JEE Mathematics Probability MCQs Set 07

Answer: (b) \( \frac{4}{35} \)

Explanation:
1. Total ways to choose 3 seats \( = \binom73 = 35 \).
2. The middle seat 4 is taken; the others cannot be 3 or 5, so they come from \( \{1, 2, 6, 7\} \) and must not be adjacent to each other: 6 pairs minus \( \{1,2\} \) and \( \{6,7\} \) = 4.
3. Probability \( = \frac{4}{35} \).

Teacher's Note:
1. Choose seats, not persons.
2. Remove adjacent pairs from the allowed seats.

Answer: (c) \( \frac{3}{7} \)

Explanation:
1. \( P(\text{black}) = \frac12\cdot\frac37 + \frac12\cdot\frac47 = \frac12 \).
2. By Bayes: \( P(\text{first} \mid \text{black}) = \frac{\frac12 \cdot \frac37}{\frac12} = \frac37 \).

Teacher's Note:
1. Use Bayes' theorem.
2. Add the paths that lead to a black ball.

Answer: (b) \( \frac{1}{36} \)

Explanation:
1. \( P(Y) = \frac12 \), \( P(R) = \frac13 \), \( P(B) = \frac16 \).
2. Independent tosses: \( \frac12 \cdot \frac13 \cdot \frac16 = \frac{1}{36} \).

Teacher's Note:
1. The order is fixed, so multiply probabilities.
2. Each toss is independent.

Answer: (b) \( \frac{11}{40} \)

Explanation:
1. \( P(\min = 4) = \frac{\binom62}{120} = \frac{15}{120} \) and \( P(\max = 8) = \frac{\binom72}{120} = \frac{21}{120} \).
2. Both: min 4, max 8, third in \( \{5,6,7\} \): \( \frac{3}{120} \).
3. Union \( = \frac{15 + 21 - 3}{120} = \frac{33}{120} = \frac{11}{40} \).

Teacher's Note:
1. Use P(A or B) = P(A) + P(B) - P(A and B).
2. Count sets, not ordered triples.

Answer: (b) \( \frac{1}{40} \)

Explanation:
1. The third number must lie strictly between 3 and 7: 4, 5 or 6, giving 3 favourable sets.
2. Total sets \( = \binom{10}{3} = 120 \). Probability \( = \frac{3}{120} = \frac{1}{40} \).

Teacher's Note:
1. Fix the minimum and maximum, then choose the third.
2. Count unordered sets.

Answer: (b) \( \frac{1}{13} \)

Explanation:
1. The ace of spades is equally likely to be at any of the 13 positions.
2. So the probability that it is 4th is \( \frac{1}{13} \).

Teacher's Note:
1. Any position is equally likely.
2. Use symmetry rather than multiplying fractions.

Answer: (c) 0.39

Explanation:
1. \( P(A \cup B) = 0.25 + 0.5 - 0.14 = 0.61 \).
2. \( P(\text{neither}) = 1 - 0.61 = 0.39 \).

Teacher's Note:
1. Neither means the complement of the union.
2. Use inclusion-exclusion.

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

Explanation:
1. Residues mod 3: four numbers each for 0, 1, 2. Sum divisible by 3 if both are residue 0 (\( \binom42 = 6 \)) or one is residue 1 and the other residue 2 (\( 4 \cdot 4 = 16 \)): 22 ways.
2. Probability \( = \frac{22}{\binom{12}{2}} = \frac{22}{66} = \frac13 \).

Teacher's Note:
1. Use residues modulo 3.
2. Pairs with sum divisible by 3 are (0,0) or (1,2).

Answer: (b) (ii), (iii) and (iv)

Explanation:
1. Independent events with non-zero probabilities cannot be mutually exclusive, so (i) is false.
2. Independence carries over to complements: (ii) and (iii) are true. (iv) is true because the conditional probabilities of an event and its complement add to 1.

Teacher's Note:
1. Independence is preserved under complementation.
2. Mutually exclusive non-zero events are dependent.

Answer: (c) 5 : 1

Explanation:
1. \( P(\text{sum} = 7) = \frac{6}{36} = \frac16 \).
2. Odds against \( = \frac{5/6}{1/6} = 5 : 1 \).

Teacher's Note:
1. Odds against = (1 - p) : p.
2. Odds are not the same as probability.

Answer: (d) none of the above implications hold

Explanation:
1. Probability ordering does not mean the events are subsets of each other.
2. For example, two different faces on a die with equal probabilities satisfy \( P(E) = P(F) \), but neither event implies the other.

Teacher's Note:
1. A smaller probability does not mean a subset.
2. Use a counterexample from a die.

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

Explanation:
1. The units digit of \( 3^n \) cycles through 3, 9, 7, 1 with period 4.
2. It is 3 when \( n \equiv 1 \pmod 4 \), i.e. for one in four values of \( n \).

Teacher's Note:
1. Units digits of powers repeat in cycles.
2. Count how many exponents in a cycle give the required digit.

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

Explanation:
1. Total arrangements \( = \frac{7!}{3!} = 840 \).
2. With TTT as one block there are 5 objects: \( 5! = 120 \) arrangements. Probability \( = \frac{120}{840} = \frac17 \).

Teacher's Note:
1. Block the repeated letters.
2. Divide by 3! for the repeated T in the total.

Answer: (b) (i) and (iii)

Explanation:
1. \( P(A \cup B) = P(A) + P(B) - P(A \cap B) \), so (iii) is an identity.
2. Since \( P(A \cup B) \leq 1 \), (iii) gives \( P(A \cap B) \geq P(A) + P(B) - 1 \), which is (i). (ii) and (iv) are false.

Teacher's Note:
1. Rearrange the inclusion-exclusion formula.
2. A union cannot exceed 1.

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

Explanation:
1. \( P(n) = kn \) and \( k(1 + 2 + \dots + 6) = 21k = 1 \), so \( k = \frac{1}{21} \).
2. \( P(3) = \frac{3}{21} = \frac17 \).

Teacher's Note:
1. Total probability must be 1.
2. Find the constant of proportionality first.

Answer: (d) \( \frac{17}{108} \)

Explanation:
1. The possible perfect-square sums are 4, 9 and 16.
2. Ways: sum 4 gives 3, sum 9 gives 25, sum 16 gives 6, total 34.
3. Probability \( = \frac{34}{216} = \frac{17}{108} \).

Teacher's Note:
1. Sums range from 3 to 18.
2. Count ordered triples for each square.

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

Explanation:
1. For \( x_{20} \) to be the middle one, 2 of the chosen numbers must be smaller (from \( x_1, \dots, x_{19} \)) and 2 larger (from \( x_{21}, \dots, x_{50} \), which are 30 numbers).
2. Favourable \( = \binom{19}{2}\binom{30}{2} \); total \( = \binom{50}{5} \).

Teacher's Note:
1. Split the remaining numbers into smaller and larger.
2. The middle number of five has two on each side.

Answer: (d) \( \frac{1}{16} \)

Explanation:
1. For the fifth toss to differ from all of the first four, the first four must be identical: probability \( \frac{2}{16} = \frac18 \).
2. Then the fifth toss must be the opposite: \( \frac12 \). Probability \( = \frac18 \cdot \frac12 = \frac{1}{16} \).

Teacher's Note:
1. All four earlier results must be the same.
2. Then one specific outcome is needed on toss five.

Answer: (c) \( \frac{1}{35} \)

Explanation:
1. Total circular arrangements \( = 7! \).
2. Alternate seating: seat the gentlemen in \( 3! \) ways, then the ladies in the 4 gaps in \( 4! \) ways: \( 6 \cdot 24 = 144 \).
3. Probability \( = \frac{144}{5040} = \frac{1}{35} \).

Teacher's Note:
1. Fix one person to handle circular symmetry.
2. Alternate seating means one group fills the gaps.

Answer: (b) \( \frac{90}{2^{10}} \)

Explanation:
1. Probabilities per draw: heart \( \frac14 \), diamond \( \frac14 \), black \( \frac12 \).
2. Number of arrangements \( = \frac{6!}{2!\,2!\,2!} = 90 \). Probability \( = 90 \cdot \left(\frac14\right)^2\left(\frac14\right)^2\left(\frac12\right)^2 = \frac{90}{2^{10}} \).

Teacher's Note:
1. Use the multinomial formula.
2. Black cards combine spades and clubs, so p = 1/2.

Answer: (b) \( \frac{44}{85 \times 49} \)

Explanation:
1. Four cards of the same suit: \( 4 \times \binom{13}{4} = 2860 \) ways out of \( \binom{52}{4} = 270725 \).
2. Probability \( = \frac{2860}{270725} = \frac{44}{4165} = \frac{44}{85 \times 49} \).

Teacher's Note:
1. Together they draw four cards.
2. Choose the suit, then four cards of it.

Answer: (a) \( \frac{5}{12} \)

Explanation:
1. A hexagon has \( \binom62 - 6 = 9 \) diagonals, so two can be chosen in \( \binom92 = 36 \) ways.
2. Every set of 4 vertices gives exactly one intersecting pair of diagonals: \( \binom64 = 15 \). Probability \( = \frac{15}{36} = \frac{5}{12} \).

Teacher's Note:
1. Four vertices decide one crossing pair.
2. Number of diagonals of an n-gon is n(n-3)/2.

Answer: (c) \( 1 - \frac{(2n)!}{(n!)^2}\cdot\frac{1}{4^n} \)

Explanation:
1. Equal heads and tails means exactly \( n \) heads: \( \binom{2n}{n}\left(\frac12\right)^{2n} = \frac{(2n)!}{(n!)^2}\cdot\frac{1}{4^n} \).
2. The required probability is the complement.

Teacher's Note:
1. Use the complement.
2. Exactly n heads in 2n tosses is a binomial term.