JEE Mathematics Probability MCQs Set 17

Mathematics Objective Questions and Answers: Probability

Access targeted multiple-choice questions for Probability designed to align with the latest JEE academic syllabus for JEE Mathematics. These objective practice sets help students evaluate their conceptual understanding and improve exam readiness.

Download Probability MCQs with Answers

Access the complete set of multiple-choice questions for Probability below. This focused format allows students to isolate specific topics for thorough review and uninterrupted practice alongside official JEE textbooks.

Multiple Choice Questions

Question 1: An unbiased die is tossed until a number greater than 4 appears. The probability that an even number of tosses is needed is
(a) 2/3
(b) 1/2
(c) 2/5
(d) 1/5
Show Answer & Explanation

Answer: (c) 2/5

Explanation:
1. P(success) = 1/3 and P(failure) = 2/3.
2. P(even number of tosses) = (2/3)(1/3)/(1 - (2/3)^2) = (2/9)/(5/9) = 2/5.

Teacher's Note:
1. The first success must come on the second, fourth, ... toss.

Question 2: The probability of India winning a test match against West Indies is 1/2. Assuming independence from match to match, the probability that in a 5-match series India's second win occurs at the third test is
(a) 2/3
(b) 1/8
(c) 1/2
(d) 1/4
Show Answer & Explanation

Answer: (d) 1/4

Explanation:
1. Exactly one win in the first two tests and a win in the third.
2. P = C(2,1)(1/2)(1/2) x (1/2) = 2/8 = 1/4.

Teacher's Note:
1. Count the arrangements WLW and LWW.

Question 3: A bag contains 2 white, 3 black and 4 green balls. Two balls are drawn one after another with replacement. The probability that the 1st one is white and the second one is black is
(a) 2/27
(b) 4/27
(c) 1/9
(d) 5/27
Show Answer & Explanation

Answer: (a) 2/27

Explanation:
1. P(white) = 2/9 and P(black) = 3/9.
2. The draws are independent: (2/9)(3/9) = 6/81 = 2/27.

Teacher's Note:
1. With replacement the draws are independent.

Question 4: A bag contains 3 white, 2 black and 4 red balls. Three balls are drawn one after another with replacement at random. The probability of drawing a white, a black and a red ball in succession in that order is
(a) 8/243
(b) 1/21
(c) 5/21
(d) 2/243
Show Answer & Explanation

Answer: (a) 8/243

Explanation:
1. P(white) = 3/9, P(black) = 2/9, P(red) = 4/9.
2. P = (3/9)(2/9)(4/9) = 24/729 = 8/243.

Teacher's Note:
1. Multiply independent probabilities for each draw in the required order.

Question 5: There are 6 red and 5 black balls in a bag. Two balls are drawn at random one after another with replacement. The probability that both the balls drawn may be red is
(a) 30/121
(b) 25/121
(c) 18/121
(d) 36/121
Show Answer & Explanation

Answer: (d) 36/121

Explanation:
1. P(red) = 6/11 on each draw.
2. P(both red) = (6/11)^2 = 36/121.

Teacher's Note:
1. Replacement keeps the probability the same for each draw.

Question 6: An urn A contains 8 black and 5 white balls. A second urn B contains 6 black and 7 white balls. A blindfolded person is asked to draw a ball selecting one of the urns at random. The probability that the ball drawn is white is
(a) 7/13
(b) 6/13
(c) 9/13
(d) 5/13
Show Answer & Explanation

Answer: (b) 6/13

Explanation:
1. P(white) = (1/2)(5/13) + (1/2)(7/13) = 6/13.

Teacher's Note:
1. Use total probability over the two urns.

Question 7: One bag contains 5 white and 3 black balls and another contains 4 white and 5 red balls. Two balls are drawn from one of them, chosen at random. The probability that they are of different colours is
(a) 275/624
(b) 275/504
(c) 5/18
(d) 15/56
Show Answer & Explanation

Answer: (b) 275/504

Explanation:
1. First bag: P(different) = (5 x 3)/C(8,2) = 15/28. Second bag: (4 x 5)/C(9,2) = 20/36 = 5/9.
2. P = (1/2)(15/28 + 5/9) = (1/2)(275/252) = 275/504.

Teacher's Note:
1. Compute the probability for each bag and average.

Question 8: A fair coin is tossed. If the result is a head, a pair of fair dice is rolled and the score is noted. If the result is a tail, a card from a well-shuffled pack of eleven cards numbered 1, 2, 3, ..., 11 is picked and the number on the card is noted. Then the probability that the number noted is either 7 or 8 is
(a) 193/1024
(b) 45/512
(c) 243/792
(d) 193/792
Show Answer & Explanation

Answer: (d) 193/792

Explanation:
1. Head: sum 7 or 8 with two dice: 6/36 + 5/36 = 11/36.
2. Tail: 7 or 8 from 11 cards: 2/11.
3. P = (1/2)(11/36) + (1/2)(2/11) = 11/72 + 1/11 = (121 + 72)/792 = 193/792.

Teacher's Note:
1. Use total probability over the coin result.

Question 9: If a randomly chosen year has 53 Sundays, then the probability that it is a non-leap year is
(a) 2/5
(b) 4/5
(c) 3/5
(d) 1/5
Show Answer & Explanation

Answer: (c) 3/5

Explanation:
1. P(non-leap) = 3/4 and P(53 Sundays
2. non-leap) = 1/7; P(leap) = 1/4 and P(53 Sundays
3. leap) = 2/7.
4. P(53 Sundays) = 3/28 + 2/28 = 5/28.
5. P(non-leap
6. 53 Sundays) = (3/28)/(5/28) = 3/5.

Teacher's Note:
1. Apply Bayes' theorem with leap and non-leap years as the hypotheses (using 1 leap year in 4).

Question 10: The chance that Doctor A will diagnose disease X correctly is 60%. The chance that a patient will die by his treatment after correct diagnosis is 40% and the chance of death after wrong diagnosis is 70%. A patient of Doctor A who had disease X died. The probability that his disease was diagnosed correctly is
(a) 5/13
(b) 7/13
(c) 2/13
(d) 6/13
Show Answer & Explanation

Answer: (d) 6/13

Explanation:
1. P(correct and died) = 0.6 x 0.4 = 0.24. P(wrong and died) = 0.4 x 0.7 = 0.28.
2. P(correct
3. died) = 0.24/(0.24 + 0.28) = 6/13.

Teacher's Note:
1. Bayes' theorem with correct and wrong diagnosis as the two hypotheses.

Question 11: A bag contains 6 balls. Two balls are drawn and found to be red. Assuming that every possible number of red balls (0 to 6) in the bag was equally likely, the probability that five balls in the bag are red is
(a) 2/7
(b) 5/6
(c) 12/17
(d) 1/3
Show Answer & Explanation

Answer: (a) 2/7

Explanation:
1. For k red balls, P(two red drawn) is proportional to C(k,2): 0, 0, 1, 3, 6, 10, 15 for k = 0, ..., 6.
2. Total = 35.
3. P(k = 5) = 10/35 = 2/7.

Teacher's Note:
1. Bayes' theorem with equal priors: the posterior is proportional to the likelihood.

Question 12: A man is known to speak the truth 2 out of 3 times. He throws a die and reports that it is a six. Assume that when he does not speak the truth he reports 'six' whatever the number on the die. The probability that it is actually a five is
(a) 2/3
(b) 3/8
(c) 4/5
(d) 1/7
Show Answer & Explanation

Answer: (d) 1/7

Explanation:
1. P(reports six and actual 5) = (1/6)(1/3) = 1/18.
2. P(reports six) = (1/6)(2/3) + (5/6)(1/3) = 7/18.
3. P(five
4. reports six) = (1/18)/(7/18) = 1/7.

Teacher's Note:
1. Bayes' theorem: only a lie can make him report six when the die shows five.

Question 13: Let E be an event which is neither a certainty nor an impossibility. If probabilities are such that \( P(E)=1+\lambda+\lambda^2 \) and \( P(\bar E)=(1+\lambda)^2 \) in terms of an unknown \( \lambda \), then P(E) is
(a) 1/2
(b) 3/4
(c) 1
(d) 1/4
Show Answer & Explanation

Answer: (b) 3/4

Explanation:
1. P(E) + P(not E) = 1: 1 + lambda + lambda^2 + 1 + 2 lambda + lambda^2 = 1, so 2 lambda^2 + 3 lambda + 1 = 0.
2. lambda = -1/2 or lambda = -1.
3. lambda = -1 gives P(E) = 1 (a certainty), which is excluded.
4. lambda = -1/2 gives P(E) = 1 - 1/2 + 1/4 = 3/4.

Teacher's Note:
1. Use the condition that the two probabilities add up to 1 and discard the root that gives a certainty.

Question 14: There are 10 pairs of shoes in a cupboard, out of which 4 shoes are picked out one by one randomly. The probability that there is at least one correct pair is
(a) 69/323
(b) 89/323
(c) 99/323
(d) 79/323
Show Answer & Explanation

Answer: (c) 99/323

Explanation:
1. P(no pair) = (20/20)(18/19)(16/18)(14/17) = 224/323.
2. P(at least one pair) = 1 - 224/323 = 99/323.

Teacher's Note:
1. Choose shoes one by one, each time excluding the partner of the shoes already picked.

Question 15: If \( \frac{1+3p}3,\ \frac{1-p}4 \) and \( \frac{1-2p}2 \) are the probabilities of three mutually exclusive events, then p lies in
(a) [1/3, 1/2]
(b) [0, 1]
(c) [0, 1/2]
(d) [1/3, 1]
Show Answer & Explanation

Answer: (a) [1/3, 1/2]

Explanation:
1. Each probability must be between 0 and 1, and their sum must be at most 1.
2. (1 - 2p)/2 at least 0 gives p at most 1/2.
3. The sum is (4(1 + 3p) + 3(1 - p) + 6(1 - 2p))/12 = (13 - 3p)/12, which is at most 1 when p is at least 1/3.
4. So 1/3 is at most p, which is at most 1/2.

Teacher's Note:
1. Apply the condition 0 to 1 for each probability and the sum at most 1.

Question 16: There are 5 balls numbered 1 to 5 and 5 boxes numbered 1 to 5. The balls are kept in the boxes, one in each box. The probability that exactly 2 balls are kept in the corresponding numbered boxes and the remaining 3 balls are in the wrong boxes is
(a) 1/12
(b) 1/5
(c) 1/6
(d) 1/10
Show Answer & Explanation

Answer: (c) 1/6

Explanation:
1. Choose the 2 balls in correct boxes: C(5,2) = 10.
2. The other 3 must be deranged: 2 ways.
3. P = 20/5! = 20/120 = 1/6.

Teacher's Note:
1. Use the number of derangements of 3 objects, which is 2.

Question 17: A 2 x 2 matrix is formed with entries from the set {0, 1}. The probability that it is singular is
(a) 5/8
(b) 3/16
(c) 1/8
(d) 5/16
Show Answer & Explanation

Answer: (a) 5/8

Explanation:
1. There are 16 matrices.
2. Non-singular (determinant nonzero): ad - bc = 1 gives 3 and ad - bc = -1 gives 3: 6 matrices.
3. Singular: 10. P = 10/16 = 5/8.

Teacher's Note:
1. Count the matrices with nonzero determinant and subtract.

Question 18: A determinant is chosen at random from the set of all determinants of order 2 with elements 0 or 1 only. The probability that the determinant is negative is
(a) 3/8
(b) 3/16
(c) 7/8
(d) 5/8
Show Answer & Explanation

Answer: (b) 3/16

Explanation:
1. ad - bc = -1 requires ad = 0 and bc = 1: b = c = 1 and (a, d) not both 1: 3 cases.
2. P = 3/16.

Teacher's Note:
1. By symmetry the number of negative determinants equals the number of positive ones.

Question 19: A determinant is chosen at random from the set of all determinants of order 2 with elements 0 or 1 only. The probability that the determinant is non-zero is
(a) 7/8
(b) 3/8
(c) 5/8
(d) 3/16
Show Answer & Explanation

Answer: (b) 3/8

Explanation:
1. Non-zero determinants: 3 positive and 3 negative = 6 of 16.
2. P = 6/16 = 3/8.

Teacher's Note:
1. Add the probabilities of positive and negative determinants.

Question 20: A letter is known to have come from CHENNAI, JAIPUR, NAINITAL or MUMBAI. On the post mark only two consecutive letters AI are legible. The probability that it came from Mumbai is
(a) 39/191
(b) 42/149
(c) 38/191
(d) 39/190
Show Answer & Explanation

Answer: (b) 42/149

Explanation:
1. Pairs of consecutive letters: CHENNAI 6 (AI once), JAIPUR 5 (once), NAINITAL 7 (once), MUMBAI 5 (once).
2. P(AI
3. town) = 1/6, 1/5, 1/7, 1/5.
4. P(Mumbai
5. AI) = (1/5)/(1/6 + 1/5 + 1/7 + 1/5) = (1/5)/(149/210) = 42/149.

Teacher's Note:
1. Equal priors: the posterior is proportional to the chance of seeing AI.

Question 21: Four friends put their car keys on a table. When they leave the place, they pick up their keys at random. The probability that no person picks his own key is
(a) 1/4
(b) 3/4
(c) 3/8
(d) 5/8
Show Answer & Explanation

Answer: (c) 3/8

Explanation:
1. The number of derangements of 4 objects is 9.
2. P = 9/24 = 3/8.

Teacher's Note:
1. Use the derangement count D(4) = 9.

Question 22: If A and B are two independent events such that \( P(A)=\frac13 \) and \( P(B)=\frac34 \), then \( P\left(\frac B{A\cup B}\right)= \)
(a) 8/10
(b) 6/10
(c) 7/10
(d) 9/10
Show Answer & Explanation

Answer: (d) 9/10

Explanation:
1. P(A or B) = 1/3 + 3/4 - (1/3)(3/4) = 5/6.
2. P(B
3. A or B) = P(B)/P(A or B) = (3/4)/(5/6) = 9/10.

Teacher's Note:
1. Because B is contained in A or B, the intersection is just P(B).

Question 23: A bag contains 6 white balls and 4 black balls. A ball is drawn and is put back in the bag along with 5 balls of the same colour as that of the ball drawn. A ball is drawn again at random. The probability that the ball drawn now is white is
(a) 2/5
(b) 1/5
(c) 4/5
(d) 3/5
Show Answer & Explanation

Answer: (d) 3/5

Explanation:
1. If the first ball is white (6/10), the bag then has 11 white and 4 black: P(white) = 11/15.
2. If black (4/10), the bag has 6 white and 9 black: P(white) = 6/15.
3. P = (6/10)(11/15) + (4/10)(6/15) = 66/150 + 24/150 = 90/150 = 3/5.

Teacher's Note:
1. Use total probability over the colour of the first ball.

Question 24: Bag A contains 4 white and 7 black balls. Bag B contains 5 white and 6 black balls. A die is rolled. If 2 or 5 turns up then bag A is chosen, otherwise bag B. If one ball is drawn at random from the selected bag, then the probability that it is black is
(a) 19/33
(b) 4/11
(c) 7/11
(d) 6/11
Show Answer & Explanation

Answer: (a) 19/33

Explanation:
1. P(bag A) = 1/3 and P(bag B) = 2/3.
2. P(black) = (1/3)(7/11) + (2/3)(6/11) = 7/33 + 12/33 = 19/33.

Teacher's Note:
1. Weight each bag's black-ball chance by the probability of choosing that bag.

Question 25: A bag contains 5 balls the colours of which are not known. Two balls are drawn and found to be red. Assuming that every possible number of red balls in the bag (0 to 5) was equally likely, the probability that all the balls in the bag are red is
(a) 1/3
(b) 1/4
(c) 1/2
(d) 1/5
Show Answer & Explanation

Answer: (c) 1/2

Explanation:
1. For k red balls, the likelihood of drawing two reds is proportional to C(k,2): 0, 0, 1, 3, 6, 10 for k = 0, ..., 5.
2. Total = 20.
3. P(k = 5) = 10/20 = 1/2.

Teacher's Note:
1. Equal priors mean the posterior is proportional to the likelihood.

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Probability Objective Questions & Solutions for JEE Mathematics

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