JEE Mathematics Probability MCQs Set 15

Multiple Choice Questions (MCQs) for JEE Mathematics: Probability

Review structured MCQ sets for JEE Mathematics Probability. Built according to official JEE guidelines, these downloadable questions support daily revision and core concept reinforcement.

Practice Probability MCQs for JEE Mathematics

View or download the dedicated Probability MCQ resource below. Practicing these 50 objective questions regularly builds familiarity with standard exam patterns and helps secure higher marks in final Mathematics evaluations.

Multiple Choice Questions

Question 1: For k = 1, 2, 3 the box \( B_k \) contains k red balls and (k + 1) white balls. Let \( P(B_1)=\frac12 \), \( P(B_2)=\frac13 \), \( P(B_3)=\frac16 \). A box is selected at random and a ball is drawn from it. If a red ball is drawn, the probability that it has come from box \( B_2 \) is
(a) 14/39
(b) 35/78
(c) 12/13
(d) 10/13
Show Answer & Explanation

Answer: (a) 14/39

Explanation:
1. P(red
2. B1) = 1/3, P(red
3. B2) = 2/5, P(red
4. B3) = 3/7.
5. P(red) = (1/2)(1/3) + (1/3)(2/5) + (1/6)(3/7) = 1/6 + 2/15 + 1/14 = 78/210.
6. P(B2
7. red) = (2/15)/(78/210) = 28/78 = 14/39.

Teacher's Note:
1. Apply Bayes' theorem with the three boxes as the hypotheses.

Question 2: A letter is known to have come from TATANAGAR or CALCUTTA. On the envelope just two consecutive letters TA are visible. The probability that the letter has come from CALCUTTA is
(a) 7/11
(b) 4/11
(c) 21/22
(d) 1/22
Show Answer & Explanation

Answer: (b) 4/11

Explanation:
1. TATANAGAR has 9 letters and 8 pairs of consecutive letters; TA occurs twice: 2/8.
2. CALCUTTA has 8 letters and 7 pairs; TA occurs once: 1/7.
3. P(CALCUTTA
4. TA) = (1/7)/(2/8 + 1/7) = (1/7)/(11/28) = 4/11.

Teacher's Note:
1. Each town is equally likely; weight by the chance of seeing TA at a random position.

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

Answer: (a) 5/7

Explanation:
1. P(reports Five) = (1/6)(2/3) + (5/6)(1/3) = 2/18 + 5/18 = 7/18.
2. P(not Five and reports Five) = 5/18.
3. P = (5/18)/(7/18) = 5/7.

Teacher's Note:
1. Use Bayes' theorem: the false reports come from the five numbers other than 5.

Question 4: What is the probability that the birthdays of six people will fall in exactly two calendar months?
(a) C(12,2)(2^6 - 2)/12^6
(b) C(12,2) 2^6/12^6
(c) C(12,2)(2^6 - 4)/12^6
(d) C(12,2)(2^6 + 2)/12^6
Show Answer & Explanation

Answer: (a) C(12,2)(2^6 - 2)/12^6

Explanation:
1. Total outcomes = 12^6.
2. Choose the two months: C(12,2). Each of 6 people has 2 choices but both months must be used: 2^6 - 2.
3. P = C(12,2)(2^6 - 2)/12^6.

Teacher's Note:
1. Subtract the cases where all six birthdays fall in a single month.

Question 5: The odds that a book will be reviewed favourably by three independent critics are 5 to 2, 4 to 3 and 3 to 4 respectively. The probability that a majority of the reviews will be favourable is
(a) 120/343
(b) 60/343
(c) 209/343
(d) 135/343
Show Answer & Explanation

Answer: (c) 209/343

Explanation:
1. Probabilities of a favourable review: 5/7, 4/7, 3/7.
2. Majority means at least two favourable: (5)(4)(4) + (5)(3)(3) + (2)(4)(3) + (5)(4)(3) all over 7^3 = (80 + 45 + 24 + 60)/343.
3. P = 209/343.

Teacher's Note:
1. Add the three 'exactly two favourable' cases and the 'all three' case.

Question 6: Set S has 4 elements, A and B are subsets of S. The probability that A and B are not disjoint is
(a) 173/256
(b) 45/64
(c) 85/128
(d) 175/256
Show Answer & Explanation

Answer: (d) 175/256

Explanation:
1. Each element is in both A and B with probability 1/4, so it is not in both with probability 3/4.
2. P(disjoint) = (3/4)^4 = 81/256.
3. P(not disjoint) = 1 - 81/256 = 175/256.

Teacher's Note:
1. Treat each element independently and use the complement.

Question 7: The sum of two natural numbers is 20. The chance that their product is less than 50 is
(a) 3/19
(b) 4/19
(c) 1/19
(d) 2/19
Show Answer & Explanation

Answer: (b) 4/19

Explanation:
1. The pairs (a, 20 - a) with a = 1, ..., 19 give 19 equally likely cases.
2. Product a(20 - a) is 19, 36, 51 for a = 1, 2, 3, and symmetric on the other side.
3. So the product is below 50 for a = 1, 2, 18, 19: 4 cases.
4. P = 4/19.

Teacher's Note:
1. List the products for small a, then use the symmetry a and 20 - a.

Question 8: Three newly wedded couples are dancing at a function. If the partners are selected at random, the chance that at least one husband is not dancing with his own wife is
(a) 5/6
(b) 1/6
(c) 2/3
(d) 1/3
Show Answer & Explanation

Answer: (a) 5/6

Explanation:
1. The only way no husband is away from his own wife is the single perfect pairing: probability 1/3! = 1/6.
2. P(at least one husband not with own wife) = 1 - 1/6 = 5/6.

Teacher's Note:
1. Use the complement of the perfect matching.

Question 9: Team A plays with 5 other teams exactly once. Assuming that for each match the probabilities of a win, draw and loss are equal, the probability that A wins and loses an equal number of matches is
(a) 34/81
(b) 17/81
(c) 32/81
(d) 16/81
Show Answer & Explanation

Answer: (b) 17/81

Explanation:
1. Each match has 3 equally likely results: 3^5 = 243 outcomes.
2. k wins and k losses: k = 0: 1; k = 1: 5 x 4 = 20 (with 3 draws); k = 2: 5!/(2! 2! 1!) = 30.
3. Total = 51, so P = 51/243 = 17/81.

Teacher's Note:
1. Count the arrangements of wins, losses and draws for k = 0, 1, 2.

Question 10: Thirty-two players ranked 1 to 32 are playing a knockout tournament. Assume that in every match between any two players the better ranked player wins. The probability that ranked 1 and ranked 2 players are winner and runner-up respectively is
(a) 18/31
(b) 1/2
(c) 16/31
(d) 17/31
Show Answer & Explanation

Answer: (c) 16/31

Explanation:
1. Players 1 and 2 must not meet before the final, so they must be in different halves.
2. Given player 1's half (16 places), player 2 is placed in one of the 31 other places, 16 of which are in the other half.
3. P = 16/31.

Teacher's Note:
1. Place the two top players in the draw: the second must land in the other half.

Question 11: In a department there are two Professors, four Readers and 6 Lecturers. A committee of three persons is to be formed out of the staff of the department. The probability that the committee consists of at least two lecturers is
(a) 1/4
(b) 1/2
(c) 1/3
(d) 1/8
Show Answer & Explanation

Answer: (b) 1/2

Explanation:
1. Total = C(12,3) = 220.
2. Exactly two lecturers: C(6,2) x 6 = 90; three lecturers: C(6,3) = 20.
3. P = 110/220 = 1/2.

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

Question 12: If 5 biscuits are distributed among 3 beggars (each biscuit being given to a beggar at random), the chance that a particular beggar will get 2 biscuits is
(a) 30/125
(b) 2/15
(c) 3/10
(d) 80/243
Show Answer & Explanation

Answer: (d) 80/243

Explanation:
1. Each biscuit goes to any of 3 beggars: 3^5 = 243 outcomes.
2. The particular beggar gets exactly 2 biscuits: C(5,2) x 2^3 = 80.
3. P = 80/243.

Teacher's Note:
1. Choose which two biscuits go to him and distribute the rest between the others.

Question 13: From a box containing 10 cards numbered 1 to 10, four cards are drawn together. The probability that their sum is odd is
(a) 10/21
(b) 1/21
(c) 11/21
(d) 1/2
Show Answer & Explanation

Answer: (a) 10/21

Explanation:
1. The sum is odd when the number of odd cards is odd: 1 or 3.
2. Odd cards: 5; even cards: 5.
3. Favourable = C(5,1)C(5,3) + C(5,3)C(5,1) = 50 + 50 = 100.
4. P = 100/C(10,4) = 100/210 = 10/21.

Teacher's Note:
1. Parity of the sum depends only on the count of odd numbers.

Question 14: Three persons A, B and C are to speak at a function along with 5 other persons. If the persons speak in random order, the probability that C speaks before A and A speaks before B is
(a) 3/8!
(b) 1/8!
(c) 1/3!
(d) C(8,3)/8!
Show Answer & Explanation

Answer: (c) 1/3!

Explanation:
1. Among the three persons A, B, C there are 3! equally likely relative orders.
2. Exactly one of them is the order C, A, B.
3. P = 1/3! = 1/6.

Teacher's Note:
1. The other five speakers do not affect the relative order of A, B and C.

Question 15: Out of the first 25 natural numbers, 2 are chosen at random. The probability for one of the numbers to be a multiple of 3 and the other to be a multiple of 5 is
(a) 13/100
(b) 1/15
(c) 1/5
(d) 2/15
Show Answer & Explanation

Answer: (a) 13/100

Explanation:
1. Multiples of 3: 8 (3, ..., 24), of 5: 5 (5, ..., 25), of both: 1 (15). Multiples of 3 only: 7; of 5 only: 4.
2. Pairs with a 3-only and a 5-only number: 7 x 4 = 28.
3. Pairs containing 15 and another multiple of 3 or 5: 7 + 4 = 11.
4. Total = 39, P = 39/300 = 13/100.

Teacher's Note:
1. Treat 15 separately because it is a multiple of both 3 and 5.

Question 16: If the letters of the word FLOWER are arranged at random, the probability that the order of the consonants is not changed is
(a) 50/720
(b) 30/720
(c) 20/720
(d) 40/270
Show Answer & Explanation

Answer: (b) 30/720

Explanation:
1. There are 6! = 720 arrangements.
2. The 4 consonants F, L, W, R have 4! equally likely relative orders; only one keeps the original order.
3. Favourable = 720/24 = 30.
4. P = 30/720 = 1/24.

Teacher's Note:
1. The relative order of any four distinct letters is equally likely among 4! orders.

Question 17: 7 balls are thrown into 4 bags numbered serially 1, 2, 3 and 4. Then the probability that none of them is found in bag number 2 is
(a) 1/4
(b) 1 - (1/4)^7
(c) (3/4)^7
(d) 3/4
Show Answer & Explanation

Answer: (c) (3/4)^7

Explanation:
1. Each ball avoids bag 2 with probability 3/4.
2. The balls are independent: (3/4)^7.

Teacher's Note:
1. The balls act independently, so multiply the individual probabilities.

Question 18: Eight persons numbered 1, 2, 3, ..., 8 are seated round a circular table at random. The probability that the person numbered 1 sits between 2 and 3 is
(a) 1/56
(b) 1/2
(c) 1/42
(d) 1/21
Show Answer & Explanation

Answer: (d) 1/21

Explanation:
1. Total circular arrangements = 7!.
2. Treat 2, 1, 3 as a block (2 orders, 2! ways): the block and 5 others arranged in a circle: 5! x 2.
3. P = 2 x 5!/7! = 240/5040 = 1/21.

Teacher's Note:
1. Glue the three persons into one block, remembering both orders.

Question 19: Twenty persons, among whom are A and B, sit at random around a round table. The probability that there are exactly 6 persons between A and B (on one side) is
(a) 2/19
(b) C(18,6) x 2! x 12!/19!
(c) C(18,6) x 2!/19!
(d) 17/19
Show Answer & Explanation

Answer: (a) 2/19

Explanation:
1. Fix A. B can sit in any of the other 19 places.
2. The places with exactly 6 persons between A and B counted around the table in one direction: 2 places (one on each side of A).
3. P = 2/19.

Teacher's Note:
1. After fixing A, count the seats at distance 7 on either side.

Question 20: If four letters are placed into 4 addressed envelopes at random, the probability that exactly two letters will go wrong is
(a) 1/2
(b) 0
(c) 1/4
(d) 1/3
Show Answer & Explanation

Answer: (c) 1/4

Explanation:
1. Exactly two wrong means the other two are correct and the two wrong ones are swapped.
2. Choose the 2 correct letters: C(4,2) = 6 ways.
3. P = 6/4! = 6/24 = 1/4.

Teacher's Note:
1. Exactly two wrong letters must have swapped envelopes.

Question 21: In the quadratic equation \( ax^2+bx+c=0 \), the coefficients a, b, c take distinct values from the set {1, 2, 3}. The probability that the roots of the equation are real is
(a) 1/4
(b) 2/5
(c) 2/3
(d) 1/3
Show Answer & Explanation

Answer: (d) 1/3

Explanation:
1. (a, b, c) is a permutation of 1, 2, 3: 6 outcomes.
2. Real roots need b^2 at least 4ac. If b = 3: 9 at least 8 (ac = 2) works for (1,3,2) and (2,3,1). If b = 1 or 2, b^2 is at most 4 while 4ac is at least 12.
3. P = 2/6 = 1/3.

Teacher's Note:
1. Test the three possible values of b.

Question 22: Let A = {1, 3, 5, 7, 9} and B = {2, 4, 6, 8}. An element (a, b) of their cartesian product \( A\times B \) is chosen at random. The probability that a + b = 9 is
(a) 1/3
(b) 1/5
(c) 2/3
(d) 2/5
Show Answer & Explanation

Answer: (b) 1/5

Explanation:
1. A x B has 5 x 4 = 20 elements.
2. Pairs with a + b = 9: (1,8), (3,6), (5,4), (7,2) = 4.
3. P = 4/20 = 1/5.

Teacher's Note:
1. List the pairs in the cartesian product that add to 9.

Question 23: A set P contains n elements. A function from P to P is picked up at random. The probability that this function is onto is
(a) (n - 1)!/n^n
(b) (n^n - n!)/n^n
(c) (n - 1)!/n!
(d) n!/n^n
Show Answer & Explanation

Answer: (d) n!/n^n

Explanation:
1. Total functions from P to P = n^n.
2. An onto function from a finite set to itself is a bijection: n! of them.
3. P = n!/n^n.

Teacher's Note:
1. For a finite set mapped to itself, onto and one-one are equivalent.

Question 24: From a heap containing 10 pairs of shoes, 8 shoes are selected. The probability that 4 correct pairs are in the selected shoes is
(a) C(10,2)/C(20,4)
(b) C(10,4)/C(20,4)
(c) C(10,4)/C(20,8)
(d) C(10,6) C(8,2)/C(20,8)
Show Answer & Explanation

Answer: (c) C(10,4)/C(20,8)

Explanation:
1. Total ways = C(20,8).
2. Four complete pairs and no other shoes: choose 4 of the 10 pairs: C(10,4).
3. P = C(10,4)/C(20,8).

Teacher's Note:
1. With 8 shoes forming 4 pairs, no shoe can be left unmatched.

Question 25: In the figure, rectangle ABCD has AB = 4, BC = 15 and E is a point on AD with BE = 5. The probability that a randomly chosen point in ABCD lies inside the triangle ABE is
BCADE1545 (a) 1/3
(b) 1/5
(c) 1/10
(d) 1/15
Show Answer & Explanation

Answer: (c) 1/10

Explanation:
1. In the right triangle ABE, AE = sqrt(5^2 - 4^2) = 3.
2. Area of triangle ABE = (1/2)(3)(4) = 6.
3. Area of rectangle ABCD = 4 x 15 = 60.
4. P = 6/60 = 1/10.

Teacher's Note:
1. For a point chosen at random in a region, the probability is the ratio of areas.

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Multiple Choice Questions (MCQs) for JEE Mathematics Probability

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