JEE Mathematics Probability MCQs Set 19

Answer: (c) \( \frac{3n}{4n^2-1} \)

Explanation:
1. Three numbers are in A.P. if the first and last have the same parity (the middle is then their mean).
2. Pairs of the same parity from n + 1 odd and n even numbers: \( {}^{n+1}C_2 + {}^nC_2 = n^2 \).
3. P = \( n^2/{}^{2n+1}C_3 = \frac{6n^2}{(2n+1)(2n)(2n-1)} = \frac{3n}{4n^2-1} \).

Teacher's Note:
1. Fix the end terms first; the middle term is then determined. Count only pairs with the same parity.

Answer: (b) \( \frac{3}{2(2n-1)} \)

Explanation:
1. The end terms must have the same parity: \( {}^nC_2 + {}^nC_2 = n(n-1) \) choices.
2. Total \( {}^{2n}C_3 = \frac{2n(2n-1)(2n-2)}{6} \).
3. P = n(n-1)/{}^{2n}C_3 = 3/(2(2n-1)).

Teacher's Note:
1. Among 2n consecutive integers there are n odd and n even numbers. The A.P. count is n(n-1).

Answer: (d) 7 : 2

Explanation:
1. Fix one of the persons; the other can take any of the 9 remaining seats, of which 2 are adjacent.
2. P(together) = 2/9.
3. Odds against = 7 : 2.

Teacher's Note:
1. Odds against = P(not event) : P(event). Fixing one person removes the rotational symmetry.

Answer: (c) \( \frac{2}{11} \)

Explanation:
1. Fix A; B can occupy any of the 11 other seats.
2. Exactly 3 persons between them: B is 4 seats away on either side, so 2 favourable seats.
3. P = 2/11.

Teacher's Note:
1. In a circle two persons can be separated by r persons in two directions, giving 2 of the n - 1 positions.

Answer: (b) \( \frac{96\times95}{99\times98} \)

Explanation:
1. Number of ways of choosing 3 non-adjacent persons from n in a circle = n(n-4)(n-5)/6.
2. Total = n(n-1)(n-2)/6.
3. P = (n-4)(n-5)/((n-1)(n-2)) = 96 x 95/(99 x 98).

Teacher's Note:
1. The standard result for k non-adjacent chosen from a circle is useful. For three it reduces to (n-4)(n-5)/((n-1)(n-2)).

Answer: (a) \( \frac{11}{36} \)

Explanation:
1. Let the arrival times be x and y (in hours after 2 pm), both in [0, 1].
2. They meet if |x - y| ≤ 1/6.
3. The area of the region where they do not meet is (5/6)² = 25/36.
4. P = 1 - 25/36 = 11/36.

Teacher's Note:
1. In a unit square, the non-meeting region consists of two right triangles. Their combined area is (1 - t)² where t is the waiting time in hours.

Answer: (d) \( \frac{1}{12^{66}} \)

Explanation:
1. The number of binary operations is \( 12^{144} \).
2. A commutative one is fixed by the 12 x 13/2 = 78 values on unordered pairs, so there are \( 12^{78} \).
3. P = 12^{78}/12^{144} = 1/12^{66}.

Teacher's Note:
1. A commutative operation is determined by the values f(a, b) with a ≤ b only. There are n(n+1)/2 such pairs.

Answer: (a) 0

Explanation:
1. An impossible event has no favourable outcome, so n(A) = 0.
2. P = 0/n(S) = 0.

Teacher's Note:
1. The probability of any event lies between 0 and 1. The impossible event gives 0 and the sure event gives 1.

Answer: (d) \( \frac{{}^nC_r}{2^n} \)

Explanation:
1. Total outcomes = 2^n.
2. Choose the r coins that show heads: \( {}^nC_r \).
3. P = \( {}^nC_r/2^n \).

Teacher's Note:
1. This is the binomial probability with p = 1/2. Each outcome is equally likely.

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

Explanation:
1. Number of sequences with an odd number of heads = \( {}^nC_1 + {}^nC_3 + \dots = 2^{n-1} \).
2. P = 2^{n-1}/2^n = 1/2.

Teacher's Note:
1. The sums of the odd-indexed and even-indexed binomial coefficients are equal, each being 2^(n-1).

Answer: (b) \( \le P(A)+P(B) \)

Explanation:
1. P(A union B) = P(A) + P(B) - P(A intersect B).
2. Since P(A intersect B) ≥ 0, P(A union B) ≤ P(A) + P(B).

Teacher's Note:
1. Equality holds only when the events are mutually exclusive. In general the union cannot exceed the sum.

Answer: (c) 0

Explanation:
1. If \( A\subset B \), then A has no element outside B.
2. So \( A\cap B^c=\varnothing \) and its probability is 0.

Teacher's Note:
1. Draw a Venn diagram with A inside B. The region of A that lies outside B is empty.

Answer: (c) \( P(B)\le P(\bar A) \)

Explanation:
1. Since A and B are disjoint, \( B\subseteq \bar A \).
2. Therefore \( P(B)\le P(\bar A) \).

Teacher's Note:
1. Mutually exclusive events lie inside each other's complements. Equality holds when A and B together make up the whole sample space.

Answer: (a) \( P(\bar E_1)-P(E_2) \)

Explanation:
1. \( P(\bar E_1\cap\bar E_2)=1-P(E_1\cup E_2) \).
2. Since the events are disjoint, \( =1-P(E_1)-P(E_2) \).
3. \( =P(\bar E_1)-P(E_2) \).

Teacher's Note:
1. De Morgan's law converts the intersection of complements to the complement of a union. Then use the sum rule for disjoint events.

Answer: (b) \( P(A\cup B)-P(A\cap B) \)

Explanation:
1. The two events are disjoint, so the probability is \( P(A\cap\bar B)+P(\bar A\cap B) \).
2. \( =[P(A)-P(A\cap B)]+[P(B)-P(A\cap B)] \).
3. \( =P(A\cup B)-P(A\cap B) \).

Teacher's Note:
1. This is the probability that exactly one of A and B occurs. It is also P(A) + P(B) - 2P(A intersect B).

Answer: (c) \( P(A)>P(B) \)

Explanation:
1. \( P(A\cap C)>P(B\cap C) \) and \( P(A\cap\bar C)>P(B\cap\bar C) \).
2. Adding the two inequalities: \( P(A)>P(B) \).

Teacher's Note:
1. Multiply each conditional inequality by the positive P(C) or P(not C). Then add using the total probability rule.

Answer: (d) \( P(A|B)<P(A) \)

Explanation:
1. \( P(B|A)=\frac{P(A\cap B)}{P(A)}<P(B) \), so \( P(A\cap B)<P(A)P(B) \).
2. Dividing by P(B): \( P(A|B)<P(A) \).

Teacher's Note:
1. If A makes B less likely, then B makes A less likely as well. The dependence is symmetric.

Answer: (b) \( P(C|D)\ge P(C) \)

Explanation:
1. Since \( C\subset D \), \( C\cap D=C \).
2. \( P(C|D)=\frac{P(C)}{P(D)}\ge P(C) \) because \( 0<P(D)\le1 \).

Teacher's Note:
1. Dividing by a number at most 1 cannot decrease the probability. Equality holds when P(D) = 1 or P(C) = 0.

Answer: (a) 0 or 1

Explanation:
1. Independence of A with itself: P(A intersect A) = P(A)P(A).
2. P(A) = [P(A)]², so P(A) = 0 or 1.

Teacher's Note:
1. An event that is independent of itself is either impossible or certain.

Answer: (b) \( \frac1{n+1} \)

Explanation:
1. \( P(\bar A_k)=\frac{k}{k+1} \).
2. By independence the product is \( \frac12\cdot\frac23\cdots\frac n{n+1} \).
3. The product telescopes to \( \frac1{n+1} \).

Teacher's Note:
1. Independence of events means the complements are also independent. Write the product and cancel the telescoping terms.

Answer: (c) \( \frac{w+b}{w+2b} \)

Explanation:
1. p = w/(w+b), q = b/(w+b).
2. P(Q wins) = p/(1 - q²) = \( \frac{w/(w+b)}{1-b^2/(w+b)^2} = \frac{w(w+b)}{(w+b)^2-b^2} \).
3. \( =\frac{w(w+b)}{w(w+2b)}=\frac{w+b}{w+2b} \).

Teacher's Note:
1. For alternate draws the first player wins with p/(1 - q²). Factor the denominator as a difference of squares.

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

Explanation:
1. Mutually exclusive and exhaustive: P(A) + P(B) = 1.
2. P(A) + 2P(A) = 1, so P(A) = 1/3.

Teacher's Note:
1. When A and B are disjoint and their union is S, the two probabilities add to 1. Then use the given ratio.

Answer: (b) \( \frac{1}{14} \)

Explanation:
1. Pairs with sum more than 13: (6,8) and (7,8).
2. Total pairs \( {}^8C_2 = 28 \).
3. P = 2/28 = 1/14.

Teacher's Note:
1. List the pairs with the large sums systematically starting from the largest number. Do not forget that the order does not matter.

Answer: (b) \( \frac{1}{33} \)

Explanation:
1. Choose 2 couples out of 6: \( {}^6C_2 = 15 \).
2. Total ways \( {}^{12}C_4 = 495 \).
3. P = 15/495 = 1/33.

Teacher's Note:
1. Count the groups made of whole couples. A selection of 4 persons forms two couples only if they come in pairs.

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

Explanation:
1. Vowels: O, E; consonants: H, M.
2. P(no vowel) = 1/\( {}^4C_2 \) = 1/6.
3. P(at least one vowel) = 5/6.

Teacher's Note:
1. 'At least one' is easiest by complement. Here there is only one way to pick both consonants.