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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]?

[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]

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

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

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

Sagot :

GinnyAnsweridn
To determine the probability distribution [tex]\( P_X(x) \)[/tex] for the number of times blue occurs when a spinner with two equal parts (red and blue) is spun twice, we follow these steps:

1. List all possible outcomes:
When the spinner is spun twice, the set of all possible outcomes [tex]\( S \)[/tex] is [tex]\( \{RR, RB, BR, BB\} \)[/tex]. This represents the following:
- [tex]\( RR \)[/tex]: Red on both spins.
- [tex]\( RB \)[/tex]: Red on the first spin, blue on the second spin.
- [tex]\( BR \)[/tex]: Blue on the first spin, red on the second spin.
- [tex]\( BB \)[/tex]: Blue on both spins.

2. Count the occurrences of blue [tex]\( X \)[/tex] for each outcome:
- For [tex]\( RR \)[/tex]: Number of blues = 0
- For [tex]\( RB \)[/tex]: Number of blues = 1
- For [tex]\( BR \)[/tex]: Number of blues = 1
- For [tex]\( BB \)[/tex]: Number of blues = 2

3. Determine the frequency of each possible value of [tex]\( X \)[/tex]:
- [tex]\( X = 0 \)[/tex]: Occurs 1 time (only in [tex]\( RR \)[/tex])
- [tex]\( X = 1 \)[/tex]: Occurs 2 times (in [tex]\( RB \)[/tex] and [tex]\( BR \)[/tex])
- [tex]\( X = 2 \)[/tex]: Occurs 1 time (only in [tex]\( BB \)[/tex])

4. Calculate the total number of outcomes:
Since each outcome is equally likely, we have a total of 4 outcomes.

5. Compute the probabilities:
- Probability [tex]\( P_X(0) \)[/tex] (no blue occurrences): [tex]\( \frac{1}{4} = 0.25 \)[/tex]
- Probability [tex]\( P_X(1) \)[/tex] (one blue occurrence): [tex]\( \frac{2}{4} = 0.50 \)[/tex]
- Probability [tex]\( P_X(2) \)[/tex] (two blue occurrences): [tex]\( \frac{1}{4} = 0.25 \)[/tex]

Thus, the probability distribution [tex]\( P_X(x) \)[/tex] is given by:

[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]

Comparing this with the given options:
- The first table matches our calculated distribution perfectly.

Therefore, the correct probability distribution [tex]\( P_X(x) \)[/tex] is given by:
[tex]\[ \begin{tabular}{|c|c|} \hline \multicolumn{2}{|c|}{} \\ \hline 0 & 0.25 \\ \hline 1 & 0.5 \\ \hline 2 & 0.25 \\ \hline \end{tabular} \][/tex]
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