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One professional basketball player typically attempts eight free throws per game. Let [tex][tex]$X$[/tex][/tex] represent the number of free throws made out of eight. The distribution for [tex][tex]$X$[/tex][/tex] is shown in the table.

\begin{tabular}{|c|c|c|c|c|c|c|c|c|c|}
\hline
\begin{tabular}{c}
Number of Free \\
Throws Made
\end{tabular} & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 \\
\hline
Probability & 0.002 & 0.008 & 0.04 & 0.12 & 0.23 & 0.28 & 0.21 & 0.09 & 0.02 \\
\hline
\end{tabular}

What is the probability that the basketball player will make at least six free throws out of the eight attempts?

Sagot :

GinnyAnswer
To determine the probability that the basketball player will make at least six free throws out of eight attempts, we need to consider the sum of the probabilities for making 6, 7, and 8 free throws.

Given the probabilities in the table:

[tex]\[ \begin{array}{|c|c|} \hline \text{Number of Free Throws Made} & \text{Probability} \\ \hline 0 & 0.002 \\ 1 & 0.008 \\ 2 & 0.04 \\ 3 & 0.12 \\ 4 & 0.23 \\ 5 & 0.28 \\ 6 & 0.21 \\ 7 & 0.09 \\ 8 & 0.02 \\ \hline \end{array} \][/tex]

We focus on the probabilities for making 6, 7, and 8 free throws:

- Probability of making 6 free throws: 0.21
- Probability of making 7 free throws: 0.09
- Probability of making 8 free throws: 0.02

To find the combined probability of making at least 6 free throws, we sum these probabilities:

[tex]\[ \text{Probability(at least 6 free throws)} = \text{Probability(6 free throws)} + \text{Probability(7 free throws)} + \text{Probability(8 free throws)} \][/tex]

Substituting in the values, we get:

[tex]\[ \text{Probability(at least 6 free throws)} = 0.21 + 0.09 + 0.02 = 0.32 \][/tex]

Therefore, the probability that the basketball player will make at least six free throws out of eight attempts is [tex]\( \boxed{0.32} \)[/tex].
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