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A spinner is divided into two equal parts, one red and one blue. The set of possible outcomes when the spinner is spun twice is [tex] S = \{RR, RB, BR, BB\} [/tex]. Let [tex] X [/tex] represent the number of times blue occurs. Which of the following is the probability distribution [tex] P_x(x) [/tex]?

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

\begin{tabular}{|c|c|}
\hline [tex]$X$[/tex] & [tex]$P_x(x)$[/tex] \\
\hline 0 & 0 \\
\hline
\end{tabular}

Sagot :

GinnyAnswer
To determine the correct probability distribution [tex]\( P_X(X) \)[/tex] for the number of times blue occurs when the spinner is spun twice, follow these steps:

1. Identify the Possible Outcomes:
The possible outcomes when the spinner is spun twice are:
[tex]\[ S = \{RR, RB, BR, BB\} \][/tex]
Here, each outcome represents the result of two spins:
- [tex]\( RR \)[/tex]: Red on both spins
- [tex]\( RB \)[/tex]: Red on the first spin and Blue on the second spin
- [tex]\( BR \)[/tex]: Blue on the first spin and Red on the second spin
- [tex]\( BB \)[/tex]: Blue on both spins

2. Define the Random Variable [tex]\( X \)[/tex]:
Let [tex]\( X \)[/tex] be the number of times blue occurs in two spins.

3. Determine the Values of [tex]\( X \)[/tex]:
The possible values for [tex]\( X \)[/tex] are 0, 1, or 2.
- [tex]\( X = 0 \)[/tex] means no blue outcomes.
- [tex]\( X = 1 \)[/tex] means one blue outcome.
- [tex]\( X = 2 \)[/tex] means two blue outcomes.

4. Count the Outcomes for Each Value of [tex]\( X \)[/tex]:
- [tex]\( X = 0 \)[/tex]: The outcome is [tex]\( RR \)[/tex]. There is 1 such outcome.
- [tex]\( X = 1 \)[/tex]: The outcomes are [tex]\( RB \)[/tex] and [tex]\( BR \)[/tex]. There are 2 such outcomes.
- [tex]\( X = 2 \)[/tex]: The outcome is [tex]\( BB \)[/tex]. There is 1 such outcome.

5. Calculate the Probability for Each Value of [tex]\( X \)[/tex]:
Each outcome is equally likely, and there are 4 possible outcomes in total.

- For [tex]\( X = 0 \)[/tex]:
[tex]\[ P_X(0) = \frac{\text{Number of outcomes with } X = 0}{\text{Total outcomes}} = \frac{1}{4} = 0.25 \][/tex]

- For [tex]\( X = 1 \)[/tex]:
[tex]\[ P_X(1) = \frac{\text{Number of outcomes with } X = 1}{\text{Total outcomes}} = \frac{2}{4} = 0.5 \][/tex]

- For [tex]\( X = 2 \)[/tex]:
[tex]\[ P_X(2) = \frac{\text{Number of outcomes with } X = 2}{\text{Total outcomes}} = \frac{1}{4} = 0.25 \][/tex]

6. Construct the Probability Distribution Table:
Based on the calculations, 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]

Hence, the correct probability distribution, [tex]\( P_X(X) \)[/tex], is given by the first table:
[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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