To solve the given problem, we need to determine the probability distribution of the random variable [tex]\(X\)[/tex], which represents the number of times blue occurs when a spinner is spun twice.
Step-by-Step Solution:
1. Identify the sample space:
The sample space [tex]\(S\)[/tex] given is [tex]\(\{RR, RB, BR, BB\}\)[/tex], where 'R' stands for red and 'B' stands for blue. This represents all possible outcomes when the spinner is spun twice.
2. Count the occurrences of blue (B) in each outcome:
- For outcome [tex]\(RR\)[/tex]: There are 0 occurrences of blue.
- For outcome [tex]\(RB\)[/tex]: There is 1 occurrence of blue.
- For outcome [tex]\(BR\)[/tex]: There is 1 occurrence of blue.
- For outcome [tex]\(BB\)[/tex]: There are 2 occurrences of blue.
3. Calculate the probabilities for each value of [tex]\(X\)[/tex]:
- Probability that [tex]\(X = 0\)[/tex] (0 blue occurrences):
Only the outcome [tex]\(RR\)[/tex] has 0 blue occurrences. There is 1 such outcome out of 4 total outcomes. So,
[tex]\[
P(X = 0) = \frac{1}{4} = 0.25
\][/tex]
- Probability that [tex]\(X = 1\)[/tex] (1 blue occurrence):
The outcomes [tex]\(RB\)[/tex] and [tex]\(BR\)[/tex] each have 1 blue occurrence. There are 2 such outcomes out of 4 total outcomes. So,
[tex]\[
P(X = 1) = \frac{2}{4} = 0.5
\][/tex]
- Probability that [tex]\(X = 2\)[/tex] (2 blue occurrences):
Only the outcome [tex]\(BB\)[/tex] has 2 blue occurrences. There is 1 such outcome out of 4 total outcomes. So,
[tex]\[
P(X = 2) = \frac{1}{4} = 0.25
\][/tex]
4. Summarize the probability distribution:
Based on the calculated probabilities, the probability distribution of [tex]\(X\)[/tex] is:
[tex]\[
\begin{array}{|c|c|}
\hline
X & P(X) \\
\hline
0 & 0.25 \\
\hline
1 & 0.5 \\
\hline
2 & 0.25 \\
\hline
\end{array}
\][/tex]
Comparing this distribution with the given options, the correct probability distribution table matches the first option:
[tex]\[
\begin{array}{|c|c|}
\hline
X & P_{C}(x) \\
\hline
0 & 0.25 \\
\hline
1 & 0.5 \\
\hline
2 & 0.25 \\
\hline
\end{array}
\][/tex]
Thus, the correct probability distribution is the one from the first table.
