To solve this problem, let's find the correct probability distribution for the number of times yellow appears (denoted as [tex]\(X\)[/tex]) when the spinner is spun twice.
The set of outcomes for two spins is given:
[tex]\[ S = \{ RB, RG, RY, RR, BR, BG, BY, BB, GR, GB, GY, GG, YR, YB, YG, YY \} \][/tex]
There are 16 possible outcomes. We need to determine the probabilities for the following cases:
1. [tex]\(X = 0\)[/tex]: Zero yellow sections
2. [tex]\(X = 1\)[/tex]: One yellow section
3. [tex]\(X = 2\)[/tex]: Two yellow sections
### Step-by-Step Breakdown
1. Calculate the number of outcomes with zero yellow sections ([tex]\(X=0\)[/tex]):
- Outcomes without any 'Y': \{RR, RB, RG, BR, BB, BG, GR, GB, GG\}
- There are 9 outcomes that do not contain yellow.
2. Calculate the number of outcomes with one yellow section ([tex]\(X=1\)[/tex]):
- Outcomes with exactly one 'Y': \{RY, BY, GY, YR, YB, YG\}
- There are 6 outcomes that contain exactly one yellow.
3. Calculate the number of outcomes with two yellow sections ([tex]\(X=2\)[/tex]):
- Outcomes with exactly two 'Y's: \{YY\}
- There is 1 outcome that contains exactly two yellows.
The total number of outcomes is 16. Hence, the probabilities are:
- [tex]\( P(X=0) = \frac{9}{16} = 0.5625 \)[/tex]
- [tex]\( P(X=1) = \frac{6}{16} = 0.375 \)[/tex]
- [tex]\( P(X=2) = \frac{1}{16} = 0.0625 \)[/tex]
### Construct the Probability Distribution Table
Based on our calculations, the correct probability distribution for the number of yellows (X) is:
[tex]\[
\begin{tabular}{|c|c|}
\hline
Yellow: $X$ & Probability: $P_X(x)$ \\
\hline
0 & 0.5625 \\
\hline
1 & 0.375 \\
\hline
2 & 0.0625 \\
\hline
\end{tabular}
\][/tex]
Thus, the first table is the correct probability distribution.
