JEE Mathematics Probability MCQs Set 14

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Chapter-wise Objective Questions: Probability

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Multiple Choice Questions

Question 1: One ticket is selected randomly from the set of 100 tickets numbered 00, 01, 02, ..., 98, 99. \( E_1 \) and \( E_2 \) denote the sum and the product of the digits of the number on the selected ticket. The value of \( P(E_1=9\mid E_2=0) \) is
(a) 3/19
(b) 1/18
(c) 2/19
(d) 1/19
Show Answer & Explanation

Answer: (c) 2/19

Explanation:
1. Product 0 means one of the digits is 0: 10 + 10 - 1 = 19 tickets.
2. Among these, digit sum 9: 09 and 90.
3. P = 2/19.

Teacher's Note:
1. Restrict to tickets whose digit product is 0, then count those with digit sum 9.

Question 2: A piece of equipment will function when all the three components A, B, C are working. The probability of A failing during one year is 0.15, that of B is 0.05 and that of C is 0.10. The probability that the equipment will fail before the end of the year is
(a) 0.95
(b) 1
(c) 0.72675
(d) 0.27325
Show Answer & Explanation

Answer: (d) 0.27325

Explanation:
1. P(all three work) = (0.85)(0.95)(0.90) = 0.72675.
2. P(equipment fails) = 1 - 0.72675 = 0.27325.

Teacher's Note:
1. The equipment fails if at least one component fails, so use the complement.

Question 3: A box contains 10 mangoes out of which 4 are rotten. Two mangoes are taken together. If one of them is found to be good, the probability that the other is rotten is
(a) 8/13
(b) 7/13
(c) 9/13
(d) 5/13
Show Answer & Explanation

Answer: (a) 8/13

Explanation:
1. Pairs with at least one good: C(10,2) - C(4,2) = 45 - 6 = 39.
2. Pairs with one good and one rotten: 6 x 4 = 24.
3. P = 24/39 = 8/13.

Teacher's Note:
1. Condition on at least one good mango and count the mixed pairs.

Question 4: Let A and B be two events such that \( P(\overline{A\cup B})=\frac16 \), \( P(A\cap B)=\frac14 \) and \( P(\bar A)=\frac14 \). Then events A and B are
(a) mutually exclusive and independent
(b) equally likely, but not independent
(c) independent but not equally likely
(d) equally likely and mutually exclusive
Show Answer & Explanation

Answer: (c) independent but not equally likely

Explanation:
1. P(A or B) = 5/6 and P(A) = 3/4.
2. P(B) = P(A or B) - P(A) + P(A and B) = 5/6 - 3/4 + 1/4 = 1/3.
3. P(A) P(B) = (3/4)(1/3) = 1/4 = P(A and B), so they are independent.
4. P(A) is not P(B), so they are not equally likely.

Teacher's Note:
1. Check independence by comparing P(A)P(B) with P(A and B).

Question 5: A candidate takes three tests in succession and the probability of passing the first test is p. The probability of passing each succeeding test is p or p/2 according as he passes or fails in the preceding one. The candidate is selected if he passes at least two tests. The probability that the candidate is selected is
(a) p(2 - p)
(b) p^2(1 - p)
(c) p^2(2 - p)
(d) p + p^2 + p^3
Show Answer & Explanation

Answer: (c) p^2(2 - p)

Explanation:
1. Cases (P = pass, F = fail): PPP = p^3; PPF = p^2(1 - p); PFP = p(1 - p)(p/2); FPP = (1 - p)(p/2)(p).
2. Sum = p^3 + p^2(1 - p) + p^2(1 - p)/2 + p^2(1 - p)/2 = p^3 + 2p^2(1 - p).
3. = 2p^2 - p^3 = p^2(2 - p).

Teacher's Note:
1. Write the four ways to pass at least two tests and use the changed probability after a failure.

Question 6: A coin is biased so that the probability of falling head when tossed is 1/4. If the coin is tossed 5 times, the probability of obtaining 2 heads and 3 tails, with heads occurring in succession, is
(a) 3^3/4^5
(b) 5 x 3^3/4^5
(c) 3^3/5^4
(d) 2 (3/8)^3
Show Answer & Explanation

Answer: (d) 2 (3/8)^3

Explanation:
1. Heads must be together: treat HH as a block among 3 tails: 4 positions.
2. Each pattern has probability (1/4)^2 (3/4)^3 = 27/1024.
3. P = 4 x 27/1024 = 27/256 = 2 (3/8)^3.

Teacher's Note:
1. Glue the two heads into a block and count its positions among the tails.

Question 7: The probability that a teacher will give a surprise test during any class meeting is 3/5. If a student is absent on two days, then the probability that he will miss at least one test is
(a) 4/25
(b) 9/25
(c) 21/25
(d) 13/25
Show Answer & Explanation

Answer: (c) 21/25

Explanation:
1. P(no test on a given day) = 2/5.
2. P(no test on both days) = (2/5)^2 = 4/25.
3. P(at least one test missed) = 1 - 4/25 = 21/25.

Teacher's Note:
1. 'At least one' is solved by the complement.

Question 8: A person draws a card from a well shuffled pack of 52 playing cards. Replaces it and shuffles the pack. He continues doing so until he draws a spade. The chance that he fails the first two times is
(a) 9/16
(b) 9/64
(c) 9/32
(d) 1/16
Show Answer & Explanation

Answer: (a) 9/16

Explanation:
1. Each draw is a spade with probability 1/4 and not a spade with probability 3/4.
2. Failing on each of the first two draws: (3/4)^2 = 9/16.

Teacher's Note:
1. The draws are independent because of replacement.

Question 9: A symmetrical die is thrown first and secondly two symmetrical dice are thrown. The probability that the first throw was a face with 6 points upward and the second throw was a sum of 6 is
(a) 1/36
(b) 5/216
(c) 5/36
(d) 1/216
Show Answer & Explanation

Answer: (b) 5/216

Explanation:
1. P(first throw shows 6) = 1/6.
2. P(two dice sum to 6) = 5/36.
3. The throws are independent: (1/6)(5/36) = 5/216.

Teacher's Note:
1. Multiply the probabilities of independent throws.

Question 10: A box contains 2 black, 3 white and 4 red balls. One ball is drawn at random. From the remaining balls another ball is drawn and kept aside, the first. The process is repeated till all the balls are drawn from the box. The probability that the balls drawn are in the sequence of 2 black, 3 white, 4 red is
(a) C(2,2) C(3,3) C(4,3)/C(9,3)
(b) 1/180
(c) 1/1260
(d) 1/630
Show Answer & Explanation

Answer: (c) 1/1260

Explanation:
1. The colour sequence BBWWWRRRR can be formed in 2! 3! 4! ways out of 9! total orderings of the balls.
2. P = 2! 3! 4!/9! = 288/362880 = 1/1260.

Teacher's Note:
1. Think of the drawing order as a random arrangement of all nine balls.

Question 11: Two fair dice are tossed. Let X be the event that the first die shows an even number and Y be the event that the second die shows an odd number. The two events X and Y are
(a) dependent
(b) independent
(c) mutually exclusive
(d) independent and mutually exclusive
Show Answer & Explanation

Answer: (b) independent

Explanation:
1. P(X) = 1/2 and P(Y) = 1/2.
2. P(X and Y) = 9/36 = 1/4 = P(X) P(Y).
3. They can occur together, so they are not mutually exclusive, and they are independent.

Teacher's Note:
1. The two dice do not influence each other, so independence is expected.

Question 12: A six-faced fair die is thrown until 1 comes. Then the probability that 1 comes in an even number of trials is
(a) 7/9
(b) 5/11
(c) 3/4
(d) 3/11
Show Answer & Explanation

Answer: (b) 5/11

Explanation:
1. P(1 first appears on trial 2k) = (5/6)^(2k-1)(1/6).
2. Sum = (5/36) (1 + (25/36) + (25/36)^2 + ...) = (5/36)/(1 - 25/36) = 5/11.

Teacher's Note:
1. Sum the infinite geometric series with ratio (5/6)^2.

Question 13: A and B alternately throw with a pair of dice. Whoever gets a sum of 7 points first will win the game. If A starts the game, the probability of his winning is
(a) 6/11
(b) 3/11
(c) 5/11
(d) 4/11
Show Answer & Explanation

Answer: (a) 6/11

Explanation:
1. P(sum 7) = 1/6 and the failure probability is q = 5/6.
2. P(A wins) = (1/6)(1 + q^2 + q^4 + ...) = (1/6)/(1 - 25/36) = 6/11.

Teacher's Note:
1. Add the chances that A wins on his first, second, third turn and so on.

Question 14: A pair of dice is rolled together till a sum of either 5 or 7 is obtained. Then the probability that 5 comes before 7 is
(a) 1/5
(b) 3/5
(c) 2/5
(d) 4/5
Show Answer & Explanation

Answer: (c) 2/5

Explanation:
1. P(sum 5) = 4/36 and P(sum 7) = 6/36.
2. Only these two sums matter: P(5 before 7) = 4/(4 + 6) = 2/5.

Teacher's Note:
1. Ignore all other rolls and compare the two competing outcomes.

Question 15: A man alternately tosses a coin and throws a die beginning with the coin. The probability that he gets a head before he gets 5 or 6 on the die is
(a) 1/2
(b) 3/4
(c) 1/8
(d) 1/4
Show Answer & Explanation

Answer: (b) 3/4

Explanation:
1. P(head) = 1/2 and P(5 or 6) = 1/3.
2. P = 1/2 + (1/2)(2/3)(1/2) + (1/2)(2/3)(1/2)(2/3)(1/2) + ... = (1/2)/(1 - 1/3) = 3/4.

Teacher's Note:
1. Each full round that fails has probability (1/2)(2/3) = 1/3.

Question 16: A and B alternately cut a card each from a pack of cards with replacement and the pack is shuffled after each cut. If A starts the game and the game is continued till one cuts a spade, the respective probabilities of A and B cutting a spade are
(a) 3/4, 1/4
(b) 1/3, 2/3
(c) 3/7, 4/7
(d) 4/7, 3/7
Show Answer & Explanation

Answer: (d) 4/7, 3/7

Explanation:
1. P(spade) = 1/4.
2. P(A) = (1/4)/(1 - (3/4)^2) = (1/4)/(7/16) = 4/7.
3. P(B) = 1 - 4/7 = 3/7.

Teacher's Note:
1. The first player has the advantage of cutting first.

Question 17: There are four machines and it is known that exactly two of them are faulty. They are tested one by one in a random order till both the faulty machines are identified. Then the probability that only two tests are needed is
(a) 1/3
(b) 1/4
(c) 1/2
(d) 1/6
Show Answer & Explanation

Answer: (a) 1/3

Explanation:
1. Two tests suffice if the first two tested machines are both faulty or both good.
2. P(both faulty) = (2/4)(1/3) = 1/6; P(both good) = (2/4)(1/3) = 1/6.
3. P = 1/3.

Teacher's Note:
1. If both good machines are found first, the remaining two are known to be faulty.

Question 18: A box contains 24 identical balls of which 12 are white and 12 are black. The balls are drawn at random from the box one at a time with replacement. The probability that a white ball is drawn for the 4th time on the 7th draw is
(a) 27/32
(b) 5/32
(c) 5/64
(d) 1/2
Show Answer & Explanation

Answer: (b) 5/32

Explanation:
1. In the first 6 draws exactly 3 are white: C(6,3)(1/2)^6 = 20/64.
2. The 7th draw must be white: 1/2.
3. P = 20/128 = 5/32.

Teacher's Note:
1. Use the binomial probability for the first six draws and then a white seventh ball.

Question 19: A biased coin with probability p, 0 < p < 1, of heads is tossed until a head appears for the first time. If the probability that the number of tosses required is even is 2/5, then p =
(a) 3/5
(b) 2/5
(c) 2/3
(d) 1/3
Show Answer & Explanation

Answer: (d) 1/3

Explanation:
1. With q = 1 - p, P(even) = qp(1 + q^2 + q^4 + ...) = qp/(1 - q^2) = q/(1 + q).
2. q/(1 + q) = 2/5 gives 5q = 2 + 2q, so q = 2/3.
3. p = 1/3.

Teacher's Note:
1. Sum the geometric series and solve the simple equation for q.

Question 20: An urn contains 2 white and 2 black balls. A ball is drawn at random. If it is white it is not replaced into the urn. Otherwise it is replaced along with another ball of the same colour. The process is repeated. The probability that the third ball drawn is black is
(a) 23/30
(b) 15/29
(c) 7/30
(d) 11/30
Show Answer & Explanation

Answer: (a) 23/30

Explanation:
1. Start with (2 white, 2 black). After a white draw the urn is (1, 2); after a black draw it is (2, 3).
2. From (1, 2): next white (1/3) gives (0, 2) and then black has probability 1; next black (2/3) gives (1, 3) and then black has probability 3/4. From (2, 3): next white (2/5) gives (1, 3), black 3/4; next black (3/5) gives (2, 4), black 2/3.
3. P = (1/2)(1/3 + 1/2) + (1/2)(3/10 + 2/5) = 5/12 + 7/20 = 46/60 = 23/30.

Teacher's Note:
1. Use a tree diagram and add the branches whose third draw is black.

Question 21: H is one of the 6 horses entered for a race and is to be ridden by one of the two jockeys A and B. It is 2 to 1 that A rides H, in which case all the horses are likely to win. If B rides H, his chance is trebled. Then the odds against H winning are
(a) 4 to 13
(b) 13 to 7
(c) 13 to 5
(d) 13 to 4
Show Answer & Explanation

Answer: (c) 13 to 5

Explanation:
1. P(A rides) = 2/3 and P(B rides) = 1/3.
2. P(H wins
3. A) = 1/6 and P(H wins
4. B) = 3/6 = 1/2.
5. P(H wins) = (2/3)(1/6) + (1/3)(1/2) = 1/9 + 1/6 = 5/18.
6. Odds against = 13 : 5.

Teacher's Note:
1. Odds against are (1 - p) : p.

Question 22: There are two bags; the first contains 3 black and 4 white balls, and the second 4 black and 3 white balls. A die is cast; if face 1 or 3 turns up a ball is taken from the first bag, and if any other face turns up a ball is taken from the second bag. The probability of choosing a black ball is
(a) 5/21
(b) 10/21
(c) 6/21
(d) 11/21
Show Answer & Explanation

Answer: (d) 11/21

Explanation:
1. P(first bag) = 2/6 = 1/3 and P(second bag) = 2/3.
2. P(black) = (1/3)(3/7) + (2/3)(4/7) = 3/21 + 8/21 = 11/21.

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

Question 23: Urn A contains 6 red and 4 black balls and urn B contains 4 red and 6 black balls. One ball is drawn at random from A and placed in B. Then one ball is drawn at random from B and placed in A. If one ball is now drawn from A, then the probability that it is found to be red is
(a) 32/55
(b) 32/63
(c) 33/55
(d) 25/66
Show Answer & Explanation

Answer: (a) 32/55

Explanation:
1. Go through the four cases for the colours moved: R then R has probability (6/10)(5/11) and A then has 6 red of 10; R then B has (6/10)(6/11) and 5 red; B then R has (4/10)(4/11) and 7 red; B then B has (4/10)(7/11) and 6 red.
2. P(red) = [30 x 6 + 36 x 5 + 16 x 7 + 28 x 6]/(110 x 10) = 640/1100 = 32/55.

Teacher's Note:
1. Treat the two transfers as stages of a tree and compute the final red fraction for each branch.

Question 24: Two-thirds of the students of a class are boys and the rest girls. It is given that the probability of a girl getting 1st class is 0.25 and the same for a boy is 0.28. A student is selected at random from the class. The probability that the student is a 1st class student is
(a) 0.26
(b) 0.27
(c) 0.25
(d) 0.28
Show Answer & Explanation

Answer: (b) 0.27

Explanation:
1. P(1st class) = (2/3)(0.28) + (1/3)(0.25) = 0.18667 + 0.08333 = 0.27.

Teacher's Note:
1. Use total probability over boys and girls.

Question 25: A bag contains (2n + 1) coins. It is known that n of these have a head on both the sides, whereas the remaining (n + 1) coins are fair. A coin is picked up at random from the bag and tossed. If the probability that the toss results in a head is 31/42, then the value of n is
(a) 6
(b) 25
(c) 10
(d) 8
Show Answer & Explanation

Answer: (c) 10

Explanation:
1. P(head) = [n(1) + (n + 1)(1/2)]/(2n + 1) = (3n + 1)/(2(2n + 1)).
2. Setting this equal to 31/42: 21(3n + 1) = 31(2n + 1), so 63n + 21 = 62n + 31 and n = 10.

Teacher's Note:
1. Write P(head) with the law of total probability and solve the linear equation.

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