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Sagot :
To determine the probability that the basketball player will make six free throws out of the eight attempts, we refer to the provided probability distribution.
From the distribution table:
[tex]\[ \begin{tabular}{|c|c|c|c|c|c|c|c|c|c|} \hline \text{Number of Free Throws Made} & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 \\ \hline \text{Probability} & 0.002 & 0.008 & 0.004 & 0.12 & 0.23 & 0.20 & 0.11 & 0.00 & 0.02 \\ \hline \end{tabular} \][/tex]
Given the table, we can see that the probability of making exactly six free throws ([tex]\(X = 6\)[/tex]) is [tex]\(0.11\)[/tex].
Thus, the probability that the basketball player will make six free throws out of the eight attempts is:
[tex]\[ \boxed{0.11} \][/tex]
From the distribution table:
[tex]\[ \begin{tabular}{|c|c|c|c|c|c|c|c|c|c|} \hline \text{Number of Free Throws Made} & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 \\ \hline \text{Probability} & 0.002 & 0.008 & 0.004 & 0.12 & 0.23 & 0.20 & 0.11 & 0.00 & 0.02 \\ \hline \end{tabular} \][/tex]
Given the table, we can see that the probability of making exactly six free throws ([tex]\(X = 6\)[/tex]) is [tex]\(0.11\)[/tex].
Thus, the probability that the basketball player will make six free throws out of the eight attempts is:
[tex]\[ \boxed{0.11} \][/tex]
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