JEE Mathematics Probability MCQs Set 16

Practice MCQs for JEE Mathematics 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.

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Multiple Choice Questions

Question 1: A bag contains 12 two-rupee coins, 7 one-rupee coins and 4 half-rupee coins. If three coins are selected at random, p = probability that the sum of the three coins is maximum, q = probability that the sum of the three coins is minimum, r = probability that each coin is of a different value. Arrange p, q and r in increasing order of magnitude.
(a) q, p, r
(b) p, q, r
(c) r, q, p
(d) r, p, q
Show Answer & Explanation

Answer: (a) q, p, r

Explanation:
1. Total = C(23,3) = 1771.
2. Maximum sum: three two-rupee coins: C(12,3) = 220. Minimum sum: three half-rupee coins: C(4,3) = 4. Different values: 12 x 7 x 4 = 336.
3. So q = 4/1771 < p = 220/1771 < r = 336/1771.

Teacher's Note:
1. Count favourable selections for each event over a common denominator.

Question 2: A and B are to throw 2 dice each. If A throws a sum of 9 points, then B's chance of throwing a higher sum is
(a) 1/3
(b) 1/6
(c) 1/2
(d) 5/9
Show Answer & Explanation

Answer: (b) 1/6

Explanation:
1. A higher sum means 10, 11 or 12.
2. Ways: 3 + 2 + 1 = 6 out of 36.
3. P = 6/36 = 1/6.

Teacher's Note:
1. Count the outcomes of two dice with sum above 9.

Question 3: A six-faced die is so biased that it is twice as likely to show an even number as an odd number when thrown. It is thrown twice. The probability that the sum of the numbers thrown is even is
(a) 5/9
(b) 1/3
(c) 6/9
(d) 4/9
Show Answer & Explanation

Answer: (a) 5/9

Explanation:
1. P(even) = 2/3 and P(odd) = 1/3.
2. The sum is even if both throws are even or both are odd.
3. P = (2/3)^2 + (1/3)^2 = 5/9.

Teacher's Note:
1. The parity of the sum is even when both throws have the same parity.

Question 4: The chance of throwing a sum of 6 points with 4 dice is
(a) 6/6^4
(b) 10/6^4
(c) 5/6^4
(d) 15/6^4
Show Answer & Explanation

Answer: (b) 10/6^4

Explanation:
1. Sum 6 with four dice: partitions 3+1+1+1 (4 arrangements) and 2+2+1+1 (6 arrangements).
2. Total = 10 outcomes out of 6^4.
3. P = 10/6^4.

Teacher's Note:
1. List partitions of 6 into four parts, each at most 6, and count their arrangements.

Question 5: Two symmetric dice are rolled once. The probability that both the dice will show 4 is p. The probability for the sum to be 8 is q. Then p : q is
(a) 1 : 5
(b) 5 : 4
(c) 5 : 1
(d) 4 : 5
Show Answer & Explanation

Answer: (a) 1 : 5

Explanation:
1. p = 1/36.
2. Sum 8: (2,6), (3,5), (4,4), (5,3), (6,2) = 5 outcomes, so q = 5/36.
3. p : q = 1 : 5.

Teacher's Note:
1. Compare the counts of favourable outcomes out of 36.

Question 6: It is given that there are 53 Fridays in a leap year. Then the probability that it will have 53 Thursdays is
(a) 1/7
(b) 1/2
(c) 2/7
(d) 4/7
Show Answer & Explanation

Answer: (b) 1/2

Explanation:
1. The two extra days contain a Friday: (Thu, Fri) or (Fri, Sat).
2. 53 Thursdays happens only for (Thu, Fri).
3. P = 1/2.

Teacher's Note:
1. Condition on the two possible extra pairs containing Friday.

Question 7: In a game of bridge, the probability of a particular player having only one ace is
(a) C(4,1) C(48,12)/C(52,13)
(b) C(48,12)/C(52,13)
(c) C(52,12)/C(52,13)
(d) C(4,1)/C(52,13)
Show Answer & Explanation

Answer: (a) C(4,1) C(48,12)/C(52,13)

Explanation:
1. A hand is 13 cards from 52.
2. Exactly one ace: choose the ace in C(4,1) ways and the other 12 cards from the 48 non-aces.
3. P = C(4,1) C(48,12)/C(52,13).

Teacher's Note:
1. Split the hand into ace cards and non-ace cards.

Question 8: In a game of bridge, player A has received two aces. The probability that his partner has been dealt the other two aces is
(a) C(2,2) C(37,11)/C(52,13)
(b) C(2,2) C(48,11)/C(52,13)
(c) C(2,2) C(48,11)/C(39,13)
(d) C(2,2) C(37,11)/C(39,13)
Show Answer & Explanation

Answer: (d) C(2,2) C(37,11)/C(39,13)

Explanation:
1. The other 39 cards (not in A's hand) contain exactly 2 aces.
2. The partner receives 13 of these 39 cards.
3. He gets both aces: C(2,2) C(37,11)/C(39,13).

Teacher's Note:
1. Only the 39 cards outside A's hand matter for the partner.

Question 9: 5 cards are drawn at random from a well-shuffled pack of 52 playing cards. If it is known that there will be at least 3 hearts, the probability that all the 5 are hearts is
(a) C(13,5)/(C(13,3) + C(13,4) + C(13,5))
(b) C(13,5)/C(52,5)
(c) C(13,5)/(C(13,3)C(13,4)C(13,5))
(d) C(13,5)/(C(13,3)C(39,2) + C(13,4)C(39,1) + C(13,5))
Show Answer & Explanation

Answer: (d) C(13,5)/(C(13,3)C(39,2) + C(13,4)C(39,1) + C(13,5))

Explanation:
1. Given at least 3 hearts, the sample space has C(13,3)C(39,2) + C(13,4)C(39,1) + C(13,5) hands.
2. All 5 hearts: C(13,5).
3. P = C(13,5)/(C(13,3)C(39,2) + C(13,4)C(39,1) + C(13,5)).

Teacher's Note:
1. Restrict the sample space to hands with 3, 4 or 5 hearts.

Question 10: If five cards are drawn at random from a pack of cards, the probability that they belong to different denominations is
(a) 2112/4165
(b) 3223/4165
(c) 1221/4165
(d) 2332/4165
Show Answer & Explanation

Answer: (a) 2112/4165

Explanation:
1. Total hands = C(52,5) = 2598960.
2. Different denominations: choose 5 of 13 values and one of 4 suits for each: C(13,5) x 4^5 = 1287 x 1024 = 1317888.
3. P = 1317888/2598960 = 2112/4165.

Teacher's Note:
1. Choose the five values, then an independent suit for each.

Question 11: If 4 squares are selected at random on a chess board having \( 8\times8 \) squares, then the probability that they will be in a diagonal line is
(a) 2 SUM(n=4 to 8) C(n,4)/C(64,4)
(b) 4 SUM(n=4 to 8) C(n,4)/C(64,4)
(c) (4 SUM(n=4 to 7) C(n,4) + 2 C(8,4))/C(64,4)
(d) (2 SUM(n=4 to 7) C(n,4) + C(8,4))/C(64,4)
Show Answer & Explanation

Answer: (c) (4 SUM(n=4 to 7) C(n,4) + 2 C(8,4))/C(64,4)

Explanation:
1. A diagonal of length n contains C(n,4) choices of 4 squares.
2. In each direction the diagonals have lengths 4, 5, 6, 7 (two of each) and 8 (one).
3. Total = 2 x [2(C(4,4) + C(5,4) + C(6,4) + C(7,4)) + C(8,4)] = 4 SUM(n=4 to 7) C(n,4) + 2 C(8,4) = 364.
4. P = (4 SUM(n=4 to 7) C(n,4) + 2 C(8,4))/C(64,4).

Teacher's Note:
1. Add the combinations over all diagonals of length at least 4, in both directions.

Question 12: Two numbers X and Y are chosen at random from the set {1, 2, ..., 3n}. The probability that \( X^2-Y^2 \) is divisible by 3 is
(a) 5n/(3n - 1)
(b) (5n + 3)/(3(3n - 1))
(c) (5n - 3)/(3(3n + 1))
(d) (5n - 3)/(3(3n - 1))
Show Answer & Explanation

Answer: (d) (5n - 3)/(3(3n - 1))

Explanation:
1. X^2 - Y^2 = (X - Y)(X + Y) is divisible by 3 if X and Y are both multiples of 3 or if they are non-multiples (then X + Y or X - Y is a multiple of 3).
2. Total pairs = C(3n,2) = 3n(3n - 1)/2.
3. Favourable = C(n,2) + C(2n,2) = n(n - 1)/2 + n(2n - 1) = n(5n - 3)/2.
4. P = (5n - 3)/(3(3n - 1)).

Teacher's Note:
1. Any non-multiple of 3 has a square leaving remainder 1 modulo 3.

Question 13: If A, B and C are three events such that P(A) = 0.3, P(B) = 0.4, P(C) = 0.8, P(A and B) = 0.12, P(A and C) = 0.28, P(A and B and C) = 0.09 and P(A or B or C) is at least 0.75, then the limits of P(B and C) are
(a) [0.32, 0.52]
(b) [0.21, 0.49]
(c) [0.23, 0.48]
(d) [0.19, 0.44]
Show Answer & Explanation

Answer: (d) [0.19, 0.44]

Explanation:
1. P(A or B or C) = 0.3 + 0.4 + 0.8 - 0.12 - 0.28 - P(B and C) + 0.09 = 1.19 - P(B and C).
2. The union is at least 0.75, so P(B and C) is at most 0.44.
3. The union is at most 1, so P(B and C) is at least 0.19.
4. So P(B and C) lies in [0.19, 0.44].

Teacher's Note:
1. Use inclusion and exclusion and apply the bounds 0.75 and 1 for the union.

Question 14: A and B are two candidates seeking admission in IIT. The probability that A is selected is 0.5 and the probability that both A and B are selected is at most 0.3. The probability of B getting selected cannot exceed
(a) 0.7
(b) 0.8
(c) 0.9
(d) 0.6
Show Answer & Explanation

Answer: (b) 0.8

Explanation:
1. P(A or B) is at most 1, so P(A) + P(B) - P(A and B) is at most 1.
2. P(B) is at most 1 - 0.5 + P(A and B) = 0.5 + P(A and B), which is at most 0.5 + 0.3 = 0.8.

Teacher's Note:
1. Use P(A or B) = P(A) + P(B) - P(A and B) at most 1 to bound P(B).

Question 15: It is given that the events A and B are such that \( P(A)=\frac14 \), \( P(A\mid B)=\frac12 \) and \( P(B\mid A)=\frac23 \). Then P(B) =
(a) 1/2
(b) 1/6
(c) 1/3
(d) 2/3
Show Answer & Explanation

Answer: (c) 1/3

Explanation:
1. P(A and B) = P(B
2. A) P(A) = (2/3)(1/4) = 1/6.
3. P(B) = P(A and B)/P(A
4. B) = (1/6)/(1/2) = 1/3.

Teacher's Note:
1. Find the intersection from one conditional probability and use it in the other.

Question 16: Two symmetrical dice are thrown. The probability that the sum of the numbers appearing is 11, if 5 appears on the first die, is
(a) 1/18
(b) 1/3
(c) 1/9
(d) 1/6
Show Answer & Explanation

Answer: (d) 1/6

Explanation:
1. Given the first die shows 5, the second die has 6 equally likely values.
2. Sum 11 needs 6 on the second die.
3. P = 1/6.

Teacher's Note:
1. Condition on the first die and look only at the second.

Question 17: A symmetrical die is thrown 3 times and the sum of points thrown is found to be 15. The chance that the 1st throw was a four is
(a) 1/3
(b) 1/4
(c) 1/6
(d) 1/5
Show Answer & Explanation

Answer: (d) 1/5

Explanation:
1. Outcomes with sum 15: 10 ordered triples.
2. With the first throw 4, the other two sum to 11: (5,6) and (6,5), so 2 triples.
3. P = 2/10 = 1/5.

Teacher's Note:
1. Count the ordered triples for the given sum and those with a 4 first.

Question 18: A couple has 2 children. The probability that both are boys, if it is known that the elder child is a boy, is
(a) 3/4
(b) 1/3
(c) 1/2
(d) 2/3
Show Answer & Explanation

Answer: (c) 1/2

Explanation:
1. Given the elder is a boy, the sample space is {BB, BG}.
2. P(BB) = 1/2.

Teacher's Note:
1. The younger child's sex is independent of the elder's.

Question 19: A and B are two independent events such that \( P(\bar A)P(B)=\frac8{25} \) and \( P(A)P(\bar B)=\frac3{25} \). Then P(A) =
(a) 1/2
(b) 3/5
(c) 1/5 or 3/5
(d) 1/5
Show Answer & Explanation

Answer: (c) 1/5 or 3/5

Explanation:
1. Let P(A) = a and P(B) = b. (1 - a)b = 8/25 and a(1 - b) = 3/25.
2. Subtract: b - a = 1/5, so b = a + 1/5.
3. Then a(4/5 - a) = 3/25, i.e. a^2 - (4/5)a + 3/25 = 0, giving a = 1/5 or a = 3/5.
4. Both give a valid b (2/5 or 4/5) and satisfy both equations.

Teacher's Note:
1. Subtract the two equations to eliminate the product ab.

Question 20: A monkey seated before a typewriter with 26 keys denoting the English alphabet types three letters at random. The probability that the monkey types the word 'SIR' is
(a) (1/26)^3
(b) 1/C(26,3)
(c) C(26,3)/26^3
(d) 3/26
Show Answer & Explanation

Answer: (a) (1/26)^3

Explanation:
1. Each key press is one of 26 letters independently.
2. P(S) P(I) P(R) = (1/26)(1/26)(1/26) = (1/26)^3.

Teacher's Note:
1. Each letter must be right independently.

Question 21: A die is loaded so that six turns up twice as often as one and three times as often as any other number. The probability of getting an even number on the die if it is rolled once is
(a) 10/17
(b) 15/17
(c) 6/17
(d) 2/17
Show Answer & Explanation

Answer: (a) 10/17

Explanation:
1. Let P(6) = 6x. Then P(1) = 3x and P(2) = P(3) = P(4) = P(5) = 2x. Total = 17x = 1.
2. P(even) = P(2) + P(4) + P(6) = 2x + 2x + 6x = 10x = 10/17.

Teacher's Note:
1. Express every probability through one unit and normalise.

Question 22: The Intermediate Board has to select an examiner from a list of 100 persons. 40 of them are women and 60 men; 50 of them know Telugu and 50 do not; 75 of them are teachers and the remaining are not. Assuming that these three characteristics are independent, the probability that the board selects a Telugu-knowing woman teacher is
(a) 17/20
(b) 1/20
(c) 2/20
(d) 3/20
Show Answer & Explanation

Answer: (d) 3/20

Explanation:
1. P(woman) = 0.4, P(Telugu) = 0.5, P(teacher) = 0.75.
2. By independence: 0.4 x 0.5 x 0.75 = 0.15 = 3/20.

Teacher's Note:
1. Multiply the three independent probabilities.

Question 23: A, B and C toss a coin one after another in that order. Whoever gets a head first will win the game. If A starts the game, the probability of A's winning is
(a) 2/7
(b) 1/7
(c) 3/7
(d) 4/7
Show Answer & Explanation

Answer: (d) 4/7

Explanation:
1. A wins on tosses 1, 4, 7, ...: P = 1/2 + (1/2)^4 + (1/2)^7 + ...
2. = (1/2)/(1 - 1/8) = 4/7.

Teacher's Note:
1. Sum the geometric series with ratio 1/8 (a full round of three tails).

Question 24: A fair coin is tossed repeatedly. If tail appears on the first four tosses, then the probability of head appearing on the fifth toss is
(a) 1/32
(b) 1/5
(c) 31/32
(d) 1/2
Show Answer & Explanation

Answer: (d) 1/2

Explanation:
1. Tosses are independent, so the earlier tails do not change the fifth toss.
2. P(head) = 1/2.

Teacher's Note:
1. Coin tosses have no memory.

Question 25: In a shooting test, the probabilities for 3 persons A, B, C to hit the target are 1/2, 2/3 and 3/4. If all of them simultaneously aim at the target, the probability for exactly two persons hitting the target is
(a) 1/384
(b) 11/24
(c) 13/24
(d) 1/12
Show Answer & Explanation

Answer: (b) 11/24

Explanation:
1. A and B only: (1/2)(2/3)(1/4) = 1/12. A and C only: (1/2)(1/3)(3/4) = 1/8. B and C only: (1/2)(2/3)(3/4) = 1/4.
2. Sum = 2/24 + 3/24 + 6/24 = 11/24.

Teacher's Note:
1. Add the three 'exactly two' cases.

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Multiple Choice Questions (MCQs) for JEE Mathematics Probability

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