Question

Family A has 6 members of which 4 are males and 2 are females & family B has 5 members consisting at least one male and one female. (Assume it is equiprobable for a member of unknown gender to be a male or female). 2 members are selected randomly either from family A or family B. If both the members are female, then the probability that they belong to family A is equal to p. The value of 9p is equal to

• A. 1
• B. 2
• C. 3
• D. 4

Answer 1

Answer: (B)

2

Answer 2

Let A be the event that the two selected members belong to family A.
Let B be the event that the two selected members belong to family B.
Since the selection is made randomly either from family A or family B, we assume P(A) = P(B) = 1/2.

Let E be the event that both selected members are female.
We want to find p = P(A|E). Using Bayes' theorem:

$p = P(A|E) = \frac{P(E|A) P(A)}{P(E|A) P(A) + P(E|B) P(B)}$

First, calculate P(E|A).
Family A has 6 members: 4 males and 2 females.
The number of ways to select 2 members from 6 is $\binom{6}{2} = \frac{6 \times 5}{2} = 15$.
The number of ways to select 2 females from 2 is $\binom{2}{2} = 1$.
So, $P(E|A) = \frac{\binom{2}{2}}{\binom{6}{2}} = \frac{1}{15}$.

Next, calculate P(E|B).
Family B has 5 members, consisting of at least one male and one female.
The possible compositions of family B (Males, Females) are (1, 4), (2, 3), (3, 2), (4, 1).
The statement "Assume it is equiprobable for a member of unknown gender to be a male or female" suggests a process where each of the 5 members' gender is determined independently with P(Male)=1/2 and P(Female)=1/2. A composition with k females and 5-k males has a probability of $\binom{5}{k} (1/2)^k (1/2)^{5-k} = \binom{5}{k} (1/2)^5$.
The constraint is "at least one male and one female", which means the number of females k must be between 1 and 4 (inclusive).
The probability of getting a composition with k females, given the constraint, is $P(k \text{ females } | 1 \le k \le 4) = \frac{P(k \text{ females and } 1 \le k \le 4)}{P(1 \le k \le 4)}$.
If $1 \le k \le 4$, the condition $1 \le k \le 4$ is met. So, $P(k \text{ females } | 1 \le k \le 4) = \frac{P(k \text{ females})}{P(1 \le k \le 4)}$.
$P(k \text{ females}) = \binom{5}{k} (1/2)^5 = \binom{5}{k}/32$.
$P(1 \le k \le 4) = 1 - P(k=0) - P(k=5) = 1 - \binom{5}{0}/32 - \binom{5}{5}/32 = 1 - 1/32 - 1/32 = 30/32$.
So, the probability of family B having k females (where $1 \le k \le 4$) is $\frac{\binom{5}{k}/32}{30/32} = \frac{\binom{5}{k}}{30}$.

Let $C_k$ be the event that family B has k females (and 5-k males).
$P(C_1) = \binom{5}{1}/30 = 5/30 = 1/6$ (4M, 1F)
$P(C_2) = \binom{5}{2}/30 = 10/30 = 1/3$ (3M, 2F)
$P(C_3) = \binom{5}{3}/30 = 10/30 = 1/3$ (2M, 3F)
$P(C_4) = \binom{5}{4}/30 = 5/30 = 1/6$ (1M, 4F)

Now, we calculate P(E|B) using the law of total probability: $P(E|B) = \sum_{k=1}^{4} P(E|C_k) P(C_k)$.
If family B has k females, the probability of selecting 2 females is $\frac{\binom{k}{2}}{\binom{5}{2}}$.
$\binom{5}{2} = 10$.
$P(E|C_1) = \frac{\binom{1}{2}}{10} = 0/10 = 0$.
$P(E|C_2) = \frac{\binom{2}{2}}{10} = 1/10$.
$P(E|C_3) = \frac{\binom{3}{2}}{10} = 3/10$.
$P(E|C_4) = \frac{\binom{4}{2}}{10} = 6/10 = 3/5$.

$P(E|B) = P(E|C_1) P(C_1) + P(E|C_2) P(C_2) + P(E|C_3) P(C_3) + P(E|C_4) P(C_4)$
$P(E|B) = (0)(1/6) + (1/10)(1/3) + (3/10)(1/3) + (3/5)(1/6)$
$P(E|B) = 0 + 1/30 + 3/30 + 3/30 = 7/30$.

Now substitute the values into Bayes' theorem formula for p:
$p = \frac{P(E|A) P(A)}{P(E|A) P(A) + P(E|B) P(B)}$
$p = \frac{(1/15)(1/2)}{(1/15)(1/2) + (7/30)(1/2)}$
$p = \frac{1/30}{1/30 + 7/60}$
The denominator is $1/30 + 7/60 = 2/60 + 7/60 = 9/60 = 3/20$.
$p = \frac{1/30}{3/20} = \frac{1}{30} \times \frac{20}{3} = \frac{20}{90} = \frac{2}{9}$.

The value of p is 2/9.
We are asked to find the value of 9p.
$9p = 9 \times \frac{2}{9} = 2$.