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]
