CUET-UG SERIES

Mathematics

Probability

54 previous year questions.

Volume: 54 Ques

Yield: High

High-Yield Trend

2025

2024

2023

2022

2021

Chapter Questions
54 MCQs

PYQ 2021

medium

mathematics ID: cuet-ug-

A card is drawn at random from a pack of 52 cards. What is the probability that the card drawn is either Ace or a King?

None of the above

PYQ 2021

medium

mathematics ID: cuet-ug-

Let and be two independent events. If and, then is ________.

PYQ 2022

hard

mathematics ID: cuet-ug-

In a game. a child will win Rs 5 if he gets all heads or all tails when three coins are tossed simultaneously and he will lose Rs 3 for all other cases. The expected amount to lose in the game is

Rs. 0

Rs. 0.8

Rs. 1

Rs. 2

PYQ 2023

medium

mathematics ID: cuet-ug-

The mean of the number of heads in a simultaneous toss of three coins is :

1

2

Official Solution

Correct Option: (3)

To determine the mean number of heads when three coins are tossed simultaneously, we consider the possible outcomes and their respective probabilities. The possible outcomes of tossing three coins are as follows, where H represents heads and T represents tails: HHH HHT HTH THH HTT THT TTH TTT The probability of each outcome is since there are possible outcomes. Let's determine the number of heads in each outcome: HHH: 3 heads HHT: 2 heads HTH: 2 heads THH: 2 heads HTT: 1 head THT: 1 head TTH: 1 head TTT: 0 heads To find the mean, compute the expected value of the number of heads. The formula for the mean () is given by: Number of Heads (x) Probability P(x) 3210 Calculate the mean: Therefore, the mean number of heads is .

Number of Heads (x)	Probability P(x)
3	$%5Cfrac%7B1%7D%7B8%7D$
2	$%5Cfrac%7B3%7D%7B8%7D$
1	$%5Cfrac%7B3%7D%7B8%7D$
0	$%5Cfrac%7B1%7D%7B8%7D$

PYQ 2023

medium

mathematics ID: cuet-ug-

If, and P(A \cup B) =, then is equal to :

PYQ 2023

medium

mathematics ID: cuet-ug-

If A and B are events such that and then the value of is:

1

PYQ 2023

medium

mathematics ID: cuet-ug-

A coin is tossed 5 times. The probability of getting atleast one head is:

PYQ 2023

medium

mathematics ID: cuet-ug-

A and B throw a die alternatively till one of them gets a number more than 4 and wins the game. Then the probability of winning the game by B, if A starts first:

PYQ 2023

medium

mathematics ID: cuet-ug-

A telephone exchange receives on an average 5 calls per minute. The probability of receiving 3 or less calls per minute is :

PYQ 2023

medium

mathematics ID: cuet-ug-

The probability of getting a correct answer to a question is . If the probability of not getting the correct answer is, then what is the value of x ?

2

3

4

6

PYQ 2023

medium

mathematics ID: cuet-ug-

If P(E)=0.05, what is the probability of 'not E'?

0.05

0.45

0.95

0.55

PYQ 2023

medium

mathematics ID: cuet-ug-

A die is thrown. Find the probability of getting: (A) A number greater then 6 and (B) A number less than 3.

PYQ 2023

medium

mathematics ID: cuet-ug-

A bag contains 12 white and 18 red balls. Two balls are drawn in succession without replacement. The probability that the first is red and second is white is:

PYQ 2023

medium

mathematics ID: cuet-ug-

In a box containing 100 bulbs, 10 are defective. The probability that out of a sample of 5 bulbs none is defective, is :

PYQ 2023

medium

mathematics ID: cuet-ug-

Three coins are tossed once. Probability of getting no head is :

PYQ 2023

medium

mathematics ID: cuet-ug-

A bag contains 6 red, 4 blue and 10 white balls. A ball is picked from the bag at random What is the probability that it is neither white nor blue?

PYQ 2023

medium

mathematics ID: cuet-ug-

A die is thrown n times. A random variable X denotes the number of times, the number on the dice is greater than 4 and P(X = 1) = 2P(X = 2). The value of n is :

2

3

4

5

PYQ 2023

medium

mathematics ID: cuet-ug-

A card is drawn at random from a well shuffled pack of 52 cards. What is the probability of getting a ‘two’ of hearts or a ‘three’ of diamonds ?

PYQ 2023

medium

mathematics ID: cuet-ug-

Choose odd one out from the following:

Copper

Iron

Aluminium

Bromine

PYQ 2023

medium

mathematics ID: cuet-ug-

Study the following sequence and then answer the question given below: 9 H 7 Z A 9 2 B C H 3 1 X W V 5 T 4 M 6 0 Q 8 F J. If every fourth letter/number in the above sequence starting from 7 is eliminated, which of the following will be fifth letter/number to the left of the 12th position from your right?

H

Z

A

C

PYQ 2023

hard

mathematics ID: cuet-ug-

Three dice are thrown. The probability of getting a sum of numbers which is a perfect cube, is:

PYQ 2023

medium

mathematics ID: cuet-ug-

A pair of dice is thrown 3 times. If getting a doublet is considered a success, then the probability of two successes is :

PYQ 2023

medium

mathematics ID: cuet-ug-

In a box, consisting 100 bulbs, 10 are defective. The probability that out of a sample of 5 bulbs, none is defective is :

10 -1

PYQ 2023

medium

mathematics ID: cuet-ug-

Two balls are chosen randomly from an urn containing 8 white balls and 4 black balls by a player. Suppose that he wins ₹30 for each black ball selected and loses ₹15 for each white ball selected. The expected value of winning amount is :

₹12.72

₹14.72

₹15

₹00

PYQ 2023

medium

mathematics ID: cuet-ug-

Match LIST I with LIST II List-I List-II A Randomization is used in I Unbiased sampling B Randomization is not used in. II Probability sampling C If every element in the population has an equal chance to be part of the selected sample, then the sampling process is called III Biased sampling D If a sampling process systematically favours certain outcomes over others, then it is called IV Non-probability sampling Choose the correct answer from the options given below

A-IV, B-II, C-I, D-III

A-II, B-IV. C-III, D-I

A-III, B-I, C-IV, D-II

A-II, B-IV,C-I, D-III

PYQ 2023

medium

mathematics ID: cuet-ug-

If, and, then is equal to:

PYQ 2023

medium

mathematics ID: cuet-ug-

Anita and Bikram are two students. Their chances of solving a problem correctly are and respectively. If their probability of making a common error is and they both obtain same answer then the probability that their answer is correct, is :

PYQ 2023

medium

mathematics ID: cuet-ug-

A pair of dice is thrown and sum of the numbers on two tosses is observed. Which of the statements are correct

Probability of getting sum as 11 is

Probability of getting sum as 5 is

Probability of getting sum as 7 is

Probability of getting sum as 6 is

PYQ 2023

medium

mathematics ID: cuet-ug-

A bag contains 6 white and 4 black ball. Two balls are drawn at random. Then the probability of getting balls of the same colour is ___.

PYQ 2023

medium

mathematics ID: cuet-ug-

Ram and Shyam are playing a game by throwing a die alternatively till one of them gets a ' and wins the game. The probabilities of winning by Ram and Shyam respectively if Ram starts first, is

PYQ 2024

easy

mathematics ID: cuet-ug-

Three defective bulbs are mixed with 8 good ones. If three bulbs are drawn one by one with replacement, the probabilities of getting exactly 1 defective, more than 2 defective, no defective, and more than 1 defective respectively are:

32

PYQ 2024

medium

mathematics ID: cuet-ug-

$Two events and are such that,, and the probability of occurrence of at least one event is 0.8. If the events are independent, then the value of is:$

1

2

3

4

Official Solution

Correct Option: (3)

$To find the value of , we need to use the formula for the probability of the union of two independent events and : Given: , , .Plugging the values into the formula:To simplify, multiply the equation by 3 to eliminate the fractions:This results in:Subtract 1 from both sides:Divide by 2 to solve for :Thus, the correct option is .$

33

PYQ 2024

medium

mathematics ID: cuet-ug-

$The probability of a boy winning a game is . Let denote the least number of times he must play the game so that the probability of winning the game at least once is more than 90%, and denotes that number of times he wins the game. Hence, mean, variance, and standard deviation of the random variable are respectively: (A) 3 (B) (C) 2 (D) 0.81 Choose the correct answer from the options given below:$

1

$(A), (C), (B), (D)$

2

$(A), (B), (C), (D)$

3

$(B), (A), (D), (C)$

4

$(C), (A), (D), (B)$

Official Solution

Correct Option: (1)

$The problem involves finding the least number of times the boy must play such that he wins at least once with a probability greater than 90%. The probability of winning one game is given by . Consequently, the probability of losing one game is . Thus, the probability of losing all games is . We want the probability of winning at least one game, which can be described as: Solving this inequality: To find, take the logarithm of both sides: Now solve for : Calculating the values using base-10 logarithms: Then: Thus, the smallest integer is 3. For a binomial distribution : Mean Variance Standard deviation The final results are:, mean = 2, variance =, standard deviation . Hence, the correct answer is (A), (C), (B), (D).$

34

PYQ 2024

medium

mathematics ID: cuet-ug-

$A die is thrown three times. Events A and B are defined as below: A: 6 on the third throw B: 4 on the first and 5 on the second throw The probability of A given that B has already occurred, is:$

1

2

3

4

Official Solution

Correct Option: (1)

$To solve this problem, we'll calculate the conditional probability of event A given that event B has occurred. We need to determine . The formula for conditional probability is: Step 1: Calculate Event B occurs when a 4 is rolled on the first throw and a 5 on the second throw. The probability of rolling a specific number on a fair six-sided die is . Thus, Step 2: Calculate Event occurs when a 6 is rolled on the third throw in addition to event B. For this, Step 3: Calculate Substitute the values found into the conditional probability formula: To simplify, multiply by the reciprocal: The conditional probability of event A given B is . Thus, the correct answer is the option .$

35

PYQ 2024

medium

mathematics ID: cuet-ug-

$A coin is tossed twice. Then Match List-I with List-II: List-I List-II (Adverbs) (A) P(exactly 2 heads) (I) (B) P(at least 1 head) (II) (C) P(at most 2 heads) (III) (D) P(exactly 1 head) (IV) Choose the correct answer from the options given below :$

1

$(A)-(I), (B)-(III), (C)-(II), (D)-(IV)$

2

$(A)-(I), (B)-(III), (C)-(IV), (D)-(II)$

3

$(A)-(III), (B)-(IV), (C)-(I), (D)-(II)$

4

$(A)-(I), (B)-(II), (C)-(III), (D)-(IV)$

Official Solution

Correct Option: (1)

$To solve the problem of matching List-I (probability events related to coin tosses) with List-II (values of probabilities), let's calculate each probability: (A) P(exactly 2 heads) : The sample space when a coin is tossed twice is: {HH, HT, TH, TT}. The event "exactly 2 heads" is {HH}. Thus, P(A) =, which matches with (I). (B) P(at least 1 head) : The favorable outcomes are {HH, HT, TH}. Thus, P(B) =, which matches with (III). (C) P(at most 2 heads) : This includes all outcomes in the sample space {HH, HT, TH, TT}, thus P(C) =, which matches with (II). (D) P(exactly 1 head) : The favorable outcomes are {HT, TH}. Thus, P(D) =, which matches with (IV). Hence, the correct matches are: (A)-(I), (B)-(III), (C)-(II), (D)-(IV).$

36

PYQ 2024

easy

mathematics ID: cuet-ug-

$A die is rolled thrice. What is the probability of getting a number greater than in the first and second throws and a number less than in the third throw?$

1

2

3

4

Official Solution

Correct Option: (4)

$When a die is rolled, the outcomes are 1, 2, 3, 4, 5, 6. The probabilities for the given events are as follows: A number greater than 4 includes {5, 6}. The probability of this event is: A number less than 4 includes {1, 2, 3}. The probability of this event is: The probability of the required outcome (a number greater than 4 on the first and second throws, and a number less than 4 on the third throw) is the product of the probabilities: Thus, the probability of the required outcome is .$

37

PYQ 2024

hard

mathematics ID: cuet-ug-

$A coin is tossed K times. If the probability of getting 3 heads is equal to the probability of getting 7 heads, then the probability of getting 8 tails is:$

1

2

3

4

Official Solution

Correct Option: (3)

$The number of tosses satisfies the condition: . The probability of heads in tosses is: . Equating : . . From symmetry of binomial coefficients: . The probability of getting 8 tails (or 2 heads) is: . . Thus, the probability is .$

38

PYQ 2024

medium

mathematics ID: cuet-ug-

$If, and, then is:$

1

$0.96$

2

$0.72$

3

$0.36$

4

$0.42$

Official Solution

Correct Option: (2)

$To find, we can use the formula of probability for the union of two events: . First, we need to find, which is the intersection of events and . We use the conditional probability formula: . Given and, we compute as follows: Now we can find : Therefore, the correct answer is 0.72.$

39

PYQ 2024

medium

mathematics ID: cuet-ug-

$In a series of 4 trials, the probability of getting two successes is equal to the probability of getting three successes. The probability of getting at least one success is:$

1

2

3

4

Official Solution

Correct Option: (1)

$The problem involves a binomial distribution for which the probability is used, where is the number of trials, is the number of successes, and is the probability of success. Here, . Given that the probability of 2 successes equals the probability of 3 successes: Solving: Simplify by eliminating common terms: Rearrange to find : The probability of getting at least one success is calculated as: Using : Thus, the probability of getting at least one success is .$

40

PYQ 2024

medium

mathematics ID: cuet-ug-

$There are two bags. Bag-1 has 4 white and 6 black balls, and Bag-2 has 5 white and 5 black balls. A die is rolled, and if it shows a multiple of 3, a ball is drawn from Bag-1; otherwise, from Bag-2. If the ball drawn is not black, the probability it was not drawn from Bag-2 is:$

1

2

3

4

Official Solution

Correct Option: (3)

$Let represent the event that the ball is not black, and represent the event that the ball was drawn from Bag-2. We need to find, which is the probability that the ball was not drawn from Bag-2, given that it is not black. We can use Bayes' theorem: First, calculate, the total probability of drawing a non-black ball: From Bag-1, the probability of drawing a non-black ball is . From Bag-2, the probability of drawing a non-black ball is . Now, we compute using the law of total probability: The probability of drawing from Bag-1 is (since the die shows a number divisible by 3), and the probability of drawing from Bag-2 is . Thus: Next, calculate, which is the probability of drawing a non-black ball from Bag-1: Now, apply Bayes' theorem: Thus, the correct answer is:$

41

PYQ 2024

easy

mathematics ID: cuet-ug-

$There are 6 cards numbered 1 to 6, one number on one card. Two cards are drawn at random without replacement. Let X denote the sum of the numbers on the two cards drawn. Then P(X > 3) is:$

1

2

3

4

Official Solution

Correct Option: (1)

$The total number of ways to choose 2 cards from 6 is: The event means the sum of the numbers on the two cards is greater than 3. The only pair with a sum is, which occurs in 1 way. Thus, the number of favorable outcomes for is: The probability is:$

42

PYQ 2024

medium

mathematics ID: cuet-ug-

$Let and be two independent events such that and . List-I (Sentences) List-II (Adverbs) (A) (I) (B) (II) (C) (III) (D) (IV) Choose the correct answer from the options given below :$

1

$(A-II), (B-III), (C-I), (D-IV)$

2

$(A-III), (B-II), (C-IV), (D-I)$

3

$(A-II), (B-III), (C-IV), (D-I)$

4

$(A-II), (B-IV), (C-I), (D-III)$

Official Solution

Correct Option: (2)

$The Correct Answer is$

43

PYQ 2024

easy

mathematics ID: cuet-ug-

$The probability of not getting 53 Tuesdays in a leap year is:$

1

2

3

$0$

4

Official Solution

Correct Option: (4)

$In a leap year, there are 366 days, which consist of 52 complete weeks plus 2 extra days. These 2 extra days can be: Sunday and Monday Monday and Tuesday Tuesday and Wednesday Wednesday and Thursday Thursday and Friday Friday and Saturday Saturday and Sunday To have 53 Tuesdays in a leap year, one of the two extra days must be a Tuesday. Thus, favorable outcomes are: Monday and Tuesday Tuesday and Wednesday So, the probability of getting 53 Tuesdays is: Thus, the probability of not getting 53 Tuesdays is: This confirms the probability of not getting 53 Tuesdays in a leap year is:$

44

PYQ 2024

medium

mathematics ID: cuet-ug-

$The probability of a shooter hitting a target is 3/4 How many minimum number of times must he fire so that the probability of hitting the target at least once is more than 90%?$

1

$1$

2

$2$

3

$3$

4

$4$

Official Solution

Correct Option: (2)

$The probability of hitting the target at least once is the complement of the probability of missing the target every time. Let the number of shots fired be n . The probability of missing the target in a single shot is: The probability of missing the target in all n shots is: The probability of hitting the target at least once is: We need Simplify: Take the logarithm (base 10) of both sides: Using and : Divide by (note the sign change): Approximate : Since n must be an integer, the minimum n is: Thus, the shooter must fire at least 2 times.$

45

PYQ 2024

hard

mathematics ID: cuet-ug-

$A pair of dice is rolled . If the two numbers appearing on them are different the probability that Match List-I with List-II LIST-I(EVENT) LIST-II(PROBABILITY) (A) The sum of the number is greater than 11 (i) 0 (B) The sum of the number is 4 or less (ii) 1/15 (C) The sum of the number is 4 (iii) 2/15 (D) The sum of the number is 4 (iv) 3/15 Choose the correct answer from the option given below$

1

$(A)-(I), (B)-(II), (C)-(III), (D)-(IV)$

2

$(A)-(I), (B)-(III), (C)-(II), (D)-(IV)$

3

$(A)-(I), (B)-(II), (C)-(IV), (D)-(III)$

4

$(A)-(III), (B)-(IV), (C)-(I), (D)-(II)$

Official Solution

Correct Option: (1)

$To solve the problem of matching List-I with List-II based on the probability of the events described, we consider the conditions for rolling a pair of dice: 1. Total possible outcomes when rolling two dice = 6 x 6 = 36. 2. We are interested in scenarios where the numbers on both dice are different. 3. Outcomes where numbers on both dice are the same (i.e., doubles) = 6 (e.g., (1,1), (2,2), ..., (6,6)). 4. Outcomes with different numbers = 36 - 6 = 30. We analyze the given events: LIST-I(EVENT) Valid Outcomes LIST-II(PROBABILITY) (A) The sum of the number is greater than 11 (5,6), (6,5) 0 (B) The sum of the number is 4 or less (1,2), (2,1), (1,3), (3,1) 1/15 (C) The sum of the number is 4 (1,3), (3,1), (2,2) 2/15 (D) The sum of the number is 4 (1,3), (3,1), (2,2) 3/15 5. Each probability is computed using the formula Number of favorable outcomes / Number of different outcomes . 6. Calculating probabilities for valid outcomes: - Event (A) probability = 0/30 = 0 (because outcomes are invalid since both numbers appear are different). - Event (B) probability = 2/30 = 1/15. - Event (C) probability = 4/30 = 2/15. - Event (D) probability = 4/30 = 3/15. Thus, the correct matching is: (A)-(I), (B)-(II), (C)-(III), (D)-(IV).$

46

PYQ 2024

easy

mathematics ID: cuet-ug-

$Two dice are thrown simultaneously. If denotes the number of fours, then the expectation of will be:$

1

2

3

4

Official Solution

Correct Option: (2)

$Each die has a probability of of showing a four. The expectation of (the number of fours) is the sum of the expectations for each die: where: Thus:$

47

PYQ 2025

medium

mathematics ID: cuet-ug-

$If A and B are any two events such that P(B) = P(A and B), then which of the following is correct$

1

$P(B|A) = 1$

2

$P(A|B) = 1$

3

$P(B|A) = 0$

4

$P(A|B) = 0$

Official Solution

Correct Option: (2)

$Step 1: Understanding the Concept: We are given a condition about the probabilities of two events, B and the intersection of A and B (A and B is the same as A B). We need to determine the value of a conditional probability. Step 2: Key Formula or Approach: The definition of conditional probability is: Step 3: Detailed Explanation: We are given the condition P(B) = P(A B). Let's use this condition to find P(A|B). From the formula for conditional probability: Substitute the given condition P(A B) = P(B) into the formula: Assuming P(B)>0 (if P(B)=0, the conditional probability P(A|B) is undefined), we get: This means that given event B has occurred, event A is certain to occur. This makes sense, as the condition P(B) = P(A B) implies that the event B is a subset of the event A. Whenever B happens, A must also happen. Let's check P(B|A): We cannot determine the value of this without knowing P(A). So, option (1) and (3) are not necessarily correct. Step 4: Final Answer: The correct statement is P(A|B) = 1 .$

48

PYQ 2025

medium

mathematics ID: cuet-ug-

$If A is any event associated with sample space and if E 1, E 2, E 3 are mutually exclusive and exhaustive events. Then which of the following are true? (A) (B) (C) (D) Choose the correct answer from the options given below:$

1

$(A) and (C) only$

2

$(A) and (D) only$

3

$(B) and (D) only$

4

$(B) and (C) only$

Official Solution

Correct Option: (4)

$Step 1: Understanding the Concept: This question is about the theorem of total probability and Bayes' theorem. The events E\textsubscript{1}, E\textsubscript{2}, E\textsubscript{3} form a partition of the sample space, as they are mutually exclusive (they cannot happen at the same time) and exhaustive (one of them must happen). Step 3: Detailed Explanation: Let's analyze each statement. (A) This is not the standard formula for the law of total probability. The formula relates P(A) to conditional probabilities of A given E\textsubscript{i}, not the other way around. From the definition of conditional probability, , which doesn't simplify nicely to P(A). So, (A) is false. (B) This is the correct statement of the Law of Total Probability. It expresses the total probability of an event A in terms of the probabilities of A conditional on each event in a partition of the sample space. Since , this formula is equivalent to , which is true because E\textsubscript{i} are mutually exclusive and exhaustive. So, (B) is true. (C) } This is the correct statement of Bayes' Theorem. The numerator, by the multiplication rule, is . The denominator, by the law of total probability (from statement B), is P(A). So, the formula becomes , which is the definition of conditional probability. So, (C) is true. (D) } This is an incorrect formulation. It seems to be a misstatement of Bayes' theorem, with the roles of A and E\textsubscript{i} confused. So, (D) is false. Step 4: Final Answer: The true statements are (B) and (C). This corresponds to option (4).$

49

PYQ 2025

easy

mathematics ID: cuet-ug-

$How many minimum number of times must a man toss a fair coin so that the probability of having at least one head is more than 90%?$

1

$3$

2

$4$

3

$5$

4

$10$

Official Solution

Correct Option: (2)

$Step 1: Understanding the Concept: The problem asks for the minimum number of coin tosses, 'n', required to make the probability of a certain event (getting at least one head) exceed a specific value (90% or 0.9). The key is to use the complement rule of probability, as calculating the probability of "at least one" is often easier by calculating 1 minus the probability of "none". Step 2: Key Formula or Approach: Let 'n' be the number of times the coin is tossed. The probability of getting a head (H) in a single toss of a fair coin is . The probability of getting a tail (T) is . The event "at least one head" is the complement of the event "no heads" (i.e., all tails). We are given the condition: Step 3: Detailed Explanation: First, let's find the probability of getting "no heads" in 'n' tosses. This means getting tails on all 'n' tosses. Since the tosses are independent events, the probability of getting n tails in a row is: Now, substitute this into our inequality: Rearrange the inequality to solve for n: To get rid of the fractions, we can take the reciprocal of both sides. When we do this, we must reverse the inequality sign: Now we need to find the smallest integer 'n' that satisfies this condition. We can test values of n: If n = 1, , which is not greater than 10. If n = 2, , which is not greater than 10. If n = 3, , which is not greater than 10. If n = 4, , which is greater than 10. Step 4: Final Answer: The minimum number of times the coin must be tossed is 4.$

50

PYQ 2025

medium

mathematics ID: cuet-ug-

$Let A and B be two events such that: Match List-I with List-II: List-I List-II (A) (I) 0.2 (B) (II) 0.32 (C) (III) 0.64 (D) (IV) 0.98$

1

$(A) - (II), (B) - (IV), (C) - (III), (D) - (I)$

2

$(A) - (II), (B) - (III), (C) - (IV), (D) - (I)$

3

$(A) - (III), (B) - (IV), (C) - (II), (D) - (I)$

4

$(A) - (III), (B) - (II), (C) - (I), (D) - (IV)$

Official Solution

Correct Option: (2)

$Step 1: Understanding the Concept: This question requires the application of basic probability formulas, including the multiplication rule for conditional probability, the formula for conditional probability, the addition rule, and the complement rule. Step 3: Detailed Explanation: Given: P(A) = 0.8, P(B) = 0.5, P(B|A) = 0.4 (A) P(A B) Using the multiplication rule: . . This matches (II) . (B) P(A|B) Using the formula for conditional probability: . . This matches (III) . (C) P(A B) Using the addition rule: . . This matches (IV) . (D) P(A') Using the complement rule: . . This matches (I) . Step 4: Final Answer: The correct matching is: (A) (II) (B) (III) (C) (IV) (D) (I) This corresponds to option (2).$

List-I	List-II
(A) $P(A%20%5Ccap%20B)$	(I) 0.2
(B) $P(A%7CB)$	(II) 0.32
(C) $P(A%20%5Ccup%20B)$	(III) 0.64
(D) $P(A')$	(IV) 0.98

51

PYQ 2025

medium

mathematics ID: cuet-ug-

$If A is any event associated with sample space and if E\textsubscript{1}, E\textsubscript{2}, E\textsubscript{3} are mutually exclusive and exhaustive events. Then which of the following are true? (A) (B) (C) } (D) } Choose the correct answer from the options given below:$

1

$(A) and (C) only$

2

$(A) and (D) only$

3

$(B) and (D) only$

4

$(B) and (C) only$

Official Solution

Correct Option: (4)

$Step 1: Understanding the Concept: This question is about the theorem of total probability and Bayes' theorem. The events E\textsubscript{1}, E\textsubscript{2}, E\textsubscript{3} form a partition of the sample space, as they are mutually exclusive (they cannot happen at the same time) and exhaustive (one of them must happen). Step 3: Detailed Explanation: Let's analyze each statement. (A) This is not the standard formula for the law of total probability. The formula relates P(A) to conditional probabilities of A given E\textsubscript{i}, not the other way around. From the definition of conditional probability, , which doesn't simplify nicely to P(A). So, (A) is false. (B) This is the correct statement of the Law of Total Probability. It expresses the total probability of an event A in terms of the probabilities of A conditional on each event in a partition of the sample space. Since , this formula is equivalent to , which is true because E\textsubscript{i} are mutually exclusive and exhaustive. So, (B) is true. (C) } This is the correct statement of Bayes' Theorem. The numerator, by the multiplication rule, is . The denominator, by the law of total probability (from statement B), is P(A). So, the formula becomes , which is the definition of conditional probability. So, (C) is true. (D) } This is an incorrect formulation. It seems to be a misstatement of Bayes' theorem, with the roles of A and E\textsubscript{i} confused. So, (D) is false. Step 4: Final Answer: The true statements are (B) and (C). This corresponds to option (4).$

52

PYQ 2025

medium

mathematics ID: cuet-ug-

$How many minimum number of times must a man toss a fair coin so that the probability of having at least one head is more than 90%?$

1

$3$

2

$4$

3

$5$

4

$10$

Official Solution

Correct Option: (2)

$Step 1: Understanding the Concept: The problem asks for the minimum number of coin tosses, 'n', required to make the probability of a certain event (getting at least one head) exceed a specific value (90% or 0.9). The key is to use the complement rule of probability, as calculating the probability of "at least one" is often easier by calculating 1 minus the probability of "none". Step 2: Key Formula or Approach: Let 'n' be the number of times the coin is tossed. The probability of getting a head (H) in a single toss of a fair coin is . The probability of getting a tail (T) is . The event "at least one head" is the complement of the event "no heads" (i.e., all tails). We are given the condition: Step 3: Detailed Explanation: First, let's find the probability of getting "no heads" in 'n' tosses. This means getting tails on all 'n' tosses. Since the tosses are independent events, the probability of getting n tails in a row is: Now, substitute this into our inequality: Rearrange the inequality to solve for n: To get rid of the fractions, we can take the reciprocal of both sides. When we do this, we must reverse the inequality sign: Now we need to find the smallest integer 'n' that satisfies this condition. We can test values of n: If n = 1, , which is not greater than 10. If n = 2, , which is not greater than 10. If n = 3, , which is not greater than 10. If n = 4, , which is greater than 10. Step 4: Final Answer: The minimum number of times the coin must be tossed is 4.$

53

PYQ 2025

medium

mathematics ID: cuet-ug-

$Let A and B be two events such that: Match List-I with List-II: List-I List-II (A) (I) 0.2 (B) (II) 0.32 (C) (III) 0.64 (D) (IV) 0.98$

1

$(A) - (II), (B) - (IV), (C) - (III), (D) - (I)$

2

$(A) - (II), (B) - (III), (C) - (IV), (D) - (I)$

3

$(A) - (III), (B) - (IV), (C) - (II), (D) - (I)$

4

$(A) - (III), (B) - (II), (C) - (I), (D) - (IV)$

Official Solution

Correct Option: (2)

$Step 1: Understanding the Concept: This question requires the application of basic probability formulas, including the multiplication rule for conditional probability, the formula for conditional probability, the addition rule, and the complement rule. Step 3: Detailed Explanation: Given: P(A) = 0.8, P(B) = 0.5, P(B|A) = 0.4 (A) P(A B) Using the multiplication rule: . . This matches (II) . (B) P(A|B) Using the formula for conditional probability: . . This matches (III) . (C) P(A B) Using the addition rule: . . This matches (IV) . (D) P(A') Using the complement rule: . . This matches (I) . Step 4: Final Answer: The correct matching is: (A) (II) (B) (III) (C) (IV) (D) (I) This corresponds to option (2).$

List-I	List-II
(A) $P(A%20%5Ccap%20B)$	(I) 0.2
(B) $P(A%7CB)$	(II) 0.32
(C) $P(A%20%5Ccup%20B)$	(III) 0.64
(D) $P(A')$	(IV) 0.98

54

PYQ 2025

medium

mathematics ID: cuet-ug-

$A black and a red die are rolled simultaneously. The probability of obtaining a sum greater than 9, given that the black die resulted in a 5 is$

1

$1/2$

2

$1$

3

$2/3$

4

$1/3$

Official Solution

Correct Option: (4)

$Step 1: Understanding the Concept: This is a problem of conditional probability. We are given some information (the outcome of the black die) which reduces our sample space, and we need to find the probability of another event within this new, smaller sample space. Step 2: Key Formula or Approach: Let A be the event "the sum is greater than 9". Let B be the event "the black die resulted in a 5". We need to find P(A|B), the probability of A given B. The formula is . Alternatively, and more simply, we can directly consider the reduced sample space. Step 3: Detailed Explanation: Let's use the reduced sample space method. The event B, "the black die resulted in a 5", has occurred. This means the possible outcomes are now only those where the first die is a 5. The reduced sample space S' is: S' = {(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6)} The total number of outcomes in this reduced sample space is n(S') = 6. Now, within this sample space, we want to find the outcomes that satisfy the event A, "the sum is greater than 9". Let's check the sum for each outcome in S': (5, 1): sum = 6 (5, 2): sum = 7 (5, 3): sum = 8 (5, 4): sum = 9 (not greater than 9) (5, 5): sum = 10 (greater than 9) (5, 6): sum = 11 (greater than 9) The favorable outcomes are {(5, 5), (5, 6)}. The number of favorable outcomes is 2. The probability is the ratio of favorable outcomes to the total outcomes in the reduced sample space. Step 4: Final Answer: The required probability is}.$

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Probability of all tails	$%5Cfrac%7B1%7D%7B32%7D$
Probability of at least one head	$%5Cfrac%7B31%7D%7B32%7D$

List-I (Sentences)	List-II (Adverbs)
(A) $P(A%20%5Ccap%20B)$	(I) $%5Cfrac%7B2%7D%7B5%7D$
(B) $P(A%20%7C%20B)$	(II) $%5Cfrac%7B4%7D%7B15%7D$
(C) $P(A'%20%7C%20B)$	(III) $%5Cfrac%7B3%7D%7B5%7D$
(D) $P(A'%20%5Ccap%20B')$	(IV) $%5Cfrac%7B2%7D%7B9%7D$

List-I		List-II
A	Randomization is used in	I	Unbiased sampling
B	Randomization is not used in.	II	Probability sampling
C	If every element in the population has an equal chance to be part of the selected sample, then the sampling process is called	III	Biased sampling
D	If a sampling process systematically favours certain outcomes over others, then it is called	IV	Non-probability sampling

List-I	List-II (Adverbs)
(A) P(exactly 2 heads)	(I) $%5Cfrac%7B1%7D%7B4%7D$
(B) P(at least 1 head)	(II) $1$
(C) P(at most 2 heads)	(III) $%5Cfrac%7B3%7D%7B4%7D$
(D) P(exactly 1 head)	(IV) $%5Cfrac%7B1%7D%7B2%7D$

LIST-I(EVENT)	LIST-II(PROBABILITY)
(A) The sum of the number is greater than 11	(i) 0
(B) The sum of the number is 4 or less	(ii) 1/15
(C) The sum of the number is 4	(iii) 2/15
(D) The sum of the number is 4	(iv) 3/15

LIST-I(EVENT)	Valid Outcomes	LIST-II(PROBABILITY)
(A) The sum of the number is greater than 11	(5,6), (6,5)	0
(B) The sum of the number is 4 or less	(1,2), (2,1), (1,3), (3,1)	1/15
(C) The sum of the number is 4	(1,3), (3,1), (2,2)	2/15
(D) The sum of the number is 4	(1,3), (3,1), (2,2)	3/15

Probability

High-Yield Trend

Chapter Questions 54 MCQs

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Chapter Questions
54 MCQs