JEE Mathematics Probability MCQs Set 22

Answer: (d) \( \frac25 \)

Explanation:
1. Given that the 4 American couples sit together: treat them as 4 blocks (2 internal orders each) with the Indian man and wife as 2 separate people: \( 5!\cdot2^4 \) circular arrangements of 6 units... counted as \( (6-1)!\cdot2^4 \).
2. With the Indian couple also adjacent: 5 blocks give \( (5-1)!\cdot2^5 \) arrangements.
3. Probability \( \frac{4!\cdot2^5}{5!\cdot2^4}=\frac25 \).

Teacher's Note:
1. Identify the standard result or condition that applies and use it directly.
2. Substitute carefully and verify the answer with a quick check.

Answer: (d) \( \frac{20}{23} \)

Explanation:
1. \( P(B\text{ green}\mid G)=\left(\frac34\right)^2+\left(\frac14\right)^2=\frac58 \) and \( P(B\text{ green}\mid R)=2\cdot\frac34\cdot\frac14=\frac38 \).
2. Bayes: \( \frac{\frac45\cdot\frac58}{\frac45\cdot\frac58+\frac15\cdot\frac38}=\frac{20}{23} \).

Teacher's Note:
1. Identify the standard result or condition that applies and use it directly.
2. Substitute carefully and verify the answer with a quick check.

Answer: (c) 5

Explanation:
1. With \( k \) fives the sum is \( 12+2k \); \( 12+2k<15 \) means \( k\le1 \).
2. Favourable outcomes: \( \binom40+\binom41=5 \) out of 16, so \( \lambda=\frac5{16} \) and \( 16\lambda=5 \).

Teacher's Note:
1. Reduce the problem to a standard result and then substitute.
2. Verify the final value with a simple case.

Answer: (b) 11

Explanation:
1. By Bayes' theorem: \( p=\frac{\frac{13}{52}\cdot\binom{12}{2}}{\frac{13}{52}\binom{12}{2}+\frac{39}{52}\binom{13}{2}} \).
2. This is \( \frac{13\cdot66}{13\cdot66+39\cdot78}=\frac{858}{3900}=\frac{11}{50} \).
3. So \( 50p=11 \).

Teacher's Note:
1. Condition on whether the lost card is a spade or not.
2. Compute likelihoods of drawing two spades in each case.

Answer: (b) \( \frac{25}{216} \)

Explanation:
1. Two non-sixes then a six: \( \left(\frac56\right)^2\cdot\frac16=\frac{25}{216} \).

Teacher's Note:
1. Identify the standard result or condition that applies and use it directly.
2. Substitute carefully and verify the answer with a quick check.

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

Explanation:
1. Given \( E \), urn \( i \) is even-numbered: \( i=2,4,\dots,n \) with equal probability \( \frac2n \) each.
2. Mean of \( i \) is \( \frac{n+2}2 \), so \( P(w|E)=\frac{(n+2)/2}{n+1}=\frac{n+2}{2(n+1)} \).

Teacher's Note:
1. Identify the standard result or condition that applies and use it directly.
2. Substitute carefully and verify the answer with a quick check.

Answer: (d) 6

Explanation:
1. P(white) = \( \frac23 \). Probability of exactly 3 whites = \( {}^nC_3\left(\frac23\right)^3\left(\frac13\right)^{n-3} \).
2. For \( n=6 \): \( 20\cdot\frac8{27}\cdot\frac1{27}=\frac{160}{729} \) which matches.
3. Check other values: n = 5 gives 80/243 (not equal), n = 7 gives 560/2187 (not equal).

Teacher's Note:
1. Use the binomial distribution.
2. Test the options directly.

Answer: (c) 5 or 10

Explanation:
1. Independence: \( \frac{|A\cap B|}{10}=\frac4{10}\cdot\frac{|B|}{10} \), so \( |A\cap B|=\frac{2|B|}{5} \).
2. So \( |B| \) is a multiple of 5: \( |B|=5 \) (with 2 common outcomes) or \( |B|=10 \) (with 4 common outcomes).

Teacher's Note:
1. Identify the standard result or condition that applies and use it directly.
2. Substitute carefully and verify the answer with a quick check.

Answer: (d) \( \frac25 \)

Explanation:
1. The winner is equally likely to be any of 5 horses.
2. The man's 2 horses cover 2 of the 5 possibilities, so the probability is \( \frac25 \).

Teacher's Note:
1. Equivalent: choose the winner first, then ask how likely the man's pair contains it.
2. Pairs containing a given horse: 4 out of 10.

Answer: (c) \( \frac23 \)

Explanation:
1. Let urns be chosen with \( P(u_i)=\frac{2i}{n(n+1)} \) and \( P(w|u_i)=\frac{i}{n+1} \).
2. \( P(w)=\sum\frac{2i^2}{n(n+1)^2}=\frac{2n(n+1)(2n+1)}{6n(n+1)^2}=\frac{2n+1}{3(n+1)} \), which tends to \( \frac23 \).

Teacher's Note:
1. Identify the standard result or condition that applies and use it directly.
2. Substitute carefully and verify the answer with a quick check.

Answer: (b) \( \frac12 \)

Explanation:
1. Allowed splits: (3,3), (4,2), (2,4).
2. \( {}^5C_3\cdot{}^5C_3=100 \); \( 2\cdot{}^5C_4\cdot{}^5C_2=100 \); total 200.
3. Probability \( =\frac{100}{200}=\frac12 \).

Teacher's Note:
1. Count only the allowed selections.
2. Equal numbers means the (3,3) split.

Answer: (b) \( \frac{25}{36} \)

Explanation:
1. Memorylessness: \( P(X>5\mid X>3)=\frac{(5/6)^5}{(5/6)^3}=\frac{25}{36} \).

Teacher's Note:
1. Identify the standard result or condition that applies and use it directly.
2. Substitute carefully and verify the answer with a quick check.

Answer: (c) Both are true but Statement 2 is not a correct explanation of Statement 1

Explanation:
1. By Bayes' rule \( P(H_i|E)=\frac{P(E|H_i)P(H_i)}{P(E)} \) and \( 0<P(E)<1 \), so \( P(H_i|E)>P(E|H_i)P(H_i) \) (when \( P(E|H_i)>0 \)): Statement 1 is true.
2. Statement 2 is true for exhaustive mutually exclusive events but is not what makes Statement 1 hold (that is \( P(E)<1 \)).

Teacher's Note:
1. Identify the standard result or condition that applies and use it directly.
2. Substitute carefully and verify the answer with a quick check.

Answer: (a) \( \frac45\ \text{or}\ \frac16 \)

Explanation:
1. Let \( P(A)=a,\ P(B)=b \): \( (1-a)b=\frac2{15} \) and \( a(1-b)=\frac16 \).
2. Subtracting: \( a-b=\frac1{30} \).
3. Substituting: \( 30b^2-29b+4=0 \), so \( b=\frac45 \) or \( \frac16 \).
4. Both give valid probabilities (\( a=\frac56 \) or \( \frac15 \)).

Teacher's Note:
1. Independence lets you multiply the probabilities of A (or its complement) and B (or its complement).
2. Check that both roots give a valid value of a.

Answer: (d) Both are true but Statement 2 is not a correct explanation of Statement 1

Explanation:
1. A unique (trivial) solution needs \( ad-bc\ne0 \): either \( ad=1,bc=0 \) (3 cases) or \( ad=0,bc=1 \) (3 cases): 6 of 16, i.e. \( \frac38 \).
2. A homogeneous system always has the trivial solution (probability 1). This does not explain the value \( \frac38 \).

Teacher's Note:
1. Identify the standard result or condition that applies and use it directly.
2. Substitute carefully and verify the answer with a quick check.

Answer: (b) \( \frac{125}{216} \)

Explanation:
1. The first three tosses are not sixes: \( \left(\frac56\right)^3=\frac{125}{216} \).

Teacher's Note:
1. Identify the standard result or condition that applies and use it directly.
2. Substitute carefully and verify the answer with a quick check.

Answer: (a) \( \frac29 \)

Explanation:
1. Each throw gives residue 0, 1 or 2 (mod 3) with probability \( \frac13 \).
2. The sum is 0 only if the powers are \( 1,\omega,\omega^2 \) in some order (three equal powers sum to \( 3\omega^k\ne0 \)): probability \( \frac{3!}{27}=\frac29 \).

Teacher's Note:
1. Identify the standard result or condition that applies and use it directly.
2. Substitute carefully and verify the answer with a quick check.

Answer: (b) \( \frac2{n+1} \)

Explanation:
1. With \( c=\frac1n \): \( P(w)=\frac1n\sum\frac{i}{n+1}=\frac12 \).
2. By Bayes: \( P(u_n|w)=\frac{\frac1n\cdot\frac n{n+1}}{1/2}=\frac2{n+1} \).

Teacher's Note:
1. Identify the standard result or condition that applies and use it directly.
2. Substitute carefully and verify the answer with a quick check.