Assume that 12 jurors are randomly selected from a population in which $85 \%$ of the people are MexicanAmericans. Refer to the probability distribution table below and find the indicated probabilities. \begin{tabular}{|r|r|} \hline$x$ & \multicolumn{1}{|c|}{$P(x)$} \\ \hline 0 & $0+$ \\ \hline 1 & $0+$ \\ \hline 2 & $0+$ \\ \hline 3 & $0+$ \\ \hline 4 & 0.0001 \\ \hline 5 & 0.0006 \\ \hline 6 & 0.004 \\ \hline 7 & 0.0193 \\ \hline 8 & 0.0683 \\ \hline 9 & 0.172 \\ \hline 10 & 0.2924 \\ \hline 11 & 0.3012 \\ \hline 12 & 0.1422 \\ \hline \end{tabular} Find the probability of exactly 5 Mexican-Americans among 12 jurors. Round your answer to four decimal places. \[ P(x=5)= \] Find the probability of 5 or fewer Mexican-Americans among 12 jurors. Round your answer to four decimal places. \[ P(x \leq 5)= \]
Solution
Step 1 ：The question asks for two probabilities. The first is the probability of exactly 5 Mexican-Americans among 12 jurors. This can be directly read from the probability distribution table provided. The second is the probability of 5 or fewer Mexican-Americans among 12 jurors. This can be calculated by summing up the probabilities of having 0, 1, 2, 3, 4, and 5 Mexican-Americans among the jurors.
Step 2 ：The probability of exactly 5 Mexican-Americans among 12 jurors is \(0.0006\).
Step 3 ：The probability of 5 or fewer Mexican-Americans among 12 jurors is calculated by summing up the probabilities of having 0, 1, 2, 3, 4, and 5 Mexican-Americans among the jurors, which is \(0+0+0+0+0.0001+0.0006=0.0007\).
Step 4 ：Final Answer: The probability of exactly 5 Mexican-Americans among 12 jurors is \(\boxed{0.0006}\). The probability of 5 or fewer Mexican-Americans among 12 jurors is \(\boxed{0.0007}\).
