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A family has two children. If B represents a boy and G represents a girl, the set of outcomes for the possible genders of the children is [tex]S=\{BB, BG, GB, GG\}[/tex], with the oldest child listed first in each pair. Let [tex]X[/tex] represent the number of times G occurs. Which of the following is the probability distribution, [tex]P_X(x)[/tex]?

\begin{tabular}{|c|c|}
\hline
X & [tex]P_X(x)[/tex] \\
\hline
0 & 0.25 \\
\hline
1 & 0.5 \\
\hline
2 & 0.25 \\
\hline
\end{tabular}

\begin{tabular}{|c|c|}
\hline
X & [tex]P_X(x)[/tex] \\
\hline
0 & 0.33 \\
\hline
1 & 0.33 \\
\hline
2 & 0.33 \\
\hline
\end{tabular}

\begin{tabular}{|c|c|}
\hline
X & [tex]P_X(x)[/tex] \\
\hline
0 & 0.25 \\
\hline
1 & 0.75 \\
\hline
2 & 0 \\
\hline
\end{tabular}

\begin{tabular}{|c|c|}
\hline
X & [tex]P_X(x)[/tex] \\
\hline
0 & 0 \\
\hline
1 & 1 \\
\hline
2 & 0 \\
\hline
\end{tabular}

Sagot :

GinnyAnsweridn
Let's carefully determine the probability distribution [tex]\( P_X(x) \)[/tex] for the random variable [tex]\( X \)[/tex], which represents the number of times a girl (G) occurs in a family with two children.

First, we list all the possible outcomes for the genders of the two children. The outcomes are:
[tex]\[ S = \{BB, BG, GB, GG\} \][/tex]

Now, we'll determine the values of [tex]\( X \)[/tex] for each outcome:
- For [tex]\( BB \)[/tex], there are 0 girls, so [tex]\( X = 0 \)[/tex].
- For [tex]\( BG \)[/tex], there is 1 girl, so [tex]\( X = 1 \)[/tex].
- For [tex]\( GB \)[/tex], there is 1 girl, so [tex]\( X = 1 \)[/tex].
- For [tex]\( GG \)[/tex], there are 2 girls, so [tex]\( X = 2 \)[/tex].

Next, we count the number of times each value of [tex]\( X \)[/tex] occurs in our sample space:
- [tex]\( X = 0 \)[/tex] occurs once (for [tex]\( BB \)[/tex]).
- [tex]\( X = 1 \)[/tex] occurs twice (for [tex]\( BG \)[/tex] and [tex]\( GB \)[/tex]).
- [tex]\( X = 2 \)[/tex] occurs once (for [tex]\( GG \)[/tex]).

The total number of outcomes is 4. Now we calculate the probability for each value of [tex]\( X \)[/tex]:
- [tex]\( P_X(0) \)[/tex] is the probability that [tex]\( X = 0 \)[/tex], which is the number of outcomes with 0 girls divided by the total number of outcomes:
[tex]\[ P_X(0) = \frac{1}{4} = 0.25 \][/tex]
- [tex]\( P_X(1) \)[/tex] is the probability that [tex]\( X = 1 \)[/tex], which is the number of outcomes with 1 girl divided by the total number of outcomes:
[tex]\[ P_X(1) = \frac{2}{4} = 0.5 \][/tex]
- [tex]\( P_X(2) \)[/tex] is the probability that [tex]\( X = 2 \)[/tex], which is the number of outcomes with 2 girls divided by the total number of outcomes:
[tex]\[ P_X(2) = \frac{1}{4} = 0.25 \][/tex]

So, the probability distribution [tex]\( P_X(x) \)[/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]

Among the given options, the correct probability distribution table that matches our findings 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]
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