JEE Mathematics Probability MCQs Set 13

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.

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Multiple Choice Questions

Question 1: Out of 5 digits 0, 3, 3, 4, 5, five-digit numbers are formed. If one number is selected at random out of them, the probability that it is divisible by 5 is
(a) 3/16
(b) 7/16
(c) 5/16
(d) 9/16
Show Answer & Explanation

Answer: (b) 7/16

Explanation:
1. Total numbers (0 not first): 4 x 4!/2! = 48.
2. Ending in 0: arrangements of 3, 3, 4, 5 = 4!/2! = 12.
3. Ending in 5: the first digit is non-zero: 4!/2! - 3!/2! = 12 - 3 = 9.
4. Favourable = 21, so P = 21/48 = 7/16.

Teacher's Note:
1. Handle the leading zero separately and count the numbers ending in 0 and in 5.

Question 2: From the first 100 natural numbers a number is chosen at random. The probability for it to be a composite number is
(a) 25/100
(b) 24/100
(c) 26/100
(d) 74/100
Show Answer & Explanation

Answer: (d) 74/100

Explanation:
1. Primes up to 100: 25. The number 1 is neither prime nor composite.
2. Composite numbers = 100 - 25 - 1 = 74.
3. P = 74/100.

Teacher's Note:
1. Remember that 1 is neither prime nor composite.

Question 3: From 101 to 1000 natural numbers a number is taken at random. The probability that the number is divisible by 17 is
(a) 53/900
(b) 58/100
(c) 58/900
(d) 53/1000
Show Answer & Explanation

Answer: (a) 53/900

Explanation:
1. Total numbers = 900.
2. Multiples of 17 up to 1000: 58; up to 100: 5.
3. So 53 multiples from 101 to 1000.
4. P = 53/900.

Teacher's Note:
1. Subtract the counts of multiples up to 1000 and up to 100.

Question 4: A magical die is so loaded that the probability of any face appearing is proportional to the number of points on its face. The probability of an odd number appearing is
(a) 4/7
(b) 2/7
(c) 3/7
(d) 5/7
Show Answer & Explanation

Answer: (c) 3/7

Explanation:
1. P(k) = k/21 for k = 1, ..., 6 (total points 21).
2. P(odd) = (1 + 3 + 5)/21 = 9/21 = 3/7.

Teacher's Note:
1. Normalise by the sum of all weights.

Question 5: A die is loaded such that 6 turning upwards is twice as often as 1 and three times as any other face. The chance that we get a face with one point when we throw such a die is
(a) 6/17
(b) 3/17
(c) 2/17
(d) 5/17
Show Answer & Explanation

Answer: (b) 3/17

Explanation:
1. Let P(6) = 6x. Then P(1) = 3x and P(2) = P(3) = P(4) = P(5) = 2x.
2. Sum: 6x + 3x + 8x = 17x = 1, so x = 1/17.
3. P(1) = 3/17.

Teacher's Note:
1. Express all probabilities with a common unit x and normalise.

Question 6: A fair die is thrown. It is known that a face with less than 5 points is obtained. The probability of obtaining not less than 2 points on the throw is
(a) 1/2
(b) 3/4
(c) 1/8
(d) 1/4
Show Answer & Explanation

Answer: (b) 3/4

Explanation:
1. Given face is 1, 2, 3 or 4: 4 equally likely faces.
2. Faces with at least 2 points: 2, 3, 4: 3 faces.
3. P = 3/4.

Teacher's Note:
1. Condition first: restrict the sample space to faces 1 to 4.

Question 7: An ordinary die has four blank faces, one face marked 2 and another marked 3. Then the probability of obtaining a total of exactly 12 in 5 throws is
(a) 10/6^4
(b) 15/6^4
(c) 5/6^4
(d) 6/6^4
Show Answer & Explanation

Answer: (c) 5/6^4

Explanation:
1. A total of 12 is possible as 3+3+3+3+0 or 2+2+2+3+3.
2. P(3,3,3,3,0 in some order) = 5 x (1/6)^4 x (4/6) = 20/6^5.
3. P(2,2,2,3,3 in some order) = C(5,3) (1/6)^5 = 10/6^5.
4. Total = 30/6^5 = 5/6^4.

Teacher's Note:
1. List the combinations of scores that add to 12 and count their arrangements.

Question 8: It is given that there are 52 Thursdays in a leap year. Then the probability that it will have 52 Fridays is
(a) 3/5
(b) 4/5
(c) 2/5
(d) 1/5
Show Answer & Explanation

Answer: (b) 4/5

Explanation:
1. A leap year has 52 weeks and 2 extra consecutive days, 7 equally likely pairs.
2. No Thursday among the extras: remove (Wed, Thu) and (Thu, Fri), leaving 5 pairs.
3. Among these, no Friday: remove (Fri, Sat), leaving 4.
4. P = 4/5.

Teacher's Note:
1. Condition on the extra pair not containing Thursday, then count pairs that also avoid Friday.

Question 9: Five cards are drawn at random from a well shuffled pack of 52 playing cards. The probability that four of them may have the same face value is
(a) C(13,1) C(4,1) C(48,1)/C(52,5)
(b) C(13,5)/C(52,5)
(c) C(13,4) C(4,1)/C(52,5)
(d) C(13,1) C(48,1)/C(52,5)
Show Answer & Explanation

Answer: (d) C(13,1) C(48,1)/C(52,5)

Explanation:
1. Choose the face value: 13 ways (all 4 cards of that value are taken).
2. The fifth card is any of the other 48 cards.
3. P = 13 x 48/C(52,5).

Teacher's Note:
1. Taking all four suits of a value leaves one free card from the other 48.

Question 10: In a hand at which the probability that 4 queens are held by a specified player is
(a) C(48,9)/C(52,13)
(b) 4 x C(48,9)/C(52,13)
(c) 4 x C(48,13)/C(52,13)
(d) 2 x C(48,13)/C(52,13)
Show Answer & Explanation

Answer: (a) C(48,9)/C(52,13)

Explanation:
1. All 4 queens are in the hand; choose the other 9 cards from the remaining 48.
2. P = C(48,9)/C(52,13).

Teacher's Note:
1. Include the four queens and choose the other nine cards freely.

Question 11: Two cards are drawn one after another at random without replacement. The probability that both of them may have the same face value is
(a) 1/17
(b) 1/19
(c) 1/221
(d) 1/169
Show Answer & Explanation

Answer: (a) 1/17

Explanation:
1. After the first card, 3 of the remaining 51 cards have the same face value.
2. P = 3/51 = 1/17.

Teacher's Note:
1. The first card is free. Only the second card must match its value.

Question 12: Two cards are drawn at random from a pack of 52 cards. The probability of getting at least a spade and an ace is
(a) 2/51
(b) 8/221
(c) 1/26
(d) 1/34
Show Answer & Explanation

Answer: (c) 1/26

Explanation:
1. Total pairs = C(52,2) = 1326.
2. Pairs with the ace of spades: the other card is any spade (12) or any other ace (3): 15.
3. Pairs without the ace of spades: a non-ace spade (12) and a non-spade ace (3): 36.
4. P = 51/1326 = 1/26.

Teacher's Note:
1. Treat the ace of spades separately because it is both a spade and an ace.

Question 13: The probability of getting different suit cards and different denomination cards when two cards are drawn from a pack is
(a) 12/17
(b) 6/17
(c) 13/17
(d) 13/34
Show Answer & Explanation

Answer: (a) 12/17

Explanation:
1. After the first card, the second must differ in suit (39 cards) and in denomination: 39 - 3 = 36 cards.
2. P = 36/51 = 12/17.

Teacher's Note:
1. Remove the cards of the same suit and the cards of the same value from the 51 left.

Question 14: There are 3 cards. Each is painted red on one side and black on the other side. The cards are randomly placed in a row. The probability that no two consecutive cards show red is
(a) 3/8
(b) 5/6
(c) 5/8
(d) 1/8
Show Answer & Explanation

Answer: (c) 5/8

Explanation:
1. Each card shows red or black with equal probability: 8 equally likely patterns.
2. Patterns with no two adjacent reds: BBB, RBB, BRB, BBR, RBR = 5.
3. P = 5/8.

Teacher's Note:
1. List the colour patterns of length 3 with no RR.

Question 15: A bag contains 10 white and 3 black balls. Balls are drawn one by one without replacement till all the black balls are drawn. The probability that the test will come to an end at the 7th draw is
(a) C(10,4) x C(3,2)/C(13,6) x 1/7
(b) C(10,4) x C(3,2)/C(13,7)
(c) C(10,5) x C(3,2)/C(13,7)
(d) C(10,4) x C(3,2)/C(13,6) x 2/7
Show Answer & Explanation

Answer: (a) C(10,4) x C(3,2)/C(13,6) x 1/7

Explanation:
1. The last black ball must appear on the 7th draw, so the first 6 draws contain exactly 2 black and 4 white balls.
2. P(first 6 draws have 2 black, 4 white) = C(10,4) C(3,2)/C(13,6).
3. The 7th ball must be the remaining black one among 7 balls: probability 1/7.
4. P = [C(10,4) C(3,2)/C(13,6)] x (1/7).

Teacher's Note:
1. Split into the composition of the first six draws and then the last ball.

Question 16: Four tickets numbered 00, 01, 10, 11 are placed in a bag. A ticket is drawn at random and replaced. Again a ticket is drawn at random. The probability that the sum of the numbers on the tickets drawn is 21 is
(a) 1/16
(b) 1/4
(c) 1/8
(d) 1/12
Show Answer & Explanation

Answer: (c) 1/8

Explanation:
1. The numbers are 0, 1, 10, 11. A sum of 21 needs 10 and 11 in either order.
2. P = 2 x (1/4)(1/4) = 1/8.

Teacher's Note:
1. Two orders of drawing are possible.

Question 17: Two squares of a chess board having \( 8\times8 \) squares are selected at random. The probability that they have a side in common is
(a) 112/C(64,2)
(b) 268/C(64,2)
(c) 168/C(64,2)
(d) 56/C(64,2)
Show Answer & Explanation

Answer: (a) 112/C(64,2)

Explanation:
1. Total pairs = C(64,2).
2. Horizontal neighbours: 8 rows x 7 = 56; vertical neighbours: 56.
3. Favourable = 112.

Teacher's Note:
1. Count neighbouring pairs in rows and in columns.

Question 18: If 16 squares of unit size are selected at random on a chess board, the probability that they form a square of \( 4\times4 \) is
(a) 36/C(64,4)
(b) 36/C(64,16)
(c) 25/C(64,16)
(d) 25/C(64,4)
Show Answer & Explanation

Answer: (c) 25/C(64,16)

Explanation:
1. Total selections = C(64,16).
2. A 4 x 4 block can be placed with its corner in 5 x 5 = 25 positions.
3. P = 25/C(64,16).

Teacher's Note:
1. Count the positions of the 4 x 4 block on the 8 x 8 board.

Question 19: Three numbers are chosen at random from {1, 2, 3, ..., 10}. The probability that the minimum of the chosen numbers is 3 or the maximum is 7 is
(a) 1/8
(b) 11/30
(c) 11/40
(d) 7/12
Show Answer & Explanation

Answer: (c) 11/40

Explanation:
1. Total = C(10,3) = 120.
2. Minimum 3: C(7,2) = 21. Maximum 7: C(6,2) = 15. Both: the middle number is 4, 5 or 6: 3.
3. P = (21 + 15 - 3)/120 = 33/120 = 11/40.

Teacher's Note:
1. Use inclusion and exclusion for the two events.

Question 20: Out of 15 persons, 10 can speak Hindi and 8 can speak English. If two persons are chosen at random, then the probability that one person speaks Hindi only and the other speaks both Hindi and English is
(a) 1/5
(b) 7/12
(c) 2/5
(d) 3/5
Show Answer & Explanation

Answer: (a) 1/5

Explanation:
1. Persons who speak both = 10 + 8 - 15 = 3; Hindi only = 7.
2. Favourable pairs = 7 x 3 = 21; total = C(15,2) = 105.
3. P = 21/105 = 1/5.

Teacher's Note:
1. First find how many speak both languages by inclusion and exclusion.

Question 21: An electric bulb will last for 150 days or more with a probability 0.7 and it will last for at most 160 days with probability 0.8. The probability that the bulb will last between 150 and 160 days is
(a) 0.1
(b) 0.4
(c) 0.5
(d) 0.3
Show Answer & Explanation

Answer: (c) 0.5

Explanation:
1. P(X at least 150) = 0.7 and P(X at most 160) = 0.8.
2. These events together cover all outcomes, so P(150 to 160) = 0.7 + 0.8 - 1 = 0.5.

Teacher's Note:
1. Use P(A and B) = P(A) + P(B) - P(A or B) with the union equal to 1.

Question 22: Two dice are thrown. The probability of getting a sum of 7 points, if it is known that the two dice are showing different numbers, is
(a) 1/5
(b) 1/4
(c) 1/6
(d) 1/8
Show Answer & Explanation

Answer: (a) 1/5

Explanation:
1. Outcomes with different numbers: 36 - 6 = 30.
2. Sum 7 always has different numbers: 6 outcomes.
3. P = 6/30 = 1/5.

Teacher's Note:
1. Restrict the sample space to the given condition.

Question 23: Two symmetric dice are rolled. If the numbers thrown up on them are different, the probability of getting an even number as the sum of the numbers is
(a) 4/5
(b) 3/5
(c) 2/5
(d) 1/5
Show Answer & Explanation

Answer: (c) 2/5

Explanation:
1. Different numbers: 30 outcomes.
2. Even sums with different numbers: sum 4: 2; sum 6: 4; sum 8: 4; sum 10: 2. Total 12.
3. P = 12/30 = 2/5.

Teacher's Note:
1. Count even sums, excluding pairs with equal numbers (such as 1,1 or 3,3).

Question 24: Three dice are rolled. If no two dice show the same face, the probability that one is an ace is
(a) 1/8
(b) 1/3
(c) 1/2
(d) 1/4
Show Answer & Explanation

Answer: (c) 1/2

Explanation:
1. Outcomes with all faces different: 6 x 5 x 4 = 120.
2. With one die showing 1: choose its position (3) and the other two from 5 values in order (5 x 4 = 20): 60.
3. P = 60/120 = 1/2.

Teacher's Note:
1. Count the ordered triples of distinct faces with and without a 1.

Question 25: In selecting 2 cards one at a time with replacement from a deck, the probability that the second card is an ace, given that the first card was a face card, is
(a) 1/13
(b) 3/13
(c) 2/13
(d) 4/13
Show Answer & Explanation

Answer: (a) 1/13

Explanation:
1. Draws are independent because of replacement.
2. P(second is ace) = 4/52 = 1/13.

Teacher's Note:
1. With replacement the first card does not affect the second.

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