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Step-by-step Solution
Step 1: Identify the events and total outcomes
• Each family has 2 children, so in total there are 4 children.
• Each child can be a boy (B) or a girl (G) with equal probability, so the total number of equally likely outcomes is $2^4 = 16$.
Step 2: Define Event A (At least two girls) and Event B (All four are girls)
• Event B: “All four children are girls” has exactly 1 favorable outcome: GGGG.
• Event A: “At least two girls” means 2, 3, or 4 girls among the children.
Step 3: Calculate $P(B)$
All children are girls in exactly 1 out of the 16 possible outcomes, hence
$$P(B) = \frac{1}{16}.$$
Step 4: Calculate $P(A)$ (Probability of at least two girls)
It is easier to subtract from 1 the probability of having fewer than two girls (i.e., 0 girls or 1 girl).
• Probability of 0 girls (i.e., 4 boys): There is $1$ way to have BBBB, so
$$P(\text{0 girls}) = \frac{1}{16}.$$
• Probability of exactly 1 girl: We can choose 1 child to be a girl in $\binom{4}{1} = 4$ ways, and the probability of each arrangement is $\frac{1}{16}$, so
$$P(\text{1 girl}) = \frac{4}{16} = \frac{1}{4}.$$
Therefore,
$$
P(A) \;=\; 1 \;-\; P(\text{0 girls}) \;-\; P(\text{1 girl})
\;=\; 1 \;-\; \frac{1}{16} \;-\; \frac{4}{16}
\;=\; 1 \;-\; \frac{5}{16}
\;=\; \frac{11}{16}.
$$
Step 5: Apply the formula for conditional probability
The conditional probability of B given A is
$$
P(B \mid A) \;=\; \frac{P(B \cap A)}{P(A)}.
$$
Since B (all four are girls) automatically satisfies “at least two girls,” we have $B \subset A$, hence $P(B \cap A) = P(B)$. Thus,
$$
P(B \mid A) \;=\; \frac{P(B)}{P(A)}
\;=\; \frac{\frac{1}{16}}{\frac{11}{16}}
\;=\; \frac{1}{11}.
$$
Final Answer
The required conditional probability is
$$\frac{1}{11}.$$
