JEE Mathematics Probability MCQs Set 27

Answer: (d) None of these

Explanation:
1. P(Amar) = 1/5, P(Bimal) = 1/6 and P(Chetan) = 2/5.
2. Only one person can win, so the events are mutually exclusive: the sum is 1/5 + 1/6 + 2/5 = 23/30.
3. 23/30 is not among the first three options.

Teacher's Note:
1. Convert odds to probabilities and add.

Answer: (b) 4/35

Explanation:
1. P₄ reaches the final only if none of P₁, P₂, P₃ is in his half of four players.
2. His three half-mates are chosen from the other 7 players; they must all come from P₅–P₈.
3. Probability = C(4, 3)/C(7, 3) = 4/35.

Teacher's Note:
1. Count the ways to choose his three half-mates.

Answer: (d) \( \frac{1}{18} \)

Explanation:
1. P(win) = 1/3, P(draw) = 1/6, P(loss) = 1/2.
2. 5 points in 3 matches is possible only as 2 + 2 + 1: two wins and one draw, in 3 orders.
3. Probability = 3 × (1/3)² × (1/6) = 1/18.

Teacher's Note:
1. List the score combinations that give 5.

Answer: (a) 5

Explanation:
1. P(at least one hit) = 1 − 0.7ⁿ, so we need 1 − 0.7ⁿ ≥ 0.8, i.e. 0.7ⁿ ≤ 0.2.
2. 0.7⁴ = 0.2401 > 0.2 but 0.7⁵ = 0.168 < 0.2.
3. So the least n is 5.

Teacher's Note:
1. Use the complement: no hit in n shots.

Answer: (d) \( \frac{1}{120} \)

Explanation:
1. The number of possible orders of the first three places is 6 × 5 × 4 = 120.
2. Only one of them has A, B, C in that order, so the probability is 1/120.

Teacher's Note:
1. Count the ordered selections for the first three places.

Answer: (d) \( \frac{12}{13} \)

Explanation:
1. Choosing 11 is the same as leaving out 4. At least 3 bowlers in the team means at most 2 bowlers are left out.
2. Favourable ways: C(10, 4) + C(5, 1)C(10, 3) + C(5, 2)C(10, 2) = 210 + 600 + 450 = 1260.
3. Total C(15, 4) = 1365, so the probability is 1260/1365 = 12/13.

Teacher's Note:
1. Count the excluded players instead of the selected ones.

Answer: (c) \( \frac{1}{10} \)

Explanation:
1. The total number of triangles is C(6, 3) = 20.
2. Equilateral triangles join alternate vertices: exactly 2 of them.
3. Probability = 2/20 = 1/10.

Teacher's Note:
1. Equilateral triangles use every second vertex.

Answer: (c) \( \frac{10}{21} \)

Explanation:
1. Total ways = C(9, 4) = 126.
2. Two children and two others from the 5 non-children: C(4, 2) × C(5, 2) = 6 × 10 = 60.
3. Probability = 60/126 = 10/21.

Teacher's Note:
1. Choose the children and the non-children separately.

Answer: (d) \( A\cap(B'\cup C')=B'\cap C' \)

Explanation:
1. A, B, C are mutually exclusive: A = {TTT}, B has three outcomes, C has four.
2. B' ∪ C' is the whole sample space, so A ∩ (B' ∪ C') = A. Also B' ∩ C' = complement of (B ∪ C) = {TTT} = A. So (d) is true.
3. (a) gives A, not B ∪ C; (b) gives A, not the whole space; (c) gives A, not A ∪ B ∪ C.

Teacher's Note:
1. Write out the three events as sets of outcomes.

Answer: (b) 7

Explanation:
1. If the coin shows a head, the die gives 6 outcomes: H1 to H6.
2. If the coin shows a tail, there is just 1 outcome: T.
3. Total = 6 + 1 = 7.

Teacher's Note:
1. Count the head and tail branches separately.

Answer: (c) \( \frac{15}{92} \)

Explanation:
1. The owner's car is in one of the middle places; the other 14 occupied places are chosen from the remaining 24 places.
2. Both neighbours empty: the 14 cars lie among the other 22 places: C(22, 14)/C(24, 14).
3. This equals (10 × 9)/(24 × 23) = 15/92.

Teacher's Note:
1. Fix the owner's car and place the other cars among the remaining places.

Answer: (b) \( \frac{11}{20} \)

Explanation:
1. x + 100/x > 50 gives x² − 50x + 100 > 0, whose roots are 25 ± 5√21 ≈ 2.08 and 47.9.
2. So x < 2.08 or x > 47.9: x = 1, 2 and x = 48, ..., 100 (53 values), a total of 55.
3. Probability = 55/100 = 11/20.

Teacher's Note:
1. Solve the quadratic inequality and count the integers.

Answer: (a) \( 1-\frac{(n-1)!}{n^{n-1}} \)

Explanation:
1. Total ways: each object can go to any of n persons, nⁿ ways. Everyone gets one object in n! ways.
2. P(everyone gets something) = n!/nⁿ = (n − 1)!/nⁿ⁻¹.
3. So the required probability is 1 − (n − 1)!/nⁿ⁻¹.

Teacher's Note:
1. Use the complement: everyone gets exactly one.

Answer: (b) \( \frac{{}^{30}C_2\times{}^{19}C_2}{{}^{50}C_5} \)

Explanation:
1. For x₂₀ to be the middle of the five, two numbers must be smaller (from the 19 below) and two larger (from the 30 above).
2. Favourable ways = C(19, 2) × C(30, 2); total ways = C(50, 5).

Teacher's Note:
1. Count the numbers below and above x₂₀.

Answer: (c) 0.84

Explanation:
1. The machine works if all three parts are non-defective: 0.98 × 0.90 × 0.95 = 0.8379.
2. This is approximately 0.84.

Teacher's Note:
1. Multiply the probabilities of each part working.