Question 1

Passage: Four dice are rolled. Question: What is the probability that getting a total of the numbers on the dice is 6 ?

Accepted Answer

Total number of outcomes when four dice are rolled is 6^4 = 1296. Let x_1, x_2, x_3, x_4 be the numbers appearing on the four dice. We need to find the number of solutions to the equation: x_1 + x_2 + x_3 + x_4 = 6 where 1 x_i 6 for i = 1, 2, 3, 4. Let x_i = y_i + 1, where y_i 0. Substituting this into the equation, we get: (y_1 + 1) + (y_2 + 1) + (y_3 + 1) + (y_4 + 1) = 6 y_1 + y_2 + y_3 + y_4 = 2 The number of non-negative integer solutions to this equation is given by ^ n+r-1 C_ r-1 , where n

Question 2

Three events A, B and C are such that A and B are disjoint, A and C are independent, B and C are independent. If 4P(A)=2P(B)=P(C) and P(A B C)=5P(A), then what is the value of P(C) ?

Accepted Answer

Let P(A) = x. From the given relation 4P(A) = 2P(B) = P(C), we have P(B) = 2x and P(C) = 4x. Since A and B are disjoint, P(A B) = 0 and P(A B C) = 0. Since A and C are independent, P(A C) = P(A)P(C) = x(4x) = 4x^2. Since B and C are independent, P(B C) = P(B)P(C) = 2x(4x) = 8x^2. Using the formula for the union of three events: P(A B C) = P(A) + P(B) + P(C) - P(A B) - P(B C) - P(C A) + P(A B C) Substituting the given values: 5x = x + 2x + 4x - 0 - 8x^2 - 4x^2 + 0 5x = 7x - 12x^2 12x^2 - 2x = 0 2

Question 3

Passage: Let A, B, C and D be mutually exclusive and exhaustive events and P(A) 2 = P(B) 3 = P(C) 5 = P(D) 8 . Question: What is [2P(A)+3P(B)] [4P(C)+5P(D)] equal to ?

Accepted Answer

Let P(A) 2 = P(B) 3 = P(C) 5 = P(D) 8 = k Then P(A) = 2k, P(B) = 3k, P(C) = 5k, and P(D) = 8k Since A, B, C, and D are mutually exclusive and exhaustive events, P(A) + P(B) + P(C) + P(D) = 1 2k + 3k + 5k + 8k = 1 18k = 1 k = 1 18 We need to evaluate 2P(A) + 3P(B) 4P(C) + 5P(D) Substituting the values in terms of k: 2(2k) + 3(3k) 4(5k) + 5(8k) = 4k + 9k 20k + 40k = 13k 60k = 13 60 Answer: 13 60

Question 4

Passage: Four dice are rolled. Question: What is the probability that getting a total of the numbers on the dice is at least 23 ?

Accepted Answer

Total number of outcomes when four dice are rolled is 6^4 = 1296. Let x_1, x_2, x_3, x_4 be the numbers on the four dice. The required condition is x_1 + x_2 + x_3 + x_4 23. Since the maximum possible sum is 24, the possible values for the sum are 23 and 24. For the sum to be 24, all four dice must show 6. Number of ways = 1 For the sum to be 23, three dice must show 6 and one die must show 5. Number of ways = 4! 3!1! = 4 Total number of favorable outcomes = 1 + 4 = 5. Required probability = 5 1

Question 5

For a Binomial distribution with mean 6 and standard deviation 2 , what is the value of P(X=0) ?

Accepted Answer

Let n be the number of trials and p be the probability of success. Mean = np = 6 Standard deviation = npq = 2 Variance = npq = 2 Dividing the variance by the mean: npq np = 2 6 q = 1 3 Since p + q = 1, we get p = 2 3 . Substituting p into the mean equation: n ( 2 3 ) = 6 n = 9 The probability of X = 0 is given by: P(X=0) = ^ n C_ 0 p^0 q^n P(X=0) = ^ 9 C_ 0 ( 2 3 )^0 ( 1 3 )^9 = ( 1 3 )^9 Answer: ( 1 3 )^9

Question 6

Passage: A 4-digit number is selected at random formed by using the digits 0, 1, 2, 3 and 4 (where repetition of digits is not allowed). Question: What is the probability that the number selected is d

Accepted Answer

The total number of 4-digit numbers that can be formed using the digits 0, 1, 2, 3, 4 without repetition is calculated by filling the thousands place in 4 ways (excluding 0) and the remaining 3 places in ^ 4 P_ 3 = 24 ways. Total numbers = 4 24 = 96 For a number to be divisible by 6, it must be divisible by both 2 and 3. For divisibility by 3, the sum of the digits must be a multiple of 3. The sum of all five digits is 0 + 1 + 2 + 3 + 4 = 10. To choose 4 digits whose sum is a multiple of 3, the

Question 7

In an entrance test there are multiple choice questions. There are four options for each question, of which only one is correct. The probability that a student knows the answer to a question is 90\%.

Accepted Answer

Let E_1 be the event that the student knows the answer and E_2 be the event that the student guesses the answer. Let A be the event that the student gets the correct answer. From the given information: P(E_1) = 90 100 = 9 10 P(E_2) = 1 - P(E_1) = 1 10 If the student knows the answer, the probability of getting it correct is 1, so P(A|E_1) = 1. If the student guesses, since there are 4 options with only one correct, P(A|E_2) = 1 4 . The probability that the student was guessing given that the ans

Question 8

Passage: Let A, B, C and D be mutually exclusive and exhaustive events and P(A) 2 = P(B) 3 = P(C) 5 = P(D) 8 . Question: If G is the geometric mean of P(A), P(B), P(C) and P(D), then what is 9G equal

Accepted Answer

Since A, B, C and D are mutually exclusive and exhaustive events, the sum of their probabilities is 1. P(A) + P(B) + P(C) + P(D) = 1 Let P(A) 2 = P(B) 3 = P(C) 5 = P(D) 8 = k. Then P(A) = 2k, P(B) = 3k, P(C) = 5k, P(D) = 8k. Substituting these into the sum equation gives: 2k + 3k + 5k + 8k = 1 18k = 1 k = 1 18 The probabilities are P(A) = 2 18 , P(B) = 3 18 , P(C) = 5 18 , P(D) = 8 18 . The geometric mean G of the four probabilities is: G = (P(A) P(B) P(C) P(D))^ 1 4 G = ( 2 18 3 18 5 18 8 18 )^

Question 9

Consider the following statements in respect of the events A, B, C : I. (A B C) ( A B C ) is an impossible event. II. (A B C) ( A B C ) is a possible event. Which of the statements given above is/are

Accepted Answer

Using De Morgan's laws, we have A B C = A B C . Substituting this in the expression for Statement I: (A B C) ( A B C ) = (A B C) (A B C) = Since the intersection of an event and its complement is the empty set , it represents an impossible event. Thus, Statement I is correct. Again, using De Morgan's laws, we have A B C = A B C . Substituting this in the expression for Statement II: (A B C) ( A B C ) = (A B C) (A B C) = This is also an impossible event. Thus, Statement II is incorrect. Therefore

Question 10

Passage: Let A, B, C and D be mutually exclusive and exhaustive events and P(A) 2 = P(B) 3 = P(C) 5 = P(D) 8 . Question: What is P(A)+P(B)+P(C) equal to ?

Accepted Answer

Let P(A) 2 = P(B) 3 = P(C) 5 = P(D) 8 = k P(A) = 2k, P(B) = 3k, P(C) = 5k, P(D) = 8k Since A, B, C and D are mutually exclusive and exhaustive events, P(A) + P(B) + P(C) + P(D) = 1 2k + 3k + 5k + 8k = 1 18k = 1 k = 1 18 P(A) + P(B) + P(C) = 2k + 3k + 5k = 10k 10k = 10 1 18 = 5 9 Answer: 5 9

Question 11

The probability that a family owns a laptop is 0.68; that it also owns a desktop is 0.56. If the probability that it owns both is 0.48, then what is the probability that a randomly selected family own

Accepted Answer

Let L be the event that a family owns a laptop and D be the event that a family owns a desktop. Given: P(L) = 0.68 P(D) = 0.56 P(L D) = 0.48 The probability that a randomly selected family owns a laptop or a desktop is given by P(L D). Using the addition theorem of probability: P(L D) = P(L) + P(D) - P(L D) Substituting the given values: P(L D) = 0.68 + 0.56 - 0.48 P(L D) = 1.24 - 0.48 = 0.76

Question 12

An urn contains 10 white and 5 red balls. If two balls are drawn at random, then what is the probability that both the balls are red?

Accepted Answer

Total number of balls in the urn = 10 + 5 = 15 Number of ways to draw 2 balls from 15 balls = ^ 15 C_ 2 Number of ways to draw 2 red balls from 5 red balls = ^ 5 C_ 2 Required probability = ^ 5 C_ 2 ^ 15 C_ 2 ^ 5 C_ 2 = 5 4 2 = 10 ^ 15 C_ 2 = 15 14 2 = 105 Required probability = 10 105 = 2 21

Question 13

An urn contains 5 white, 6 red and 4 blue balls. Three balls are drawn at random. What is the probability that a white ball, a red ball and a blue ball are drawn?

Accepted Answer

Total number of balls in the urn = 5 + 6 + 4 = 15 Total number of ways to draw 3 balls from 15 balls = ^ 15 C_ 3 = 15 14 13 3 2 1 = 455 Number of ways to draw 1 white, 1 red and 1 blue ball = ^ 5 C_ 1 ^ 6 C_ 1 ^ 4 C_ 1 = 5 6 4 = 120 Required probability = 120 455 = 24 91

Question 14

Under which of the following conditions may binomial distribution be used? I. The number of trials is infinite and not fixed. II. The trials are independent. III. Each trial has two possible outcomes.

Accepted Answer

The conditions for a binomial distribution are: 1. The number of trials is finite and fixed. 2. The trials are independent of each other. 3. Each trial has exactly two mutually exclusive outcomes, often termed as success and failure. 4. The probability of success remains constant for each trial. Statement I is incorrect because the number of trials must be finite and fixed. Statement II is correct because the trials must be independent. Statement III is correct because each trial must have exact

Question 15

Three faces of a die are black, two faces are white and one face is red. The die is tossed three times. What is the probability that the colours black, white and red appear in the first, second and th

Accepted Answer

Total number of faces on the die is 6. Number of black faces is 3. Probability of getting a black face in the first toss is P(B) = 3 6 = 1 2 . Number of white faces is 2. Probability of getting a white face in the second toss is P(W) = 2 6 = 1 3 . Number of red faces is 1. Probability of getting a red face in the third toss is P(R) = 1 6 . Since the tosses are independent events, the required probability is the product of the individual probabilities. Required probability = P(B) P(W) P(R) = 1 2

Question 16

A person X speaks the truth 4 out of 5 times and person Y speaks the truth 5 out of 6 times. What is the probability that they will contradict each other in stating the fact?

Accepted Answer

Let P(X) be the probability that person X speaks the truth and P(Y) be the probability that person Y speaks the truth. P(X) = 4 5 P(Y) = 5 6 The probability that X lies is P(X') = 1 - 4 5 = 1 5 . The probability that Y lies is P(Y') = 1 - 5 6 = 1 6 . They contradict each other when one speaks the truth and the other lies. The probability of this event is given by: P( contradiction ) = P(X)P(Y') + P(X')P(Y) P( contradiction ) = ( 4 5 ) ( 1 6 ) + ( 1 5 ) ( 5 6 ) P( contradiction ) = 4 30 + 5 30 =

Question 17

An event X can happen with probability p and event Y can happen with probability q. Further, X and Y are independent events. Which of the following statements is/are correct? I. The probability that e

Accepted Answer

Given P(X) = p and P(Y) = q. Since X and Y are independent events, P(X Y) = P(X)P(Y) = pq. The probability that exactly one of the events happens is given by P(X Y') + P(X' Y) = P(X) + P(Y) - 2P(X Y) = p + q - 2pq. Statement I claims this probability is p + q - pq, which is incorrect. The probability that at least one of the events happens is given by P(X Y) = P(X) + P(Y) - P(X Y) = p + q - pq. Statement II claims this probability is p + q - 2pq, which is also incorrect. Therefore, neither State

Question 18

A fair coin is tossed 4 times. What is the probability that two heads do not occur consecutively?

Accepted Answer

Total number of possible outcomes when a fair coin is tossed 4 times is 2^4 = 16. Let E be the event that two heads do not occur consecutively. The favorable outcomes are the sequences of length 4 that do not contain 'HH'. We can count these by the number of heads in the sequence: Case 1: 0 heads TTTT (1 outcome) Case 2: 1 head HTTT, THTT, TTHT, TTTH (4 outcomes) Case 3: 2 heads To avoid consecutive heads, the heads must be separated by tails. The possible arrangements are HTHT, HTTH, THTH (3 ou

Question 19

In a throw of three dice, what is the probability of getting one prime number, one composite number and one number which is neither prime nor composite?

Accepted Answer

The total number of outcomes when three dice are thrown is 6 6 6 = 216. The numbers on a die are 1, 2, 3, 4, 5, 6. Prime numbers are 2, 3, 5 (total 3). Composite numbers are 4, 6 (total 2). The number which is neither prime nor composite is 1 (total 1). The number of ways to get one prime, one composite, and one neither prime nor composite is the number of permutations of these three types among the three dice multiplied by the number of choices for each type. Number of favorable outcomes = 3! (

Question 20

If P(A) = 0.3, P(B) = 0.4 and P(A|B) = 0.5, then what is the value of P(B|A)?

Accepted Answer

Probability — Previous Year Questions & Solutions