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Sagot :
To determine the correct probability distribution for the random variable representing the number of times yellow (Y) appears when Joaquin spins the spinner twice, let's go through the steps systematically.
1. List the Sample Space:
The sample space [tex]\( S \)[/tex] for two spins contains 16 possible outcomes:
[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]
2. Count the Occurrences of Yellow (Y):
We need to count the outcomes based on the number of times yellow appears.
- Outcomes with 0 yellow (Y):
[tex]\[ \{R B, R G, R R, B R, B G, B B, G R, G B, G G\} \][/tex]
Number of outcomes = 9
- Outcomes with 1 yellow (Y):
[tex]\[ \{R Y, B Y, G Y, Y R, Y B, Y G\} \][/tex]
Number of outcomes = 6
- Outcomes with 2 yellow (Y):
[tex]\[ \{Y Y\} \][/tex]
Number of outcomes = 1
3. Calculate the Probabilities:
We calculate the probabilities of these counts by dividing each by the total number of outcomes (which is 16).
- Probability of 0 yellow:
[tex]\[ P(X = 0) = \frac{\text{Number of outcomes with 0 yellow}}{\text{Total number of outcomes}} = \frac{9}{16} = 0.5625 \][/tex]
- Probability of 1 yellow:
[tex]\[ P(X = 1) = \frac{\text{Number of outcomes with 1 yellow}}{\text{Total number of outcomes}} = \frac{6}{16} = 0.375 \][/tex]
- Probability of 2 yellows:
[tex]\[ P(X = 2) = \frac{\text{Number of outcomes with 2 yellows}}{\text{Total number of outcomes}} = \frac{1}{16} = 0.0625 \][/tex]
4. Compare with Given Options:
We have calculated the probabilities:
[tex]\[ (0, 0.5625), (1, 0.375), (2, 0.0625) \][/tex]
Comparing this with the given tables:
[tex]\[ \begin{tabular}{|c|c|} \hline Yellow: $X$ & Probability: $P(X)$ \\ \hline 0 & 0.5625 \\ \hline 1 & 0.375 \\ \hline 2 & 0.0625 \\ \hline \end{tabular} \][/tex]
This matches our calculations exactly. Therefore, the correct probability distribution is the first table:
[tex]\[ \begin{tabular}{|c|c|} \hline Yellow: $X$ & Probability: $P(X)$ \\ \hline 0 & 0.5625 \\ \hline 1 & 0.375 \\ \hline 2 & 0.0625 \\ \hline \end{tabular} \][/tex]
1. List the Sample Space:
The sample space [tex]\( S \)[/tex] for two spins contains 16 possible outcomes:
[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]
2. Count the Occurrences of Yellow (Y):
We need to count the outcomes based on the number of times yellow appears.
- Outcomes with 0 yellow (Y):
[tex]\[ \{R B, R G, R R, B R, B G, B B, G R, G B, G G\} \][/tex]
Number of outcomes = 9
- Outcomes with 1 yellow (Y):
[tex]\[ \{R Y, B Y, G Y, Y R, Y B, Y G\} \][/tex]
Number of outcomes = 6
- Outcomes with 2 yellow (Y):
[tex]\[ \{Y Y\} \][/tex]
Number of outcomes = 1
3. Calculate the Probabilities:
We calculate the probabilities of these counts by dividing each by the total number of outcomes (which is 16).
- Probability of 0 yellow:
[tex]\[ P(X = 0) = \frac{\text{Number of outcomes with 0 yellow}}{\text{Total number of outcomes}} = \frac{9}{16} = 0.5625 \][/tex]
- Probability of 1 yellow:
[tex]\[ P(X = 1) = \frac{\text{Number of outcomes with 1 yellow}}{\text{Total number of outcomes}} = \frac{6}{16} = 0.375 \][/tex]
- Probability of 2 yellows:
[tex]\[ P(X = 2) = \frac{\text{Number of outcomes with 2 yellows}}{\text{Total number of outcomes}} = \frac{1}{16} = 0.0625 \][/tex]
4. Compare with Given Options:
We have calculated the probabilities:
[tex]\[ (0, 0.5625), (1, 0.375), (2, 0.0625) \][/tex]
Comparing this with the given tables:
[tex]\[ \begin{tabular}{|c|c|} \hline Yellow: $X$ & Probability: $P(X)$ \\ \hline 0 & 0.5625 \\ \hline 1 & 0.375 \\ \hline 2 & 0.0625 \\ \hline \end{tabular} \][/tex]
This matches our calculations exactly. Therefore, the correct probability distribution is the first table:
[tex]\[ \begin{tabular}{|c|c|} \hline Yellow: $X$ & Probability: $P(X)$ \\ \hline 0 & 0.5625 \\ \hline 1 & 0.375 \\ \hline 2 & 0.0625 \\ \hline \end{tabular} \][/tex]
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