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Sagot :
To determine the correct probability distribution for the number of times yellow appears in two spins of the spinner, we need to calculate the probabilities for 0, 1, and 2 appearances of yellow.
The spinner has four sections: red (R), blue (B), green (G), and yellow (Y). The possible outcomes for two spins can be enumerated as follows:
[tex]\[ S = \{RR, RB, RG, RY, BR, BB, BG, BY, GR, GB, GG, GY, YR, YB, YG, YY\} \][/tex]
Next, we calculate the probabilities for each scenario:
1. Probability of 0 yellows (x = 0)
To have 0 yellows means neither spin can result in yellow. Therefore, the outcomes that don't include Y are:
[tex]\[ RR, RB, RG, BR, BB, BG, GR, GB, GG \][/tex]
There are 9 such outcomes. Since each spin is independent and the probability of not landing on yellow for one spin is [tex]\( \frac{3}{4} \)[/tex], the probability for two spins not landing on yellow is:
[tex]\[ P(x=0) = \left(\frac{3}{4}\right) \times \left(\frac{3}{4}\right) = \frac{9}{16} = 0.5625 \][/tex]
2. Probability of 1 yellow (x = 1)
To have exactly 1 yellow means one spin results in yellow and the other does not. The relevant outcomes are:
[tex]\[ RY, BY, GY, YR, YB, YG \][/tex]
There are 6 such outcomes. The probability of one spin resulting in yellow and the other not can be computed as:
[tex]\[ P(x=1) = 2 \times \left(\frac{1}{4}\right) \times \left(\frac{3}{4}\right) = 2 \times \frac{3}{16} = \frac{6}{16} = 0.375 \][/tex]
3. Probability of 2 yellows (x = 2)
To have 2 yellows means both spins result in yellow. The only relevant outcome is:
[tex]\[ YY \][/tex]
There is only 1 such outcome. The probability for both spins resulting in yellow is:
[tex]\[ P(x=2) = \left(\frac{1}{4}\right) \times \left(\frac{1}{4}\right) = \frac{1}{16} = 0.0625 \][/tex]
Comparing these calculated probabilities with the given tables, the correct probability distribution is:
\begin{tabular}{|c|c|}
\hline
Yellow: [tex]\( x \)[/tex] & Probability: [tex]\( P_2(x) \)[/tex] \\
\hline
0 & 0.5625 \\
\hline
1 & 0.375 \\
\hline
2 & 0.0625 \\
\hline
\end{tabular}
However, note that the closest matching pre-provided answer table is:
\begin{tabular}{|c|c|}
\hline
Yellow: [tex]\( x \)[/tex] & Prokability: [tex]\( P_2(x) \)[/tex] \\
\hline
0 & 0.5 \\
\hline
1 & 0.375 \\
\hline
2 & 0.125 \\
\hline
\end{tabular}
There might be a typo in the calculated probabilities in the problems provided since the exact matching set isn't present.
The spinner has four sections: red (R), blue (B), green (G), and yellow (Y). The possible outcomes for two spins can be enumerated as follows:
[tex]\[ S = \{RR, RB, RG, RY, BR, BB, BG, BY, GR, GB, GG, GY, YR, YB, YG, YY\} \][/tex]
Next, we calculate the probabilities for each scenario:
1. Probability of 0 yellows (x = 0)
To have 0 yellows means neither spin can result in yellow. Therefore, the outcomes that don't include Y are:
[tex]\[ RR, RB, RG, BR, BB, BG, GR, GB, GG \][/tex]
There are 9 such outcomes. Since each spin is independent and the probability of not landing on yellow for one spin is [tex]\( \frac{3}{4} \)[/tex], the probability for two spins not landing on yellow is:
[tex]\[ P(x=0) = \left(\frac{3}{4}\right) \times \left(\frac{3}{4}\right) = \frac{9}{16} = 0.5625 \][/tex]
2. Probability of 1 yellow (x = 1)
To have exactly 1 yellow means one spin results in yellow and the other does not. The relevant outcomes are:
[tex]\[ RY, BY, GY, YR, YB, YG \][/tex]
There are 6 such outcomes. The probability of one spin resulting in yellow and the other not can be computed as:
[tex]\[ P(x=1) = 2 \times \left(\frac{1}{4}\right) \times \left(\frac{3}{4}\right) = 2 \times \frac{3}{16} = \frac{6}{16} = 0.375 \][/tex]
3. Probability of 2 yellows (x = 2)
To have 2 yellows means both spins result in yellow. The only relevant outcome is:
[tex]\[ YY \][/tex]
There is only 1 such outcome. The probability for both spins resulting in yellow is:
[tex]\[ P(x=2) = \left(\frac{1}{4}\right) \times \left(\frac{1}{4}\right) = \frac{1}{16} = 0.0625 \][/tex]
Comparing these calculated probabilities with the given tables, the correct probability distribution is:
\begin{tabular}{|c|c|}
\hline
Yellow: [tex]\( x \)[/tex] & Probability: [tex]\( P_2(x) \)[/tex] \\
\hline
0 & 0.5625 \\
\hline
1 & 0.375 \\
\hline
2 & 0.0625 \\
\hline
\end{tabular}
However, note that the closest matching pre-provided answer table is:
\begin{tabular}{|c|c|}
\hline
Yellow: [tex]\( x \)[/tex] & Prokability: [tex]\( P_2(x) \)[/tex] \\
\hline
0 & 0.5 \\
\hline
1 & 0.375 \\
\hline
2 & 0.125 \\
\hline
\end{tabular}
There might be a typo in the calculated probabilities in the problems provided since the exact matching set isn't present.
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