In a family having three children,there may be no girl,one girl,two girls,or three girls. So,the probability of each is $\frac{1}{4}$. Is this correct? Justify your answer.

(B) No,the statement is incorrect because the outcomes are not equally likely.
Justification:
For a family with three children,the sample space $S$ consists of $2^3 = 8$ equally likely outcomes: $S = \{BBB, BBG, BGB, GBB, BGG, GBG, GGB, GGG\}$,where $B$ represents a boy and $G$ represents a girl.
Let $X$ be the number of girls. The possible values for $X$ are $0, 1, 2, 3$.
- $P(X=0) = P(\{BBB\}) = \frac{1}{8}$
- $P(X=1) = P(\{BBG, BGB, GBB\}) = \frac{3}{8}$
- $P(X=2) = P(\{BGG, GBG, GGB\}) = \frac{3}{8}$
- $P(X=3) = P(\{GGG\}) = \frac{1}{8}$
Since the probabilities are $\frac{1}{8}, \frac{3}{8}, \frac{3}{8}, \text{ and } \frac{1}{8}$ respectively,they are not equal to $\frac{1}{4}$.