JEE Mathematics Probability MCQs Set 20

Answer: (b) \( \frac{29}{30} \)

Explanation:
1. The room is dark only if all three bulbs are defective: \( {}^4C_3/{}^{10}C_3 = 4/120 = 1/30 \).
2. P(lighted) = 1 - 1/30 = 29/30.

Teacher's Note:
1. The complement of 'at least one good' is 'all defective'. There are 10 - 6 = 4 defective bulbs.

Answer: (c) \( \frac{{}^7P_5}{7^5} \)

Explanation:
1. Each person has 7 choices, so there are 7^5 outcomes.
2. Different floors for all: \( {}^7P_5 \) ways.
3. P = \( {}^7P_5/7^5 \).

Teacher's Note:
1. The persons are distinguishable, so use arrangements, not combinations. The total uses repetition, giving 7^5.

Answer: (d) \( \frac{2469}{2470} \)

Explanation:
1. Total ways \( {}^{40}C_4 = 91390 \).
2. Four consecutive numbers can start at 1, 2, ..., 37: 37 ways.
3. P(consecutive) = 37/91390 = 1/2470.
4. P(not consecutive) = 2469/2470.

Teacher's Note:
1. Count blocks of consecutive numbers by their starting number. The last possible start is 40 - 4 + 1 = 37.

Answer: (b) \( \frac{9}{14} \)

Explanation:
1. The smaller number is at least 4 if both numbers are in {4,...,8}: \( {}^5C_2 = 10 \) pairs.
2. P(smaller less than 4) = 1 - 10/28 = 18/28 = 9/14.

Teacher's Note:
1. Use the complement: both numbers must come from the larger group to make the smaller one at least 4.

Answer: (d) \( 1-\frac{{}^{31}C_2}{{}^{100}C_2} \)

Explanation:
1. The minimum is at least 70 if both numbers come from {70, 71, ..., 100}, which has 31 numbers.
2. P(minimum at least 70) = \( {}^{31}C_2/{}^{100}C_2 \).
3. Required probability = 1 - 31C2/100C2.

Teacher's Note:
1. Count the numbers from 70 to 100 inclusive: 100 - 70 + 1 = 31. Then use the complement.

Answer: (a) \( \frac{1}{3} \)

Explanation:
1. Sum 7: 6 outcomes (1,6), (2,5), (3,4), (4,3), (5,2), (6,1).
2. Those containing a 2: (2,5), (5,2).
3. P = 2/6 = 1/3.

Teacher's Note:
1. In a conditional problem count within the given event only. Here the reduced sample space has 6 outcomes.

Answer: (c) \( \frac{1}{7} \)

Explanation:
1. Let P(C) = x. P(A) = P(B) = 3x.
2. One of them must win: 3x + 3x + x = 1, so x = 1/7.

Teacher's Note:
1. If P(A) = P(B) = 3x then both are separately 3x. Add the probabilities and equate to 1.

Answer: (a) \( \frac{29}{32} \)

Explanation:
1. Total functions = 4^4 = 256.
2. Onto functions from a finite set to itself are bijections: 4! = 24.
3. Into functions = 256 - 24 = 232.
4. P = 232/256 = 29/32.

Teacher's Note:
1. For a function from a finite set to itself, onto and one-one are equivalent. So the onto functions are the 4! permutations.

Answer: (d) \( \frac{3}{38} \)

Explanation:
1. The first and last numbers must have the same parity: \( {}^{10}C_2 + {}^{10}C_2 = 90 \).
2. Total \( {}^{20}C_3 = 1140 \).
3. P = 90/1140 = 3/38.

Teacher's Note:
1. Counting the common-difference progressions gives the same 90. The middle term is the mean of the other two.

Answer: (a) \( \frac{4}{9} \)

Explanation:
1. Real roots need discriminant 16 - 4c ≥ 0, so c ≤ 4.
2. c can be 1, 2, 3, 4: 4 values out of 9.
3. P = 4/9.

Teacher's Note:
1. Use the discriminant condition and count the integer values of c in the allowed range.

Answer: (d) \( \frac{35}{76} \)

Explanation:
1. Favourable = \( {}^{15}C_2\times{}^5C_1 = 105\times5=525 \).
2. Total = \( {}^{20}C_3 = 1140 \).
3. P = 525/1140 = 35/76.

Teacher's Note:
1. Select the boys and the girl separately and multiply. Then divide by the number of ways to choose 3 students from all 20.

Answer: (a) \( \frac{7}{8} \)

Explanation:
1. P(no head) = (1/2)³ = 1/8.
2. P(at least one head) = 7/8.

Teacher's Note:
1. Whenever 'at least one' appears, compute the complement. All tails is the only case with no head.

Answer: (a) \( \frac{1}{2} \)

Explanation:
1. A's heads: 0, 1, 2 with probabilities 1/4, 1/2, 1/4. B's heads 0, 1, 2, 3: 1/8, 3/8, 3/8, 1/8.
2. P(B > A) = (1/4)(7/8) + (1/2)(4/8) + (1/4)(1/8) = 7/32 + 8/32 + 1/32 = 1/2.

Teacher's Note:
1. By symmetry, 3 coins against 2 coins: B's extra coin makes more heads and fewer heads equally likely. So the answer is 1/2.

Answer: (b) \( \frac{57}{64} \)

Explanation:
1. P(0 head) = 1/64, P(1 head) = 6/64.
2. P(at least 2) = 1 - 7/64 = 57/64.

Teacher's Note:
1. Use the complement of 0 or 1 head. The binomial coefficients 1 and 6 give the cases.

Answer: (b) \( \frac{5}{16} \)

Explanation:
1. Heads exceed tails if heads = 3 or 4.
2. Number of outcomes = 4 + 1 = 5.
3. P = 5/16.

Teacher's Note:
1. With 4 tosses, heads exceed tails only for 3 or 4 heads. The case of 2 heads is a tie and not counted.

Answer: (c) 1/2

Explanation:
1. If h heads occur, the total = 2h + (3 - h) = 3 + h.
2. The total is odd if h is even: h = 0 or 2.
3. P = (1 + 3)/8 = 1/2.

Teacher's Note:
1. Write the score in terms of the number of heads. Only the parity of h matters.

Answer: (a) \( \frac{5}{36} \)

Explanation:
1. Sum 6: (1,5), (2,4), (3,3), (4,2), (5,1).
2. P = 5/36.

Teacher's Note:
1. The number of ways for sum k with two dice is k - 1 when k is at most 7.

Answer: (b) \( \frac{2}{36} \)

Explanation:
1. Minimum 2 occurs in (1,1); maximum 12 occurs in (6,6).
2. P = 2/36.

Teacher's Note:
1. The extreme sums occur in exactly one way each. Add the two cases.

Answer: (a) \( \frac{7}{12} \)

Explanation:
1. Sums 7 to 12 occur in 6 + 5 + 4 + 3 + 2 + 1 = 21 ways.
2. P = 21/36 = 7/12.

Teacher's Note:
1. By symmetry sums 2 to 6 have 15 ways. Subtract from 36 to check.

Answer: (d) \( \frac{1}{2} \)

Explanation:
1. The sum is even if both numbers have the same parity.
2. Ways: 3 x 3 + 3 x 3 = 18.
3. P = 18/36 = 1/2.

Teacher's Note:
1. Parity depends only on whether each die shows odd or even. Equal parity gives an even sum.

Answer: (a) \( \frac{1}{9} \)

Explanation:
1. Pairs with difference 4: (1,5), (5,1), (2,6), (6,2).
2. P = 4/36 = 1/9.

Teacher's Note:
1. List the pairs with a difference of 4 in both orders. Each ordered pair counts separately.

Answer: (a) \( \frac{5}{9} \)

Explanation:
1. First die: any of 6, second: 5 choices, third: 4 choices.
2. Favourable = 6 x 5 x 4 = 120.
3. P = 120/216 = 5/9.

Teacher's Note:
1. Count by sequential choices: each die must avoid the earlier values.

Answer: (a) \( \frac{1}{54} \)

Explanation:
1. B needs a sum of 17 or 18.
2. Sum 17 occurs in 3 ways, sum 18 in 1 way.
3. P = 4/216 = 1/54.

Teacher's Note:
1. Only A's score matters for the target, and the remaining possible sums are 17 and 18. A's own outcomes do not change B's probability.

Answer: (b) \( \frac{5}{108} \)

Explanation:
1. Triples with sum 6 using values from {1, -1, 2, -2, 3, 0}: (3,3,0) in 3 orders, (3,2,1) in 6 orders, (2,2,2) in 1 order.
2. Total ordered triples = 3 + 6 + 1 = 10.
3. P = 10/216 = 5/108.

Teacher's Note:
1. Enumerate unordered triples first and then count their orderings. Each ordered triple has probability 1/216.

Answer: (c) \( \frac1{6^3} \)

Explanation:
1. Choose which 2 of the 4 throws give 3 (the other 2 give 6): \( {}^4C_2 = 6 \).
2. P = 6/6^4 = 1/6³.

Teacher's Note:
1. Each throw gives a fixed face with probability 1/6, so each favourable sequence has probability 1/6^4. Count the sequences.