To determine the probability distribution [tex]\(P_X(x)\)[/tex] for the number of times a girl (G) occurs in the family of two children, we start by enumerating all possible gender outcomes for the two children:
1. [tex]\(BB\)[/tex] (no girls)
2. [tex]\(BG\)[/tex] (one girl)
3. [tex]\(GB\)[/tex] (one girl)
4. [tex]\(GG\)[/tex] (two girls)
Given this set [tex]\(S = \{BB, BG, GB, GG\}\)[/tex], we calculate the probability for each value of [tex]\(x\)[/tex], where [tex]\(x\)[/tex] is the number of girls (G):
Step-by-Step Solution:
1. Count the occurrences for each value of girls [tex]\(x\)[/tex]:
- 0 girls (x = 0): There is only one combination where there are no girls, which is [tex]\(BB\)[/tex].
[tex]\[
P_X(0) = \frac{1}{4}
\][/tex]
- 1 girl (x = 1): There are two combinations where there is exactly one girl, which are [tex]\(BG\)[/tex] and [tex]\(GB\)[/tex].
[tex]\[
P_X(1) = \frac{2}{4}
\][/tex]
- 2 girls (x = 2): There is only one combination where there are two girls, which is [tex]\(GG\)[/tex].
[tex]\[
P_X(2) = \frac{1}{4}
\][/tex]
2. Calculate the probabilities:
- [tex]\(P_X(0) = \frac{1}{4} = 0.25\)[/tex]
- [tex]\(P_X(1) = \frac{2}{4} = 0.5\)[/tex]
- [tex]\(P_X(2) = \frac{1}{4} = 0.25\)[/tex]
Thus, the resulting probability distribution for [tex]\(X\)[/tex] is:
[tex]\[
\begin{array}{|c|c|}
\hline
x & P_X(x) \\
\hline
0 & 0.25 \\
\hline
1 & 0.5 \\
\hline
2 & 0.25 \\
\hline
\end{array}
\][/tex]
After comparing with the given options in the question, we find that the correct probability distribution matches the first tabular option:
[tex]\[
\begin{tabular}{|c|c|}
\hline
x & P_X(x) \\
\hline
0 & 0.25 \\
\hline
1 & 0.5 \\
\hline
2 & 0.25 \\
\hline
\end{tabular}
\][/tex]
