Let us analyze the problem and find the correct probability distribution.
We are given that a spinner is divided into two equal parts: one red (R) and one blue (B). The spinner is spun twice, and the set of all possible outcomes is:
[tex]\[ S = \{RR, RB, BR, BB\} \][/tex]
Next, we define the random variable [tex]\( X \)[/tex] which represents the number of times blue (B) occurs in these two spins. The possible values for [tex]\( X \)[/tex] are 0, 1, or 2.
Now let's consider each possible value of [tex]\( X \)[/tex]:
1. [tex]\( X = 0 \)[/tex]: The outcome must have zero blue faces. The only outcome in [tex]\( S \)[/tex] that fits this criterion is [tex]\( \{RR\} \)[/tex].
2. [tex]\( X = 1 \)[/tex]: The outcome must have one blue face. The outcomes in [tex]\( S \)[/tex] that fit this criterion are [tex]\( \{RB, BR\} \)[/tex].
3. [tex]\( X = 2 \)[/tex]: The outcome must have two blue faces. The only outcome in [tex]\( S \)[/tex] that fits this criterion is [tex]\( \{BB\} \)[/tex].
We now calculate the probability distributions for each value of [tex]\( X \)[/tex]:
1. Probability that [tex]\( X = 0 \)[/tex]:
[tex]\[
P(X = 0) = \frac{\text{Number of outcomes with 0 blue}}{\text{Total number of outcomes}} = \frac{1}{4} = 0.25
\][/tex]
2. Probability that [tex]\( X = 1 \)[/tex]:
[tex]\[
P(X = 1) = \frac{\text{Number of outcomes with 1 blue}}{\text{Total number of outcomes}} = \frac{2}{4} = 0.5
\][/tex]
3. Probability that [tex]\( X = 2 \)[/tex]:
[tex]\[
P(X = 2) = \frac{\text{Number of outcomes with 2 blue}}{\text{Total number of outcomes}} = \frac{1}{4} = 0.25
\][/tex]
From the calculations above, the probability distribution [tex]\( P_X(x) \)[/tex] can be summarized as:
[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]
Comparing this with the provided tables, we see that the correct table 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]
