JEE Mathematics Probability MCQs Set 08

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

Explanation:
1. Possible splits (L, D, E): \( (2,1,1) \): \( 10 \cdot 3 \cdot 2 = 60 \); \( (1,2,1) \): \( 5 \cdot 3 \cdot 2 = 30 \); \( (1,1,2) \): \( 5 \cdot 3 \cdot 1 = 15 \). Total favourable 105.
2. Probability \( = \frac{105}{\binom{10}{4}} = \frac{105}{210} = \frac12 \).

Teacher's Note:
1. List the ways one category contributes two persons.
2. Total is C(10,4) = 210.

Answer: (b) \( \frac{3}{28} \)

Explanation:
1. No boy between girls means the three girls sit together. Treat them as one block. The block can sit in 6 positions among the 5 boys, the girls can be arranged in \( 3! \) ways and the boys in \( 5! \) ways: \( 6 \cdot 3! \cdot 5! \) favourable arrangements.
2. Total arrangements \( = 8! \). Probability \( = \frac{6 \cdot 6 \cdot 120}{40320} = \frac{3}{28} \).

Teacher's Note:
1. Glue the girls into a block.
2. Count the block positions among the boys.

Answer: (c) \( \frac{6}{5} \)

Explanation:
1. \( P(A) + P(B) = P(A \cup B) + P(A \cap B) = \frac35 + \frac15 = \frac45 \).
2. \( P(\bar A) + P(\bar B) = 2 - \frac45 = \frac65 \).

Teacher's Note:
1. Use P(A) + P(B) = P(A∪B) + P(A∩B).
2. Complements subtract from 1 each.

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

Explanation:
1. He wins more tosses if there are at least 4 heads. By symmetry, \( P(\text{heads} \geq 4) = P(\text{heads} \leq 3) \).
2. These two events cover all outcomes (7 is odd), so each has probability \( \frac12 \).

Teacher's Note:
1. Use symmetry between heads and tails.
2. An odd number of tosses cannot end in a tie.

Answer: (c) \( \frac{1}{4} \)

Explanation:
1. Closer to the centre than to the boundary means the distance from the centre is less than \( \frac r2 \).
2. Area ratio \( = \frac{\pi (r/2)^2}{\pi r^2} = \frac14 \).

Teacher's Note:
1. Convert the condition into a smaller concentric circle.
2. Probability is a ratio of areas.

Answer: (b) exactly \( \frac12 \)

Explanation:
1. Each throw shows an even number with probability \( \frac12 \). In \( 2n + 1 \) throws, the number of evens is odd or even with equal likelihood.
2. Pairing each outcome with the outcome in which every throw is replaced by its opposite parity shows the probability is exactly \( \frac12 \).

Teacher's Note:
1. With p = 1/2 and an odd number of trials, odd and even counts are equally likely.
2. Use the complement-pairing argument.

Answer: (a) \( \frac{9}{64} \)

Explanation:
1. Each draw gives a heart with probability \( \frac14 \). Not a heart in the first two draws, then a heart: \( \frac34 \cdot \frac34 \cdot \frac14 = \frac{9}{64} \).

Teacher's Note:
1. First success at the third trial is a geometric probability.
2. Draws are independent because of replacement.

Answer: (c) \( \frac{50}{663} \)

Explanation:
1. A's two cards are fixed. B must choose exactly one of them and one of the other 50 cards: \( 2 \times 50 = 100 \) ways.
2. Probability \( = \frac{100}{\binom{52}{2}} = \frac{100}{1326} = \frac{50}{663} \).

Teacher's Note:
1. Treat A's cards as fixed marked cards.
2. B chooses one marked and one unmarked card.

Answer: (d) (i) and (iii)

Explanation:
1. \( P(A) + P(B) = P(A \cup B) + P(A \cap B) \). This is \( \leq 1 + \frac38 = \frac{11}{8} \) and \( > \frac34 + \frac18 = \frac78 \), so (i) and (iii) hold.
2. (ii) fails: take \( P(A) = P(B) = \frac{11}{16} \), \( P(A \cap B) = \frac38 \), \( P(A \cup B) = 1 \); then \( P(A)P(B) = \frac{121}{256} > \frac38 \).

Teacher's Note:
1. Use P(A) + P(B) = P(A∪B) + P(A∩B).
2. Test the product claim with an extreme example.

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

Explanation:
1. \( P(A) = \frac12 \), \( P(B) = \frac13 \) for one game.
2. Alternate wins: ABA or BAB: \( \frac12\cdot\frac13\cdot\frac12 + \frac13\cdot\frac12\cdot\frac13 = \frac1{12} + \frac1{18} = \frac{5}{36} \).

Teacher's Note:
1. Estimate probabilities from the frequencies.
2. Alternating means two possible orders.

Answer: (a) \( \frac{1}{210} \)

Explanation:
1. Total distinct arrangements \( = \frac{9!}{4!\,3!\,2!} = 1260 \).
2. Same denominations together: arrange the 3 blocks in \( 3! = 6 \) ways. Probability \( = \frac{6}{1260} = \frac{1}{210} \).

Teacher's Note:
1. Coins of the same value are identical.
2. Treat each denomination as one block.

Answer: (a) \( \frac{7}{15} \)

Explanation:
1. Place the 7 white balls; this creates 8 gaps. Choose 3 gaps for the black balls: \( \binom83 = 56 \).
2. Total positions for the black balls: \( \binom{10}{3} = 120 \). Probability \( = \frac{56}{120} = \frac{7}{15} \).

Teacher's Note:
1. Use the gap method for non-adjacent objects.
2. Choose positions, not individual balls.

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

Explanation:
1. Ordered triples with sum 15: \( (6,6,3) \) in 3 orders, \( (6,5,4) \) in 6 orders, \( (5,5,5) \) in 1 order: total 10.
2. Probability \( = \frac{10}{216} = \frac{5}{108} \).

Teacher's Note:
1. List the unordered triples and count their orderings.
2. Total outcomes are 6³ = 216.

Answer: (a) \( \frac{13}{32} \)

Explanation:
1. \( P(W) = \frac34, \frac24, \frac14 \) for the three boxes.
2. WWB: \( \frac34\cdot\frac24\cdot\frac34 = \frac{18}{64} \); WBW: \( \frac34\cdot\frac24\cdot\frac14 = \frac{6}{64} \); BWW: \( \frac14\cdot\frac24\cdot\frac14 = \frac{2}{64} \).
3. Sum \( = \frac{26}{64} = \frac{13}{32} \).

Teacher's Note:
1. List every arrangement of colours across the boxes.
2. Boxes are independent, so multiply.

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

Explanation:
1. The region \( x^2 \lt y \lt \sqrt x \) lies inside the unit square.
2. Its area is \( \int_0^1(\sqrt x - x^2)\,dx = \frac23 - \frac13 = \frac13 \), which is the probability.

Teacher's Note:
1. Use area for a uniform point in the unit square.
2. The two parabolas meet at (0,0) and (1,1).

Answer: (a) \( \frac{137}{143} \)

Explanation:
1. Total distributions \( = \binom{15}{5} = 3003 \). Each person gets at least one: \( \binom95 = 126 \).
2. P(everyone gets some) \( = \frac{126}{3003} = \frac{6}{143} \), so the required probability is \( 1 - \frac{6}{143} = \frac{137}{143} \).

Teacher's Note:
1. Use the complement.
2. Distributing identical items is a stars-and-bars count.

Answer: (a) (ii) and (iii)

Explanation:
1. Let \( S_1 = m+p+c \), \( S_2 = mp+pc+cm \), \( S_3 = mpc \). Then \( P(\geq 2) = S_2 - 2S_3 = 0.5 \) and \( P(\text{exactly } 2) = S_2 - 3S_3 = 0.4 \), so \( S_3 = 0.1 \) and \( S_2 = 0.7 \).
2. \( P(\geq 1) = S_1 - S_2 + S_3 = 0.75 \), so \( S_1 = 0.75 + 0.7 - 0.1 = 1.35 = \frac{27}{20} \).

Teacher's Note:
1. Use the elementary symmetric functions of m, p, c.
2. Subtract the two equations for 'at least two' and 'exactly two'.

Answer: (a) 7

Explanation:
1. \( P(\text{at least one hit}) = 1 - 0.9^n \geq 0.5 \), so \( 0.9^n \leq 0.5 \).
2. \( n \geq \frac{\ln 0.5}{\ln 0.9} \approx 6.6 \). For \( n = 7 \) the chance is about 52%, so \( n = 7 \).

Teacher's Note:
1. Use the complement of missing every time.
2. Take the logarithm to solve for n.

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

Explanation:
1. Primes in \( A \): 2, 3, 5, 7, 11, 13, 17, 19 (8 primes).
2. Those above 10: 11, 13, 17, 19 (4 primes). Probability \( = \frac48 = \frac12 \).

Teacher's Note:
1. Reduce the sample space to primes only.
2. List the primes carefully.

Answer: (a) \( \frac{23}{50} \)

Explanation:
1. Exactly one: \( P(A)P(\bar B) + P(\bar A)P(B) = \frac{3}{10}\cdot\frac35 + \frac{7}{10}\cdot\frac25 \).
2. \( = \frac{9}{50} + \frac{14}{50} = \frac{23}{50} \).

Teacher's Note:
1. Exactly one means A only or B only.
2. Use independence to multiply.