To determine the probability distribution of [tex]\( X \)[/tex], the number of times [tex]\( G \)[/tex] (girl) occurs, we start by understanding the sample space of outcomes for the genders of the two children, given by:
[tex]\[ S = \{BB, BG, GB, GG\} \][/tex]
There are four equally likely outcomes.
Now, let's identify the probabilities for each value of [tex]\( X \)[/tex]:
1. [tex]\( X = 0 \)[/tex]: This corresponds to the outcome where there are no girls, which is [tex]\( BB \)[/tex].
- There is 1 outcome in the sample space that fits this description.
- So, the probability [tex]\( P(X = 0) \)[/tex] is calculated as:
[tex]\[
P(X = 0) = \frac{\text{Number of outcomes with no girl}}{\text{Total number of outcomes}} = \frac{1}{4} = 0.25
\][/tex]
2. [tex]\( X = 1 \)[/tex]: This corresponds to the outcomes where there is exactly one girl, which are [tex]\( BG \)[/tex] and [tex]\( GB \)[/tex].
- There are 2 outcomes that fit this description.
- So, the probability [tex]\( P(X = 1) \)[/tex] is calculated as:
[tex]\[
P(X = 1) = \frac{\text{Number of outcomes with exactly one girl}}{\text{Total number of outcomes}} = \frac{2}{4} = 0.5
\][/tex]
3. [tex]\( X = 2 \)[/tex]: This corresponds to the outcome where there are two girls, which is [tex]\( GG \)[/tex].
- There is 1 outcome that fits this description.
- So, the probability [tex]\( P(X = 2) \)[/tex] is calculated as:
[tex]\[
P(X = 2) = \frac{\text{Number of outcomes with two girls}}{\text{Total number of outcomes}} = \frac{1}{4} = 0.25
\][/tex]
Summarizing these probabilities:
[tex]\[
P_X(x) =
\begin{cases}
0.25 & \text{if } x = 0 \\
0.5 & \text{if } x = 1 \\
0.25 & \text{if } x = 2 \\
\end{cases}
\][/tex]
Therefore, the correct probability distribution for [tex]\( X \)[/tex] is:
\begin{tabular}{|c|c|}
\hline
[tex]\( X \)[/tex] & [tex]\( P(X) \)[/tex] \\
\hline
0 & 0.25 \\
\hline
1 & 0.5 \\
\hline
2 & 0.25 \\
\hline
\end{tabular}
This matches the first given table, confirming it is the correct probability distribution.
