JEE Mathematics Probability MCQs Set 12

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Multiple Choice Questions

Question 1: A positive integer is selected at random. If A is the event that it is divisible by 5 and B is the event that it has zero at the units place, then \( A\cup\bar B \) is
(a) The event that the number has a non-zero digit at the units place
(b) A certain event
(c) \( \bar A\cap B \)
(d) An impossible event
Show Answer & Explanation

Answer: (b) A certain event

Explanation:
1. A means the units digit is 0 or 5, and the complement of B means the units digit is not 0.
2. Every integer either has units digit 0 or not, and A covers 0 and 5, so every integer is in A or in the complement of B.
3. Hence the union is the whole sample space, a certain event.

Teacher's Note:
1. Check whether the two events together cover all possible units digits. If they do, the union is the sure event.

Question 2: There are 7 seats in a row. Three persons take seats at random. The probability that the middle seat is always occupied and no two persons are consecutive is
(a) 4/35
(b) 6/35
(c) 9/35
(d) 9/70
Show Answer & Explanation

Answer: (a) 4/35

Explanation:
1. Total arrangements = 7P3 = 210.
2. The middle seat 4 is occupied. The other two seats cannot be 3 or 5, and the two cannot be adjacent: choose from seats 1, 2, 6, 7 with no two adjacent: (1,6), (1,7), (2,6), (2,7) = 4 pairs.
3. Favourable = 4 x 3! = 24, so P = 24/210 = 4/35.

Teacher's Note:
1. Choose the seat sets first and then multiply by the 3! ways to place the persons.

Question 3: The probability that the birthdays of 6 girls will fall on 6 different calendar months of a year is
(a) C(12,6)/12^6
(b) 12P6/12^6
(c) 1/12^11
(d) 1/12^5
Show Answer & Explanation

Answer: (b) 12P6/12^6

Explanation:
1. Each girl's birthday can fall in any of 12 months: 12^6 equally likely outcomes.
2. For different months, the girls are assigned distinct months in order: 12P6 ways.
3. P = 12P6/12^6.

Teacher's Note:
1. Distinct assignments with order mean permutations, not combinations.

Question 4: Three integers are chosen at random without replacement from the first 20 integers. The probability that their product is odd is
(a) 1/19
(b) 3/19
(c) 4/19
(d) 2/19
Show Answer & Explanation

Answer: (d) 2/19

Explanation:
1. The product is odd only if all three are odd.
2. There are 10 odd integers among the first 20.
3. P = C(10,3)/C(20,3) = 120/1140 = 2/19.

Teacher's Note:
1. A product is odd only when every factor is odd.

Question 5: There are m persons sitting in a row. Two of them are selected at random. The probability that the two selected persons are together is
(a) (m-3)/C(m,2)
(b) (m-1)/C(m,2)
(c) C(m-1,2)/C(m,2)
(d) m/C(m,2)
Show Answer & Explanation

Answer: (b) (m-1)/C(m,2)

Explanation:
1. Total ways = C(m,2).
2. Adjacent pairs in a row of m persons: m - 1.
3. P = (m-1)/C(m,2) = 2/m.

Teacher's Note:
1. In a row there are m - 1 adjacent pairs.

Question 6: Consider a lottery that sells \( n^2 \) tickets and awards n prizes. If one buys n tickets, the probability of his winning, i.e. getting at least one prize, is
(a) 1 - C(n^2-n,n)/C(n^2,n)
(b) n/C(n^2,n)
(c) ((n^2-n)/n^2)^n
(d) 1/C(n^2,n)
Show Answer & Explanation

Answer: (a) 1 - C(n^2-n,n)/C(n^2,n)

Explanation:
1. Total ways to choose n tickets = C(n^2, n).
2. He wins nothing if all his n tickets are among the n^2 - n non-prize tickets: C(n^2-n, n).
3. P(at least one prize) = 1 - C(n^2-n,n)/C(n^2,n).

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

Question 7: Dialling a telephone number to his daughter, an old man forgets the last two digits and dialled at random remembering only that they are different. The probability that the number dialled is correct is
(a) 1/10
(b) 1/45
(c) 1/90
(d) 1/135
Show Answer & Explanation

Answer: (c) 1/90

Explanation:
1. The last two digits are different, so there are 10 x 9 = 90 ordered possibilities.
2. Only one is correct.
3. P = 1/90.

Teacher's Note:
1. Order matters in a phone number, so count ordered pairs.

Question 8: 12 balls are distributed among 3 boxes. The probability that the first box contains 3 balls is
(a) (110/9) x (2/3)^10
(b) (2/3)^10
(c) 100/9
(d) 100/(9 x 3^10)
Show Answer & Explanation

Answer: (a) (110/9) x (2/3)^10

Explanation:
1. Each ball can go to any of 3 boxes: 3^12 outcomes.
2. Choose 3 balls for box 1 and put the other 9 in the 2 remaining boxes: C(12,3) x 2^9.
3. P = 220 x 2^9/3^12 = (110/9)(2/3)^10.

Teacher's Note:
1. Treat each ball independently and use the binomial form C(12,3)(1/3)^3(2/3)^9.

Question 9: Three groups of children contain respectively 3 girls and 1 boy; 2 girls and 2 boys; 1 girl and 3 boys. One child is selected at random from each group. The chance that the three selected children consist of one girl and 2 boys is
(a) 13/32
(b) 9/32
(c) 7/32
(d) 11/32
Show Answer & Explanation

Answer: (a) 13/32

Explanation:
1. Probabilities of a girl in the groups: 3/4, 1/2, 1/4.
2. GBB = (3/4)(1/2)(3/4) = 9/32; BGB = (1/4)(1/2)(3/4) = 3/32; BBG = (1/4)(1/2)(1/4) = 1/32.
3. Total = 13/32.

Teacher's Note:
1. Add the three arrangements in which exactly one of the three children is a girl.

Question 10: A letter is taken out at random from the word ASSISTANT and another from STATISTICS. The probability that they are the same letters is
(a) 19/90
(b) 15/90
(c) 17/90
(d) 13/90
Show Answer & Explanation

Answer: (a) 19/90

Explanation:
1. ASSISTANT has 9 letters: A 2, S 3, I 1, T 2, N 1. STATISTICS has 10 letters: S 3, T 3, A 1, I 2, C 1.
2. P = (2/9)(1/10) + (3/9)(3/10) + (2/9)(3/10) + (1/9)(2/10) for A, S, T, I.
3. = (2 + 9 + 6 + 2)/90 = 19/90.

Teacher's Note:
1. Add the probability of the same letter over every common letter, multiplying frequencies in the two words.

Question 11: If a number x is selected from the first 100 natural numbers at random, then the probability that \( x+\frac{100}x>50 \) is
(a) 5/20
(b) 11/20
(c) 1/2
(d) 9/20
Show Answer & Explanation

Answer: (b) 11/20

Explanation:
1. x + 100/x > 50 gives x^2 - 50x + 100 > 0, with roots 25 plus or minus sqrt(525), about 2.09 and 47.9.
2. So x = 1, 2 or x at least 48 (x = 48, ..., 100: 53 values).
3. Favourable = 2 + 53 = 55, so P = 55/100 = 11/20.

Teacher's Note:
1. Multiply through by x (positive) and solve the quadratic inequality. Then count integers.

Question 12: Three newly wedded couples are dancing at a function. If the partners are selected at random, the chance that not all the husbands are dancing with their own wives is
(a) 5/6
(b) 1/3
(c) 2/3
(d) 1/6
Show Answer & Explanation

Answer: (a) 5/6

Explanation:
1. All husbands with their own wives is exactly one of the 3! = 6 equally likely pairings.
2. P(all with own wives) = 1/6.
3. P(not all with own wives) = 1 - 1/6 = 5/6.

Teacher's Note:
1. Use the complement of the single perfect-match arrangement.

Question 13: 2n boys are randomly divided into two subgroups containing n boys each. The probability that the two tallest boys are in different groups is
(a) n/(2n-1)
(b) 2n/(2n-1)
(c) (n-1)/(2n-1)
(d) 1/2
Show Answer & Explanation

Answer: (a) n/(2n-1)

Explanation:
1. Place the tallest boy in a group.
2. The second tallest has 2n - 1 places, of which n are in the other group.
3. P = n/(2n-1).

Teacher's Note:
1. Fix one boy and count the places left for the other.

Question 14: A bag contains apples and oranges, five in all, and at least one of each. All combinations are equally likely. If one fruit is selected at random from the bag, assuming all fruits are distinguishable, the probability that it is an orange is
(a) 1/20
(b) 1/5
(c) 1/10
(d) 1/2
Show Answer & Explanation

Answer: (d) 1/2

Explanation:
1. Possible contents (apples, oranges): (1,4), (2,3), (3,2), (4,1), each with probability 1/4.
2. P(orange) = (1/4)(4/5 + 3/5 + 2/5 + 1/5) = (1/4)(2) = 1/2.

Teacher's Note:
1. Use total probability over the possible compositions of the bag.

Question 15: A box contains 100 tickets numbered 1, 2, ..., 100. Two tickets are chosen at random. It is given that the maximum number on the two chosen tickets is not more than 10. The probability that the minimum number on them is 5 is
(a) 1/9
(b) 3/9
(c) 2/9
(d) 4/9
Show Answer & Explanation

Answer: (a) 1/9

Explanation:
1. Given the maximum is at most 10, both tickets are from 1 to 10: C(10,2) = 45 pairs.
2. Minimum 5: the other number is 6, 7, 8, 9 or 10: 5 pairs.
3. P = 5/45 = 1/9.

Teacher's Note:
1. Reduce the sample space to the given condition before counting.

Question 16: Five horses are in a race. Mr. A selects two of the horses at random and bets on them. The probability that Mr. A selected the winning horse is
(a) 3/5
(b) 2/5
(c) 4/5
(d) 1/5
Show Answer & Explanation

Answer: (b) 2/5

Explanation:
1. P(winner not chosen) = C(4,2)/C(5,2) = 6/10 = 3/5.
2. P(winner chosen) = 1 - 3/5 = 2/5.

Teacher's Note:
1. He wins when the winner is one of his two choices, so the answer is 2/5 directly.

Question 17: Two integers x and y are chosen with replacement out of the set {0, 1, 2, ..., 10}. Then the probability that \( |x-y|>5 \) is
(a) 30/121
(b) 81/121
(c) 25/121
(d) 20/121
Show Answer & Explanation

Answer: (a) 30/121

Explanation:
1. Total ordered pairs = 11 x 11 = 121.
2. Pairs with |x - y| = d: 2(11 - d) for d = 6, 7, 8, 9, 10: 10 + 8 + 6 + 4 + 2 = 30.
3. P = 30/121.

Teacher's Note:
1. Count the ordered pairs for each possible difference and add.

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 positive is
(a) 5/8
(b) 3/8
(c) 3/16
(d) 7/8
Show Answer & Explanation

Answer: (c) 3/16

Explanation:
1. There are 2^4 = 16 determinants.
2. ad - bc = 1 requires ad = 1 and bc = 0: a = d = 1 and (b, c) not both 1: 3 cases.
3. P = 3/16.

Teacher's Note:
1. With entries 0 or 1, the determinant takes only the values -1, 0, 1.

Question 19: A has 3 tickets of a lottery containing 3 prizes and 9 blanks. B has 2 tickets of another lottery containing 2 prizes and 6 blanks. The ratio of their chances of success is
(a) 34/55 : 15/28
(b) 34/55 : 13/28
(c) 32/55 : 13/28
(d) 32/55 : 15/28
Show Answer & Explanation

Answer: (b) 34/55 : 13/28

Explanation:
1. P(A succeeds) = 1 - C(9,3)/C(12,3) = 1 - 84/220 = 34/55.
2. P(B succeeds) = 1 - C(6,2)/C(8,2) = 1 - 15/28 = 13/28.
3. Ratio = 34/55 : 13/28.

Teacher's Note:
1. Success means at least one prize, so use the complement 'all blanks'.

Question 20: S = {1, 2, 3, ..., 11}. If 3 numbers are chosen at random from S, the probability that they are in G.P. is
(a) 9/C(11,3)
(b) 7/C(11,3)
(c) 5/C(11,3)
(d) 4/C(11,3)
Show Answer & Explanation

Answer: (d) 4/C(11,3)

Explanation:
1. Three-term geometric progressions (as sets) within 1 to 11: {1,2,4}, {2,4,8}, {1,3,9}, {4,6,9}.
2. There are no others (for example {1,4,16}, {3,6,12}, {2,6,18} exceed 11).
3. P = 4/C(11,3) = 4/165.

Teacher's Note:
1. List ratios 2, 3, 3/2 and 4 systematically. Only four sets fit.

Question 21: If 10 identical apples are distributed among 6 persons at random, then the probability that at least one of them will receive none is
(a) 1/143
(b) C(14,4)/C(15,5)
(c) 137/143
(d) 6/143
Show Answer & Explanation

Answer: (c) 137/143

Explanation:
1. Total distributions = C(15,5) = 3003.
2. Everyone receives at least one: C(9,5) = 126.
3. P(at least one gets none) = 1 - 126/3003 = 1 - 6/143 = 137/143.

Teacher's Note:
1. Count with stars and bars, then use the complement.

Question 22: In a game called 'odd man out', 4 persons toss a coin to decide who will buy refreshments for the entire group. A person who gets an outcome different from that of the rest of the members of the group is called the odd man out. The probability that there is a loser in any game is
(a) 1/5
(b) 1/2
(c) 1/3
(d) 1/4
Show Answer & Explanation

Answer: (b) 1/2

Explanation:
1. Total outcomes = 2^4 = 16.
2. A loser exists when exactly one person differs: 4 + 4 = 8 outcomes (one head or one tail).
3. P = 8/16 = 1/2.

Teacher's Note:
1. The odd man exists only for the splits 1 : 3 and 3 : 1.

Question 23: Two numbers a and b are chosen at random from the numbers 1, 2, ..., 30. The chance that \( a^2-b^2 \) is divisible by 3 is
(a) 15/87
(b) 47/87
(c) 9/87
(d) 12/87
Show Answer & Explanation

Answer: (b) 47/87

Explanation:
1. Total pairs = C(30,2) = 435.
2. a^2 - b^2 is divisible by 3 when a, b are both multiples of 3 or both non-multiples (squares of non-multiples are 1 mod 3).
3. C(10,2) + C(20,2) = 45 + 190 = 235.
4. P = 235/435 = 47/87.

Teacher's Note:
1. The square of a number is 0 or 1 modulo 3. The squares must leave the same remainder.

Question 24: Two numbers are selected at random from 1, 2, 3, ..., 100 and are multiplied. The probability that the product thus obtained is divisible by 3, correct to two places of decimal, is
(a) 0.45
(b) 0.55
(c) 0.35
(d) 0.25
Show Answer & Explanation

Answer: (b) 0.55

Explanation:
1. The product is not divisible by 3 only if neither number is a multiple of 3: 67 non-multiples among 100.
2. P(not divisible) = C(67,2)/C(100,2) = 2211/4950.
3. P(divisible) = 1 - 2211/4950 = 0.5533, about 0.55.

Teacher's Note:
1. Use the complement with the non-multiples of 3.

Question 25: Four-digit numbers with different digits are formed using the digits 1, 2, 3, 4, 5, 6, 7, 8. One number from them is picked up at random. The chance that the selected number contains the digit 1 is
(a) 1/2
(b) 1/4
(c) 1/8
(d) 1/16
Show Answer & Explanation

Answer: (a) 1/2

Explanation:
1. Total numbers = 8P4 = 1680.
2. Numbers without the digit 1 = 7P4 = 840.
3. P(contains 1) = 1 - 840/1680 = 1/2.

Teacher's Note:
1. Each digit position is equally likely to hold the digit 1, giving 4/8 = 1/2.

Question 26: Three numbers are chosen at random from the set {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}. The probability that the smallest of the three numbers chosen is even is
(a) 1/6
(b) 5/12
(c) 5/6
(d) 7/12
Show Answer & Explanation

Answer: (b) 5/12

Explanation:
1. Total = C(10,3) = 120.
2. Smallest 2: C(8,2) = 28; smallest 4: C(6,2) = 15; smallest 6: C(4,2) = 6; smallest 8: C(2,2) = 1.
3. Favourable = 50, so P = 50/120 = 5/12.

Teacher's Note:
1. Fix the smallest number and choose the other two from the larger numbers.

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