability that the ball drawn is :

(i) white

(ii) red

(iii) not green

(iv) red or white

The box has a total of 12 white, 16 red, and 20 green balls.

∴ The total number of balls, or possible outcomes, is 48.

(i) For the white balls, we have 12 in the box.

∴ The number of favourable outcomes for drawing a white ball is 12.

The probability of selecting a white ball is given by:

Therefore, the probability of drawing a white ball is \dfrac{1}{4}.

(ii) Similarly, there are 16 red balls available.

∴ The number of favourable outcomes for drawing a red ball is 16.

The probability of selecting a red ball is:

Thus, the probability of drawing a red ball is \dfrac{1}{3}.

(iii) To find the probability of not drawing a green ball, consider the white and red balls together, which total 28 (12 white + 16 red).

∴ The number of favourable outcomes for not drawing a green ball is 28.

The probability of not picking a green ball is:

Thus, the probability of not drawing a green ball is \dfrac{7}{12}.

(iv) Since the balls are either red, white, or green, the event of drawing a red or white ball is equivalent to not drawing a green ball.

Therefore, the probability of drawing either a red or white ball is the same as not drawing a green ball:

Hence, the probability of drawing a red or white ball is \dfrac{7}{12}.

Question 13

A card is drawn from a pack of 52 cards. Find the probability that the card drawn is :

(i) a red card

(ii) a black card

(iii) a spade

(iv) an ace

(v) a black ace

(vi) ace of diamonds

(vii) not a club

(viii) a queen or a jack.

A standard deck consists of 52 cards divided into 4 suits, each containing 13 cards.

No. of possible outcomes = 52.

(i) The deck contains 26 red cards, which include 13 hearts and 13 diamonds.

∴ No. of favourable outcomes = 26.

Probability of drawing a red card = \dfrac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \dfrac{26}{52} = \dfrac{1}{2}.

Thus, the probability of drawing a red card is \dfrac{1}{2}.

(ii) Similarly, there are 26 black cards, split into 13 clubs and 13 spades.

∴ No. of favourable outcomes = 26.

Probability of drawing a black card = \dfrac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \dfrac{26}{52} = \dfrac{1}{2}.

Thus, the probability of drawing a black card is \dfrac{1}{2}.

(iii) The deck includes 13 spade cards.

∴ No. of favourable outcomes = 13.

Probability of drawing a spade = \dfrac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \dfrac{13}{52} = \dfrac{1}{4}.

Thus, the probability of drawing a spade is \dfrac{1}{4}.

(iv) There are a total of 4 aces in the deck.

∴ No. of favourable outcomes = 4.

Probability of drawing an ace = \dfrac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \dfrac{4}{52} = \dfrac{1}{13}.

Thus, the probability of drawing an ace is \dfrac{1}{13}.

(v) The deck has 2 black aces, one from clubs and one from spades.

∴ No. of favourable outcomes = 2.

Probability of drawing a black ace = \dfrac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \dfrac{2}{52} = \dfrac{1}{26}.

Thus, the probability of drawing a black ace is \dfrac{1}{26}.

(vi) Only 1 ace of diamonds is present in the deck.

∴ No. of favourable outcomes = 1.

Probability of drawing the ace of diamonds = \dfrac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \dfrac{1}{52}.

Thus, the probability of drawing the ace of diamonds is \dfrac{1}{52}.

(vii) With 13 clubs in the deck, the number of non-club cards is 39 (52 – 13).

∴ No. of favourable outcomes = 39.

Probability of not drawing a club = \dfrac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \dfrac{39}{52} = \dfrac{3}{4}.

Thus, the probability of not drawing a club is \dfrac{3}{4}.

(viii) There are 4 queens and 4 jacks in the deck, totalling 8 cards.

∴ No. of favourable outcomes = 8.

Probability of drawing a queen or a jack = \dfrac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \dfrac{8}{52} = \dfrac{2}{13}.

Thus, the probability of drawing a queen or a jack is \dfrac{2}{13}.

Question 14

Thirty identical cards are marked with numbers 1 to 30. If one card is drawn at random, find the probability that it is :

(i) a multiple of 4 or 6.

(ii) a multiple of 3 and 5.

(iii) a multiple of 3 or 5.

There are 30 cards, each with a unique number from 1 to 30.

The total number of outcomes is 30.

(i) The cards that are multiples of either 4 or 6 are numbered 4, 6, 8, 12, 16, 18, 20, 24, 28, and 30.

∴ The number of favourable outcomes is 10.

Thus, the probability of drawing a card that is a multiple of 4 or 6 is given by:

Hence, the probability of drawing a card which is a multiple of 4 or 6 = \dfrac{1}{3}.

(ii) The cards that are multiples of both 3 and 5 are numbered 15 and 30.

∴ The number of favourable outcomes is 2.

Therefore, the probability of drawing a card that is a multiple of 3 and 5 is:

Hence, the probability of drawing a card which is a multiple of 3 and 5 = \dfrac{1}{15}.

(iii) Cards that are multiples of either 3 or 5 include those numbered 3, 5, 6, 9, 10, 12, 15, 18, 20, 21, 24, 25, 27, and 30.

∴ The number of favourable outcomes is 14.

So, the probability of drawing a card that is a multiple of 3 or 5 is:

Hence, the probability of drawing a card which is a multiple of 3 or 5 = \dfrac{7}{15}.

Question 15

In a single throw of two dice, find the probability of :

(i) a doublet

(ii) a number less than 3 on each dice

(iii) an odd number as a sum

(iv) a total of atmost 10

(v) an odd number on one dice and a number less than or equal to 4 on other dice.

Rolling two dice simultaneously gives us a total of 36 possible outcomes, as each die has 6 faces and 6 × 6 = 36.

(i) A doublet occurs when both dice show the same number. The pairs that satisfy this are: (1, 1), (2, 2), (3, 3), (4, 4), (5, 5), and (6, 6).

∴ Number of favourable outcomes = 6

The probability of rolling a doublet is given by:

Hence, the probability of getting a doublet = \dfrac{1}{6}.

(ii) To get a number less than 3 on each die, consider these outcomes: (1, 1), (1, 2), (2, 1), and (2, 2).

∴ Number of favourable outcomes = 4

The probability of each die showing a number less than 3 is:

Hence, the probability of getting a number less than 3 on each die = \dfrac{1}{9}.

(iii) To achieve an odd sum, the following outcomes are possible: (1, 2), (1, 4), (1, 6), (2, 1), (2, 3), (2, 5), (3, 2), (3, 4), (3, 6), (4, 1), (4, 3), (4, 5), (5, 2), (5, 4), (5, 6), (6, 1), (6, 3), (6, 5).

∴ Number of favourable outcomes = 18

The probability of obtaining an odd sum is:

Hence, the probability of getting an odd number as the sum = \dfrac{1}{2}.

(iv) For a total of at most 10, the outcomes are: (1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (6, 1), (6, 2), (6, 3), (6, 4).

∴ Number of favourable outcomes = 33

The probability of a total of at most 10 is:

Hence, the probability of getting a total of at most 10 = \dfrac{11}{12}.

(v) To have an odd number on one die and a number less than or equal to 4 on the other, the possible outcomes are: (1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (2, 1), (2, 3), (2, 5), (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (4, 1), (4, 3), (4, 5), (5, 1), (5, 2), (5, 3), (5, 4).

∴ Number of favourable outcomes = 20

The probability here is:

Hence, the probability of getting an odd number on one die and a number less than or equal to 4 on the other = \dfrac{5}{9}.

Test Yourself

Question 1(a)

Three coins are tossed simultaneously the probability of getting atleast two heads is :

• (a) \dfrac{1}{2}
• (b) \dfrac{1}{3}
• (c) \dfrac{1}{4}
• (d) \dfrac{3}{8}

Consider the scenario where you toss three coins at once. The possible outcomes are as follows: HHH, HHT, HTH, THH, HTT, THT, TTH, and TTT. This gives us a total of 8 outcomes.

Now, we’re interested in finding the probability of getting at least two heads. The outcomes that meet this criterion are: HHH, HHT, HTH, and THH. Thus, there are 4 favorable outcomes.

The probability of obtaining at least two heads is calculated by dividing the number of favorable outcomes by the total number of possible outcomes:

Hence, Option 1 is the correct option.

Question 1(b)

A card is drawn from a pack of a well shuffled cards. The probability of getting either a king or a queen is :

• (a) \dfrac{1}{13}
• (b) \dfrac{2}{13}
• (c) \dfrac{3}{13}
• (d) 0

Consider a standard deck of cards, which contains 52 cards in total. Among these, there are 4 kings and 4 queens, one from each suit. Therefore, the total number of favorable outcomes for drawing either a king or a queen is 4 (kings) + 4 (queens) = 8.

The probability of drawing either a king or a queen can be calculated using the formula:

Simplifying the fraction, we get:

Hence, Option 2 is the correct option.

Question 1(c)

Two dice are rolled together and the product (P) of their scores is obtained :

Event A : P is 6.

Event B : P is an odd number.

Event C : P is 35.

Which of the following event/events has probability equal to 0?

• (a) Events B and C
• (b) Events A and B
• (c) Event B
• (d) Event C

When rolling two dice, each die can land on any number from 1 to 6. This results in a total of 36 possible combinations, as shown here:

{(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}.

The highest possible product, 36, occurs when both dice show 6. The next highest product, 30, results from rolling a 5 and a 6.

For event C, where the product is 35, observe that there are no combinations of two numbers between 1 and 6 that multiply to 35. Hence, there are 0 favourable outcomes for event C.

∴ The probability of event C occurring is \dfrac{\text{No. of favourable outcomes}}{\text{Total no. of possible outcomes}} = \dfrac{0}{36} = 0.

Hence, Option 4 is the correct option.

Question 1(d)

A bag contains 20 red balls, 12 blue balls and 8 green balls. Three balls are drawn one by one with replacement, the probability that the third ball being green is :

• (a) \dfrac{1}{2}
• (b) \dfrac{2}{5}
• (c) \dfrac{1}{5}
• (d) 1.8

The total number of balls in the bag can be calculated by adding the red, blue, and green balls: 20 + 12 + 8, which equals 40.

The count of green balls is 8.

Since each ball is replaced after being drawn, the probability of drawing a green ball on any single draw remains the same.

∴ The probability that the third ball drawn is green is the same as the probability of drawing a green ball in any draw.

This probability is given by the ratio of the number of green balls to the total number of balls:

Thus, option 3 is the correct option.

Question 1(e)

Below are given the probabilities A, B, C and D of an event. P(A) : \dfrac{3}{5}, P(B) : 1.2, P(C) : -1.2 and P(D) = 1\dfrac{2}{3}. Which of the above values of the probabilities is possible :

• (a) Event A
• (b) Event B
• (c) Event C
• (d) Event D

Recall that the probability of any event must lie between 0 and 1, inclusive. This means it cannot be negative or exceed 1.

Notice that:
– P(A) = \dfrac{3}{5} is within the range [0, 1].
– P(B) = 1.2 is greater than 1, which is not possible.
– P(C) = -1.2 is negative, which is not allowed.
– P(D) = 1\dfrac{2}{3} equals 1.666…, which is also greater than 1.

∴ Only the probability of event A is valid.

Hence, Option 1 is the correct option.

Question 1(f)

A letter of English alphabet is chosen at random from English alphabets.

Assertion(A): The probability that the chosen letter is not a consonant is 5 : 52.

Reason(R): The probability of an event = \dfrac{\text{Total number of outcomes}}{\text{Number of favourable outcomes}}

• (a) A is true, R is false.
• (b) A is false, R is true.
• (c) Both A and R are true and R is correct reason for A.
• (d) Both A and R are false.

Consider the English alphabet, which contains 26 letters in total.

Out of these, 5 are vowels: a, e, i, o, u.

The rest, 21, are consonants.

The probability of any event is given by the formula:

Thus, the statement in Reason (R) is incorrect, as the formula is reversed.

Now, the probability that the chosen letter is not a consonant, which means it is a vowel, is:

This shows that the assertion (A) is incorrect since it states the probability as 5 : 52.

∴ Both the assertion and the reason are false.

Hence, option 4 is the correct option.

Question 1(g)

Number x is chosen from -3, -2, -1, 0, 1, 2 and 3. Also, x^2 ≤ 5.

Assertion(A): Probability for x^2 ≤ 5 is \dfrac{3}{7}.

Reason(R): Probability = \dfrac{\text{Favourable outcomes}}{\text{Total number of outcomes}}.

• (a) A is true, R is false.
• (b) A is false, R is true.
• (c) Both A and R are true and R is correct reason for A.
• (d) Both A and R are true and R is incorrect reason for A.

Consider the selection of number x from the set -3, -2, -1, 0, 1, 2, 3. We need to determine for which values of x the condition x^2 \leq 5 holds true.

Calculate the square of each element:

• ((-3)^2 = 9), which is not favourable.
• ((-2)^2 = 4), which is favourable.
• ((-1)^2 = 1), which is favourable.
• 0^2 = 0, which is favourable.
• 1^2 = 1, which is favourable.
• 2^2 = 4, which is favourable.
• ((3)^2 = 9), which is not favourable.

The numbers that satisfy x^2 \leq 5 are -2, -1, 0, 1, 2. Thus, there are 5 favourable outcomes.

Using the probability formula, we get:

Therefore, the assertion (A) is false because the probability is \dfrac{5}{7}, not \dfrac{3}{7}. However, the reason (R) is true as it correctly states the probability formula.

Hence, option 2 is the correct option.

Question 1(h)

Face cards of spades are remove from the pack of 52 cards and the remaining cards are well shuffled. Then a card is drawn from the pack.

Statement (1): The probability of drawing a face card is \dfrac{7}{52}.

Statement (2): Kings, queens and jacks are the three face card and so the total number of face cards in the pack of 52 card is 3 x 4 = 12.

• (a) Both the statements are true.
• (b) Both the statements are false.
• (c) Statement 1 is true, and statement 2 is false.
• (d) Statement 1 is false, and statement 2 is true.

In a deck of 52 playing cards, each of the four suits—hearts, diamonds, clubs, and spades—has 3 face cards: Jack, Queen, and King. This gives us a total of 3 \times 4 = 12 face cards.

Thus, statement 2 is accurate.

When the face cards of spades (Jack, Queen, King) are removed, we subtract these 3 cards from the total face cards:

12 - 3 = 9 face cards remain.

Additionally, removing these 3 cards from the entire deck leaves us with:

52 - 3 = 49 cards in total.

The probability of drawing a face card from this adjusted deck is calculated using the formula:

Thus, the probability of selecting a face card is:

Consequently, statement 1 is incorrect.

∴ Statement 1 is false, and statement 2 is true.

Hence, option 4 is the correct option.

Question 1(i)

In a lottery ticket, there are 20 prizes and 25 blanks.

Statement (1): Probability of not getting the prize = 1 – \dfrac{20}{45}

Statement (2): P(getting prize) + P(blank) = 1

• (a) Both the statements are true.
• (b) Both the statements are false.
• (c) Statement 1 is true, and statement 2 is false.
• (d) Statement 1 is false, and statement 2 is true.

Consider the scenario: there are 20 prize-winning tickets and 25 blanks, making a total of 45 tickets.

To find the probability of an event, we use the formula:

For the probability of winning a prize, we have:

And for drawing a blank:

Adding these probabilities gives:

This confirms that the sum of the probabilities of getting a prize and drawing a blank equals 1, validating statement 2.

Now, for the probability of not winning a prize, which is the same as drawing a blank:

Thus, statement 1 is also true since the probability of not getting a prize is indeed 1 - \dfrac{20}{45}.

∴ Both statements are true. Hence, option 1 is the correct option.

Question 2

From a well-shuffled deck of 52 cards, one card is drawn. Find the probability that the card drawn is :

(i) a face card

(ii) not a face card

(iii) a queen of black colour

(iv) a card with number 5 or 6

(v) a card with number less than 8

(vi) a card with number between 2 and 9

A standard deck contains 52 cards.

The total outcomes possible = 52.

(i) In a deck, there are 12 face cards (comprising 4 kings, 4 queens, and 4 jacks).

∴ Favourable outcomes = 12

The probability of drawing a face card is given by:

P(\text{drawing a face card}) = \dfrac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \dfrac{12}{52} = \dfrac{3}{13}.

Hence, probability of drawing a face card = \dfrac{3}{13}.

(ii) Since drawing a face card and not drawing a face card are complementary events:

∴ Probability of drawing a face card + Probability of not drawing a face card = 1

⇒ Probability of not drawing a face card = 1 – Probability of drawing a face card

⇒ Probability of not drawing a face card = 1 - \dfrac{3}{13} = \dfrac{13 - 3}{13} = \dfrac{10}{13}.

Hence, probability of not drawing a face card = \dfrac{10}{13}.

(iii) There are 2 black queens (one from clubs and one from spades).

∴ Favourable outcomes = 2

The probability of drawing a black queen is:

P(\text{drawing a queen of black colour}) = \dfrac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \dfrac{2}{52} = \dfrac{1}{26}.

Hence, probability of drawing a queen of black colour = \dfrac{1}{26}.

(iv) Each number 5 and 6 has 4 cards in the deck (one per suit).

∴ Favourable outcomes = 8

The probability of drawing a card with number 5 or 6 is:

P(\text{drawing a card with number 5 or 6}) = \dfrac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \dfrac{8}{52} = \dfrac{2}{13}.

Hence, probability of drawing a card with number 5 or 6 = \dfrac{2}{13}.

(v) Cards numbered less than 8 are {2, 3, 4, 5, 6, 7} for each suit.

∴ Favourable outcomes = 6 × 4 = 24.

The probability of drawing a card with a number less than 8 is:

P(\text{getting a card with number less than 8}) = \dfrac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \dfrac{24}{52} = \dfrac{6}{13}.

Hence, the probability of drawing a card with number less than 8 = \dfrac{6}{13}.

(vi) Cards numbered between 2 and 9 are {3, 4, 5, 6, 7, 8} for each suit.

∴ Favourable outcomes = 6 × 4 = 24.

The probability of drawing a card with a number between 2 and 9 is:

P(\text{getting a card with number between 2 and 9}) = \dfrac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \dfrac{24}{52} = \dfrac{6}{13}.

Hence, the probability of drawing a card with number between 2 and 9 = \dfrac{6}{13}.

Question 3

In a match between A and B;

(i) the probability of winning of A is 0.83. What is the probability of winning of B?

(ii) the probability of losing the match is 0.49 for B. What is the probability of winning of A?

(i) The match involves two players, A and B.

∴ The chance of A losing is the same as B winning.

We know that:

⇒ Probability of A winning + Probability of A losing = 1 (since they are complementary events)

⇒ Probability of A winning + Probability of B winning = 1

⇒ 0.83 + Probability of B winning = 1

⇒ Probability of B winning = 1 – 0.83 = 0.17

Therefore, the probability of B winning is 0.17.

(ii) Given the match is between A and B.

∴ The probability of A winning is equal to B losing.

The probability of A winning is 0.49.

Question 4

A and B are friends. Ignoring the leap year, find the probability that both friends will have:

(i) different birthdays?

(ii) the same birthday?

Consider two friends, A and B. Friend A can have a birthday on any of the 365 days of the year. Similarly, friend B can also have a birthday on any of these 365 days.

Assuming each day is equally likely for a birthday, let’s determine the probabilities.

(i) For A and B to have different birthdays, B must be born on any day except A’s birthday. This gives us 364 favourable days for B.

Thus, the probability that A and B have different birthdays is:

Therefore, the probability that both friends will have different birthdays is \dfrac{364}{365}.

(ii) For A and B to share the same birthday, we use the complement rule. The probability that they have the same birthday is:

Therefore, the probability that both friends will have the same birthday is \dfrac{1}{365}.

Question 5

A man tosses two different coins (one of ₹ 2 and another of ₹ 5) simultaneously. What is the probability that he gets :

(i) at least one head ?

(ii) at most one head ?

Consider the scenario where two coins are tossed together. The potential outcomes are: ((H, H), (H, T), (T, H), (T, T)).

∴ Total number of possible outcomes = 4.

(i) For the event ‘at least one head’, the favourable outcomes are ((H, H), (H, T), (T, H)).

∴ Number of favourable outcomes = 3.

Thus, the probability of getting at least one head is given by:

Hence, the probability that he gets at least one head = \dfrac{3}{4}.

(ii) For the event ‘at most one head’, the favourable outcomes are ((H, T), (T, H), (T, T)).

∴ Number of favourable outcomes = 3.

Therefore, the probability of getting at most one head is:

Hence, the probability that he gets at most one head = \dfrac{3}{4}.

Question 6

All the three face cards of spades are removed from a well shuffled pack of 52 cards. A card is then drawn at random from the remaining pack. Find the probability of getting :

(i) a black face card

(ii) a queen

(iii) a black card

Initially, we start with a full deck of 52 cards.

When the 3 face cards of spades are taken out, we are left with 49 cards (since 52 – 3 = 49).

∴ The total number of possible outcomes is 49.

(i) Among the remaining cards, we have 3 black face cards from the clubs.

∴ The number of favourable outcomes for a black face card is 3.

Thus, the probability of selecting a black face card is given by:

P(drawing a black face card) = \dfrac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \dfrac{3}{49}.

Hence, the probability of drawing a black face card = \dfrac{3}{49}

(ii) Initially, there are 4 queens in the deck, but with the queen of spades removed, only 3 queens remain.

∴ The number of favourable outcomes for drawing a queen is 3.

Therefore, the probability of drawing a queen is:

P(drawing a queen) = \dfrac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \dfrac{3}{49}.

Hence, the probability of drawing a queen = \dfrac{3}{49}.

(iii) Originally, there are 26 black cards in the deck. After removing the 3 black spade face cards, 23 black cards remain (since 26 – 3 = 23).

∴ The number of favourable outcomes for drawing a black card is 23.

So, the probability of drawing a black card is:

P(drawing a black card) = \dfrac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \dfrac{23}{49}.

Hence, the probability of drawing a black card = \dfrac{23}{49}.

Question 7

In a musical chairs game, a person has been advised to stop playing the music at any time within 40 seconds after its start. What is the probability that the music will stop within the first 15 seconds ?

Consider the time span from 0 seconds to 15 seconds as the favourable outcomes, and the entire duration from 0 seconds to 40 seconds as the possible outcomes.

∴ The number of favourable outcomes = 15

The number of possible outcomes = 40.

The probability that the music will stop within the first 15 seconds is calculated as:

Hence, the probability that the music will stop within the first 15 seconds = \dfrac{3}{8}.

Question 8

In a bundle of 50 shirts, 44 are good, 4 have minor defects and 2 have major defects. What is the probability that :

(i) it is acceptable to a trader who accepts only a good shirt ?

(ii) it is acceptable to a trader who rejects only a shirt with major defects ?

Consider the total number of shirts, which is 50.

∴ The total possible outcomes are 50.

(i) For a trader who accepts only good shirts, note that there are 44 good shirts available.

∴ The number of favourable outcomes is 44.

The probability that the trader will accept a shirt is given by:

Thus, the probability that a shirt is acceptable to the trader who accepts only a good shirt is \dfrac{22}{25}.

(ii) For a trader who only rejects shirts with major defects, notice that there are 2 shirts with major defects.

Hence, the number of shirts that this trader will accept is 50 – 2 = 48.

∴ The number of favourable outcomes is 48.

The probability that the trader will accept a shirt is calculated as:

Therefore, the probability that a shirt is acceptable to a trader who rejects only a shirt with major defects is \dfrac{24}{25}.

Question 9

Two dice are thrown at the same time. Find the probability that the sum of the two numbers appearing on the top of the dice is :

(i) 8

(ii) 13

(iii) less than or equal to 12

When two dice are rolled together, the total number of possible outcomes is calculated as 6 \times 6 = 36.

(i) To achieve a sum of 8, the successful pairs are: ((2, 6), (3, 5), (4, 4), (5, 3), (6, 2)).

∴ The count of successful outcomes is 5.

The probability of the sum being 8 is given by:

Hence, the probability that the sum of the two numbers appearing on the top of the dice is 8 = \dfrac{5}{36}.

(ii) It is impossible to roll a sum of 13 with two dice.

∴ The count of successful outcomes is 0.

The probability of the sum being 13 is:

Hence, the probability that the sum of the two numbers appearing on the top of the dice is 13 = 0.

(iii) Every possible outcome results in a sum that is less than or equal to 12.

∴ The count of successful outcomes is 36.

The probability of the sum being less than or equal to 12 is:

Hence, the probability that the sum of the two numbers appearing on the top of the dice is less than or equal to 12 is 1.

Question 10

The probability that two boys do not have the same birthday is 0.897. What is the probability that the two boys have the same birthday ?

Consider that the events of the two boys having the same birthday and having different birthdays are complementary. This means the sum of their probabilities is always 1.

∴ P(do not have the same birthday) + P(having the same birthday) = 1

Given that P(do not have the same birthday) = 0.897, we can substitute:

⇒ 0.897 + P(having the same birthday) = 1

To find the probability of the boys having the same birthday, rearrange the equation:

⇒ P(having the same birthday) = 1 – 0.897

⇒ P(having the same birthday) = 0.103

Therefore, the probability that the two boys have the same birthday is 0.103.

Question 11

A bag contains 10 red balls, 16 white balls and 8 green balls. A ball is drawn out of the bag at random. What is the probability that the ball drawn will be :

(i) not red ?

(ii) neither red nor green ?

(iii) white or green ?

The bag contains a total of 34 balls, calculated as 10 red, 16 white, and 8 green balls.

(i) To find the probability of drawing a ball that is not red, count the non-red balls: 24 (16 white + 8 green).

∴ The number of favourable outcomes is 24.

The probability of drawing a non-red ball is given by:

Hence, the probability of drawing a not red ball = \dfrac{12}{17}.

(ii) Since the bag only contains red, white, and green balls, drawing neither a red nor a green ball means drawing a white ball. The count of white balls is 16.

∴ The number of favourable outcomes is 16.

The probability of drawing a white ball is:

∴ The probability of drawing neither a red nor a green ball is \dfrac{8}{17}.

Hence, the probability that the ball drawn is neither red nor green is \dfrac{8}{17}.

(iii) To determine the probability of drawing either a white or green ball, note that there are 24 such balls (16 white + 8 green).

∴ The number of favourable outcomes is 24.

The probability of drawing a white or green ball is:

Hence, the probability of drawing a white or green ball = \dfrac{12}{17}.

Question 12

A bag contains twenty ₹ 5 coins, fifty ₹ 2 coins and thirty ₹ 1 coins. If it is equally likely that one of the coins will fall down when the bag is turned upside down, what is the probability that the coin :

(i) will be a ₹ 1 coin ?

(ii) will not be a ₹ 2 coin ?

(iii) will neither be a ₹ 5 coin nor be a ₹ 1 coin ?

Let’s determine the total number of coins in the bag: 20 ₹ 5 coins, 50 ₹ 2 coins, and 30 ₹ 1 coins. Adding these gives us a total of 100 coins. Thus, the total number of possible outcomes is 100.

(i) Consider the ₹ 1 coins. There are 30 of them in the bag. Therefore, the number of favourable outcomes for drawing a ₹ 1 coin is 30.

The probability of selecting a ₹ 1 coin is given by:

Hence, the probability of drawing a ₹ 1 coin = \dfrac{3}{10}.

(ii) Now, let’s calculate the probability of not drawing a ₹ 2 coin. The coins that are not ₹ 2 are the ₹ 5 and ₹ 1 coins, which together total 50 (20 + 30).

Thus, the number of favourable outcomes for this event is 50.

The probability that the coin drawn will not be a ₹ 2 coin is:

Hence, the probability of not drawing a ₹ 2 coin = \dfrac{1}{2}.

(iii) For the event where the coin is neither a ₹ 5 nor a ₹ 1 coin, we focus on the ₹ 2 coins. There are 50 ₹ 2 coins, so the favourable outcomes for drawing a ₹ 2 coin is 50.

The probability of drawing a ₹ 2 coin is:

Since there are only three types of coins in the bag, the probability of drawing neither a ₹ 5 nor a ₹ 1 coin is the same as drawing a ₹ 2 coin.

Hence, the probability of drawing neither ₹ 5 nor ₹ 1 coin = \dfrac{1}{2}.

Question 13

A game consists of spinning arrow which comes to rest pointing at one of the numbers 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12; as shown below.

If the outcomes are equally likely, find the probability that the pointer will point at:

(i) 6

(ii) an even number

(iii) a prime number

(iv) a number greater than 8

(v) a number less than or equal to 9

(vi) a number between 3 and 11.

Consider the total number of possible outcomes, which is 12.

(i) To find the probability that the pointer lands on 6, notice that there is only 1 favorable outcome, which is the number 6 itself.

Therefore, the probability is given by:

P(pointer lands on 6) = \dfrac{\text{Number of favorable outcomes}}{\text{Number of possible outcomes}} = \dfrac{1}{12}.

Hence, the probability that the pointer lands on 6 = \dfrac{1}{12}.

(ii) For an even number, the favorable outcomes are 2, 4, 6, 8, 10, and 12. This gives us 6 favorable outcomes.

Thus, the probability is:

P(pointer lands on an even number) = \dfrac{\text{Number of favorable outcomes}}{\text{Number of possible outcomes}} = \dfrac{6}{12} = \dfrac{1}{2}.

Hence, the probability that the pointer lands on an even number = \dfrac{1}{2}.

(iii) The prime numbers between 1 and 12 are 2, 3, 5, 7, and 11, giving us 5 favorable outcomes.

Therefore, the probability is:

P(pointer lands on a prime number) = \dfrac{\text{Number of favorable outcomes}}{\text{Number of possible outcomes}} = \dfrac{5}{12}.

Hence, the probability that the pointer lands on a prime number = \dfrac{5}{12}.

(iv) Considering numbers greater than 8, the favorable outcomes are 9, 10, 11, and 12, resulting in 4 favorable outcomes.

So, the probability becomes:

P(pointer lands on a number greater than 8) = \dfrac{\text{Number of favorable outcomes}}{\text{Number of possible outcomes}} = \dfrac{4}{12} = \dfrac{1}{3}.

Hence, the probability that the pointer lands on a number greater than 8 = \dfrac{1}{3}.

(v) For numbers less than or equal to 9, the favorable outcomes are 1, 2, 3, 4, 5, 6, 7, 8, and 9, totaling 9 favorable outcomes.

Thus, the probability is:

P(pointer lands on a number ≤ 9) = \dfrac{\text{Number of favorable outcomes}}{\text{Number of possible outcomes}} = \dfrac{9}{12} = \dfrac{3}{4}.

Hence, the probability that the pointer lands on a number less than or equal to 9 = \dfrac{3}{4}.

(vi) For numbers between 3 and 11, the favorable outcomes are 4, 5, 6, 7, 8, 9, and 10, which gives us 7 favorable outcomes.

Therefore, the probability is:

P(pointer lands on a number between 3 and 11) = \dfrac{\text{Number of favorable outcomes}}{\text{Number of possible outcomes}} = \dfrac{7}{12}.

Hence, the probability that the pointer lands on a number between 3 and 11 = \dfrac{7}{12}.

Question 14

One card is drawn from a well shuffled deck of 52 cards. Find the probability of getting :

(i) a queen of red color

(ii) a black face card

(iii) the jack or the queen of the hearts

(iv) a diamond

(v) a diamond or a spade

Let’s analyze the problem with a deck of 52 cards.

(i) For a red queen, we have 2 such cards: the queen of hearts and the queen of diamonds.

∴ The number of favorable outcomes is 2.

The probability of drawing a red queen is given by:

Thus, the probability of drawing a red queen is \dfrac{1}{26}.

(ii) Considering black face cards, there are 3 from clubs and 3 from spades, totaling 6.

∴ The number of favorable outcomes is 6.

The probability of drawing a black face card is:

Hence, the probability of drawing a black face card is \dfrac{3}{26}.

(iii) For the jack or queen of hearts, we have exactly 2 cards.

∴ The number of favorable outcomes is 2.

The probability of drawing either the jack or the queen of hearts is:

Thus, the probability of drawing either the jack or the queen of hearts is \dfrac{1}{26}.

(iv) There are 13 diamond cards in total.

∴ The number of favorable outcomes is 13.

The probability of drawing a diamond card is:

Therefore, the probability of drawing a diamond is \dfrac{1}{4}.

(v) For either a diamond or a spade, we have 13 diamonds and 13 spades, giving us 26 cards.

∴ The number of favorable outcomes is 26.

The probability of drawing either a diamond or a spade is:

Hence, the probability of drawing a diamond or a spade is \dfrac{1}{2}.

Question 15

From a deck of 52 cards, all the face cards are removed and then the remaining cards are shuffled. Now one card is drawn from the remaining deck. Find the probability that the card drawn is :

(i) a black card

(ii) 8 of red colour

(iii) a king of black colour.

In a standard deck of cards, there are 12 face cards. Removing these from the deck leaves us with 40 cards (52 – 12). Thus, the total number of possible outcomes is 40.

(i) Originally, there are 26 black cards in a deck. Since the face cards are removed, and there are 6 black face cards (consisting of a king, queen, and jack from both clubs and spades), we are left with 20 black cards (26 – 6).

∴ The number of favourable outcomes for drawing a black card is 20.

The probability of drawing a black card is given by:

P(drawing a black card) = \dfrac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \dfrac{20}{40} = \dfrac{1}{2}.

Hence, the probability of drawing a black card = \dfrac{1}{2}.

(ii) There are 2 red cards with the number 8, one from hearts and one from diamonds.

∴ The number of favourable outcomes for drawing an 8 of red colour is 2.

The probability of drawing an 8 of red colour is:

P(drawing an 8 of red colour) = \dfrac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \dfrac{2}{40} = \dfrac{1}{20}.

Hence, the probability of drawing an 8 of red colour = \dfrac{1}{20}.

(iii) All face cards, including kings, have been removed from the deck. Therefore, there are no black kings left.

∴ The number of favourable outcomes for drawing a king of black colour is 0.

The probability of drawing a king of black colour is:

P(drawing a king of black colour) = \dfrac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \dfrac{0}{40} = 0.

Hence, the probability of drawing a king of black colour = 0.

Question 16

Seven cards : the eight, the nine, the ten, jack, queen, king and ace of diamonds are well shuffled. One card is then picked up at random.

(i) What is the probability that the card drawn is the eight or the king ?

(ii) If the king is drawn and put aside, what is the probability that the second card picked up is :

(a) an ace ?

(b) a king ?

We begin with a total of seven cards.

∴ The total number of possible outcomes is 7.

(i) To find the probability of drawing either the eight or the king, we note there are 2 favourable outcomes.

The probability, therefore, is given by:

Hence, the probability that the card drawn is the eight or the king = \dfrac{2}{7}.

(ii) When the king is drawn and set aside, the number of remaining cards becomes 6.

∴ The total number of possible outcomes now is 6.

(a) For the probability of drawing an ace, there is 1 favourable outcome.

Thus, the probability is:

Hence, the probability that the card drawn is an ace = \dfrac{1}{6}.

(b) Since the king has already been drawn and put aside, there are no kings left among the remaining cards.

∴ The number of favourable outcomes for drawing a king is 0.

Thus, the probability becomes:

Hence, the probability that the card drawn is a king = 0.

Question 17

(i) 4 defective pens are accidentally mixed with 16 good ones. It is not possible to just look at a pen and tell whether or not it is defective. One pen is drawn at random from the lot. What is the probability that the pen is defective ?

(ii) Suppose the pen drawn in (i) is defective and is not replaced. Now one more pen is drawn at random from the rest. What is the probability that this pen is :

(a) defective ?

(b) not defective ?

(i) The total number of pens is 20, which includes 4 defective and 16 good pens.

∴ The total number of possible outcomes is 20.

Since we have 4 defective pens, the number of favorable outcomes is 4.

The probability of picking a defective pen is given by:

Hence, the probability of drawing a defective pen is \dfrac{1}{5}.

(ii) Given that the first pen drawn is defective and not replaced, we now have 19 pens remaining, with 3 being defective.

(a) The number of favorable outcomes for drawing another defective pen is 3.

The total number of possible outcomes is now 19.

Thus, the probability of drawing a defective pen is:

Therefore, the probability of drawing a defective pen is \dfrac{3}{19}.

(b) The number of favorable outcomes for drawing a good pen is 16.

The total number of possible outcomes remains 19.

Thus, the probability of drawing a good pen is:

Therefore, the probability of drawing a not defective pen is \dfrac{16}{19}.

Question 18

A bag contains 100 identical marble stones which are numbered from 1 to 100. If one stone is drawn at random from the bag, find the probability that it bears :

(i) a perfect square number.

(ii) a number divisible by 4.

(iii) a number divisible by 5.

(iv) a number divisible by 4 or 5.

(v) a number divisible by 4 and 5.

We have a total of 100 identical marble stones, each numbered from 1 to 100.

∴ The total number of possible outcomes is 100.

(i) Let’s identify the stones with numbers that are perfect squares: 1, 4, 9, 16, 25, 36, 49, 64, 81, and 100. This gives us 10 favourable outcomes.

The probability of selecting a stone with a perfect square number is given by:

Hence, the probability of drawing a stone with a perfect square number is \dfrac{1}{10}.

(ii) Next, consider stones with numbers divisible by 4: 4, 8, 12, 16, 20, 24, 28, 32, 36, 40, 44, 48, 52, 56, 60, 64, 68, 72, 76, 80, 84, 88, 92, 96, and 100. Here, there are 25 favourable outcomes.

The probability of selecting a stone with a number divisible by 4 is:

Hence, the probability of drawing a stone with a number divisible by 4 is \dfrac{1}{4}.

(iii) Now, identify stones with numbers divisible by 5: 5, 10, 15, 20, 25, 30, 35, 40, 45, 50, 55, 60, 65, 70, 75, 80, 85, 90, 95, and 100. This results in 20 favourable outcomes.

The probability of selecting a stone with a number divisible by 5 is:

Hence, the probability of drawing a stone with a number divisible by 5 is \dfrac{1}{5}.

(iv) Consider stones with numbers divisible by either 4 or 5. These numbers are: 4, 8, 12, 16, 24, 28, 32, 36, 44, 48, 52, 56, 64, 68, 72, 76, 84, 88, 92, 96, 5, 10, 15, 20, 25, 30, 35, 40, 45, 50, 55, 60, 65, 70, 75, 80, 85, 90, 95, and 100. This gives us 40 favourable outcomes.

The probability of selecting a stone with a number divisible by 4 or 5 is:

Hence, the probability of drawing a stone with a number divisible by 4 or 5 is \dfrac{2}{5}.

(v) Finally, let’s find stones with numbers divisible by both 4 and 5. These numbers are: 20, 40, 60, 80, and 100. Here, we have 5 favourable outcomes.

The probability of selecting a stone with a number divisible by both 4 and 5 is:

Hence, the probability of drawing a stone with a number divisible by both 4 and 5 is \dfrac{1}{20}.

Question 19

A circle with diameter 20 cm is drawn somewhere on a rectangular piece of paper with length 40 cm and width 30 cm. This paper is kept horizontal on table top and a dice, very small in size, is dropped on the rectangular paper without seeing towards it. If the dice falls and lands on the paper only, find the probability that it will fall and land :

(i) inside the circle.

(ii) outside the circle.

The circle has a diameter of 20 cm, so its radius is calculated as \frac{\text{Diameter}}{2} = \frac{20}{2} = 10 cm.

To find the area of the circle, use the formula \pi r^2:

The rectangular paper has a length of 40 cm and a width of 30 cm. Thus, the area of the rectangle, which represents all possible outcomes, is:

(i) The number of favorable outcomes for the dice landing inside the circle is the area of the circle, \frac{2200}{7} \text{ cm}^2.

The probability that the dice lands inside the circle is given by:

Therefore, the probability that the dice lands inside the circle is \dfrac{11}{42}.

(ii) Since the sum of probabilities for all possible events equals 1, we have:

Thus, the probability that the dice lands outside the circle is:

Hence, the probability that the dice lands outside the circle is \dfrac{31}{42}.

Question 20

Two dice (each bearing numbers 1 to 6) are rolled together. Find the probability that the sum of the numbers on the upper-most faces of two dice is :

(i) 4 or 5

(ii) 7, 8 or 9.

(iii) between 5 and 8.

(iv) more than 10.

(v) less than 6.

When rolling two dice simultaneously, each die has 6 faces, leading to a total of 6 \times 6 = 36 possible outcomes.

(i) To achieve a sum of 4 or 5 on the top faces, the possible combinations are: (1, 3), (1, 4), (2, 2), (2, 3), (3, 1), (3, 2), and (4, 1).

∴ The number of favourable outcomes is 7.

The probability of obtaining a sum of 4 or 5 is given by:

P(4 or 5) = \dfrac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \dfrac{7}{36}.

Thus, the probability of the sum being 4 or 5 is \dfrac{7}{36}.

(ii) For sums of 7, 8, or 9, the favourable pairs are: (1, 6), (2, 5), (2, 6), (3, 4), (3, 5), (3, 6), (4, 3), (4, 4), (4, 5), (5, 2), (5, 3), (5, 4), (6, 1), (6, 2), and (6, 3).

∴ There are 15 favourable outcomes.

The probability of getting a sum of 7, 8, or 9 is:

P(7, 8, or 9) = \dfrac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \dfrac{15}{36} = \dfrac{5}{12}.

Thus, the probability of the sum being 7, 8, or 9 is \dfrac{5}{12}.

(iii) For a sum between 5 and 8, the combinations are: (1, 5), (1, 6), (2, 4), (2, 5), (3, 3), (3, 4), (4, 2), (4, 3), (5, 1), (5, 2), and (6, 1).

∴ The number of favourable outcomes is 11.

The probability of getting a sum between 5 and 8 is:

P(between 5 and 8) = \dfrac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \dfrac{11}{36}.

Thus, the probability of the sum being between 5 and 8 is \dfrac{11}{36}.

(iv) For sums greater than 10, the favourable outcomes are: (5, 6), (6, 5), and (6, 6).

∴ There are 3 favourable outcomes.

The probability of a sum greater than 10 is:

P(more than 10) = \dfrac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \dfrac{3}{36} = \dfrac{1}{12}.

Thus, the probability of the sum being more than 10 is \dfrac{1}{12}.

(v) For sums less than 6, the combinations are: (1, 1), (1, 2), (1, 3), (1, 4), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2), and (4, 1).

∴ The number of favourable outcomes is 10.

The probability of a sum less than 6 is:

P(less than 6) = \dfrac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \dfrac{10}{36} = \dfrac{5}{18}.

Thus, the probability of the sum being less than 6 is \dfrac{5}{18}.

Question 21

Three coins are tossed together. Write all the possible outcomes. Now, find the probability of getting :

(i) exactly two heads.

(ii) at least two heads.

(iii) atmost two heads.

(iv) all tails

(v) at least one tail.

When you toss three coins at once, the possible outcomes are: {HHH, TTT, HHT, HTH, THH, TTH, THT, HTT}.

(i) To find the probability of getting exactly two heads, look at the outcomes HHT, HTH, and THH. These are the favourable cases.

The number of favourable outcomes is 3.

The probability of getting exactly two heads is given by:

Hence, the probability of getting exactly two heads = \dfrac{3}{8}.

(ii) For getting at least two heads, the favourable outcomes are HHT, HTH, THH, and HHH.

Here, the number of favourable outcomes is 4.

The probability of getting at least two heads is:

Hence, the probability of getting at least two heads = \dfrac{1}{2}.

(iii) For at most two heads, consider the outcomes TTT, HHT, HTH, THH, TTH, THT, and HTT.

The number of favourable outcomes is 7.

The probability of getting at most two heads is:

Hence, the probability of getting at most two heads = \dfrac{7}{8}.

(iv) To get all tails, the only favourable outcome is TTT.

The number of favourable outcomes is 1.

The probability of getting all tails is:

Hence, the probability of getting all tails = \dfrac{1}{8}.

(v) For at least one tail, the outcomes TTT, HHT, HTH, THH, TTH, THT, and HTT are favourable.

The number of favourable outcomes is 7.

The probability of getting at least one tail is:

Hence, the probability of getting at least one tail is = \dfrac{7}{8}.

Question 22

Two dice are thrown simultaneously. What is the probability that :

(i) 4 will not come up either time ?

(ii) 4 will come up at least once ?

When you roll two dice at the same time, the total number of outcomes you can get is 6 \times 6, which equals 36.

(i) To find the probability that a 4 does not appear on either die, list the outcomes where 4 is absent: (1, 1), (1, 2), (1, 3), (1, 5), (1, 6), (2, 1), (2, 2), (2, 3), (2, 5), (2, 6), (3, 1), (3, 2), (3, 3), (3, 5), (3, 6), (5, 1), (5, 2), (5, 3), (5, 5), (5, 6), (6, 1), (6, 2), (6, 3), (6, 5), (6, 6). Count these, and you’ll find there are 25 such outcomes.

The probability that 4 does not appear is given by:

Thus, the probability of not rolling a 4 on either die is \dfrac{25}{36}.

(ii) Now, to determine the probability that a 4 appears at least once, consider the outcomes where 4 shows up: (1, 4), (2, 4), (3, 4), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (5, 4), (6, 4). There are 11 such outcomes.

So, the probability that 4 appears at least once is:

Therefore, the probability of getting a 4 at least once is \dfrac{11}{36}.

Question 23

Offices in Delhi are open for five days in a week (Monday to Friday). Two employees of an office remain absent for one day in the same particular week. Find the probability that they remain absent on :

(i) the same day

(ii) consecutive day

(iii) different days.

The total number of possible outcomes when choosing days for two employees to be absent is given by 5 days for the first employee and 5 days for the second employee: 5 \times 5 = 25.

Let’s denote the weekdays as follows: Monday (M), Tuesday (T), Wednesday (W), Thursday (Th), and Friday (F).

(i) For both employees to be absent on the same day, the possible pairs are: MM, TT, WW, ThTh, and FF.

The number of favourable cases is 5.

Thus, the probability that both employees are absent on the same day is calculated as:

Hence, probability that employees remain absent on the same day = \dfrac{1}{5}.

(ii) For the employees to be absent on consecutive days, the favourable pairs are: MT, TM, TW, WT, WTh, ThW, ThF, and FTh.

This gives us 8 favourable cases.

Therefore, the probability that the employees are absent on consecutive days is:

Hence, probability that employees remain absent on consecutive day = \dfrac{8}{25}.

(iii) Since being absent on the same day and on different days are complementary events, we have:

P(absent on different days) + P(absent on same day) = 1

⇒ P(absent on different days) = 1 – P(absent on same day)

⇒ P(absent on different days) = 1 – \dfrac{1}{5} = \dfrac{4}{5}.

Hence, the probability that employees remain absent on different days = \dfrac{4}{5}.

Question 24

A box contains some black balls and 30 white balls. If the probability of drawing a black ball is two-fifths of a white ball; find the number of black balls in the box.

Assume there are x black balls in the box.

The total number of balls becomes x + 30, since there are 30 white balls present.

∴ The total possible outcomes are x + 30.

The number of favourable outcomes for selecting a black ball is x.

Thus, the probability of picking a black ball is \dfrac{x}{x + 30}.

Similarly, the number of favourable outcomes for choosing a white ball is 30.

Therefore, the probability of selecting a white ball is \dfrac{30}{x + 30}.

According to the problem, the probability of drawing a black ball is two-fifths of that of a white ball.

Hence, the number of black balls = 12.

Question 25

From a pack of 52 playing cards, all cards whose numbers are multiples of 3 are removed. A card is now drawn at random. What is the probability that the card drawn is

(i) a face card (King, Jack or Queen)

(ii) an even numbered red card ?

First, identify the cards to be removed. The numbers that are multiples of 3 include 3, 6, and 9 from each suit: hearts, clubs, diamonds, and spades.

The total number of cards removed is calculated as 3 cards per suit multiplied by 4 suits, giving us 12 cards removed.

After removing these, the deck contains 40 cards (since 52 – 12 = 40).

∴ The total number of possible outcomes is 40.

(i) In a standard deck, there are 12 face cards (Kings, Queens, and Jacks across all suits).

∴ The number of favourable outcomes for drawing a face card is 12.

The probability of drawing a face card is given by:

Hence, the probability that the card drawn is a face card = \dfrac{3}{10}.

(ii) For an even numbered red card, consider the cards 2, 4, 8, and 10 from both hearts and diamonds.

∴ The number of favourable outcomes for this scenario is 8.

The probability of drawing an even numbered red card is:

Hence, the probability that the card drawn is an even numbered red card = \dfrac{1}{5}.

Question 26

A dice has 6 faces marked by the given numbers as shown below :

The dice is thrown once. What is the probability of getting

(i) a positive integer ?

(ii) an integer greater than -3 ?

(iii) the smallest integer ?

A standard die features six distinct faces.

∴ The total number of possible outcomes is 6.

(i) For obtaining a positive integer, the favourable numbers are 1, 2, and 3.

∴ The number of favourable outcomes is 3.

The probability of rolling a positive integer is given by:

Hence, the probability of getting a positive integer = \dfrac{1}{2}.

(ii) For an integer greater than -3, the numbers -2, -1, 1, 2, and 3 are favourable.

∴ The number of favourable outcomes is 5.

Thus, the probability of rolling an integer greater than -3 is:

Hence, the probability of getting an integer greater than -3 = \dfrac{5}{6}.

(iii) The smallest integer on the die is -3.

∴ There is only 1 favourable outcome.

The probability of rolling the smallest integer is:

Hence, the probability of getting smallest integer = \dfrac{1}{6}.

Question 27

Sixteen cards are labelled as a, b, c, ………., m, n, o, p. They are put in a box and shuffled. A boy is asked to draw a card from the box. What is the probability that the card drawn is :

(i) a vowel

(ii) a consonant

(iii) none of the letters of the word median ?

The total number of cards, which represent possible outcomes, is 16.

(i) Among the letters a, b, c, …, m, n, o, p, the vowels are a, e, i, and o.

∴ The number of favourable outcomes for drawing a vowel is 4.

The probability of drawing a vowel is given by:

Thus, the probability of drawing a vowel is \dfrac{1}{4}.

(ii) With 4 vowels present, the remaining 12 cards are consonants.

∴ The number of favourable outcomes for drawing a consonant is 12.

The probability of drawing a consonant is:

Thus, the probability of drawing a consonant is \dfrac{3}{4}.

(iii) The letters in the word ‘median’ are ‘m’, ‘e’, ‘d’, ‘i’, ‘a’, and ‘n’.

Removing these 6 letters from the total leaves us with 10 letters.

∴ The number of favourable outcomes for drawing none of the letters in ‘median’ is 10.

The probability of not drawing any letter from ‘median’ is:

Thus, the probability that the card drawn contains none of the letters of the word ‘median’ is \dfrac{5}{8}.

Question 28

A box contains a certain number of balls. On each of 60% balls, letter A is marked. On each of 30% balls, letter B is marked and on each of remaining balls, letter C is marked. A ball is drawn from the box at random. Find the probability that the ball drawn is :

(i) marked C

(ii) A or B

(iii) neither B nor C.

We know that 60% of the balls have the letter A, 30% have the letter B, and the remaining 10% are marked with the letter C.

Let the total number of balls be denoted by x.

∴ Total possible outcomes = x.

The number of balls with the letter A is calculated as:

Similarly, the number of balls with the letter B is:

And the number of balls with the letter C is:

(i) To find the probability of drawing a ball marked with C:

The number of favorable outcomes is the number of C marked balls, which is \dfrac{x}{10}.

Thus, the probability is:

Hence, the probability of drawing a ball marked C = \dfrac{1}{10}.

(ii) For the probability of drawing a ball marked either A or B:

The number of balls marked A or B is:

The probability becomes:

Hence, the probability of drawing a ball marked A or B = \dfrac{9}{10}.

(iii) Since the balls are marked either A, B, or C, drawing a ball that is neither B nor C means drawing a ball marked A.

The number of favorable outcomes for A is \dfrac{3x}{5}.

Therefore, the probability is:

∴ The probability of drawing a ball marked neither B nor C is:

Hence, the probability of drawing ball marked neither B nor C = \dfrac{3}{5}.

Question 29

A box contains a certain number of balls. Some of these balls are marked A, some are marked B and the remainings are marked C. When a ball is drawn at random from the box P(A) = \dfrac{1}{3} and P(B) = \dfrac{1}{4}. If there are 40 balls in the box which are marked C, find the number of balls in the box.

Events A, B, and C are mutually exclusive, meaning they cannot happen simultaneously.

∴ P(A) + P(B) + P(C) = 1

Given that P(A) = \dfrac{1}{3} and P(B) = \dfrac{1}{4}, we proceed as follows:

The probability of picking a ball marked C is given by:

Thus, we have:

Hence, there are 96 balls in the box.