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A spinner contains four sections: red, blue, green, and yellow. Joaquin spins the spinner twice. The set of outcomes is given as [tex]S=\{RB, RG, RY, RR, BR, BG, BY, BB, GR, GB, GY, GG, YR, YB, YG, YY\}[/tex].

If the random variable is "yellow [tex](X)[/tex]," which of the following is the correct probability distribution?

\begin{tabular}{|c|c|}
\hline
Yellow: [tex]X[/tex] & Probability: [tex]P_X(x)[/tex] \\
\hline
0 & 0.5625 \\
\hline
1 & 0.375 \\
\hline
2 & 0.0625 \\
\hline
\end{tabular}

\begin{tabular}{|c|c|}
\hline
Yellow: [tex]X[/tex] & Probability: [tex]P_X(x)[/tex] \\
\hline
0 & 0.75 \\
\hline
1 & 0.25 \\
\hline
2 & 0 \\
\hline
\end{tabular}

\begin{tabular}{|c|c|}
\hline
Yellow: [tex]X[/tex] & Probability: [tex]P_X(x)[/tex] \\
\hline
0 & 0.5 \\
\hline
1 & 0.375 \\
\hline
2 & 0.125 \\
\hline
\end{tabular}

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

Sagot :

GinnyAnsweridn
To find the correct probability distribution for the number of yellows (denoted as [tex]\(X\)[/tex]) when Joaquin spins the spinner twice, we start by enumerating all possible outcomes, which you've correctly listed as:
[tex]\[ S = \{R B, R G, R Y, R R, B R, B G, B Y, B B, G R, G B, G Y, G G, Y R, Y B, Y G, Y Y\} \][/tex]

Since there are 4 colors and Joaquin spins the spinner twice, there are [tex]\(4 \times 4 = 16\)[/tex] possible outcomes.

Next, we are interested in the random variable [tex]\(X\)[/tex], which represents the number of times the result is yellow ([tex]\(Y\)[/tex]). Therefore, [tex]\(X\)[/tex] can take on the values 0, 1, or 2. We need to determine the probabilities of these values.

1. Count the occurrences of [tex]\(X=0\)[/tex] (no yellow [tex]\(Y\)[/tex] in the outcome):
[tex]\[ \{R B, R G, R R, B R, B G, B B, G R, G B, G G\} \][/tex]
There are 9 outcomes with no yellow. The probability is:
[tex]\[ P(X=0) = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{9}{16} = 0.5625 \][/tex]

2. Count the occurrences of [tex]\(X=1\)[/tex] (one yellow [tex]\(Y\)[/tex] in the outcome):
[tex]\[ \{R Y, B Y, G Y, Y R, Y B, Y G\} \][/tex]
There are 6 outcomes with exactly one yellow. The probability is:
[tex]\[ P(X=1) = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{6}{16} = 0.375 \][/tex]

3. Count the occurrences of [tex]\(X=2\)[/tex] (two yellows [tex]\(Y\)[/tex] in the outcome):
[tex]\[ \{Y Y\} \][/tex]
There is 1 outcome with exactly two yellows. The probability is:
[tex]\[ P(X=2) = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{1}{16} = 0.0625 \][/tex]

Combining these probabilities, the correct probability distribution for the yellow count 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]

Therefore, the correct option among the choices given in the question 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]
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