Question

1. bob makes 60% of his free-throw attempts. if he shoots 12 free throws, what is the probability that he makes exactly 10?

Explanation:

Step1: Identify the distribution

This is a binomial probability problem. The binomial probability formula is $P(X = k) = \binom{n}{k} p^k (1 - p)^{n - k}$, where $n$ is the number of trials, $k$ is the number of successes, $p$ is the probability of success on a single trial. Here, $n = 12$, $k = 10$, $p = 0.6$.

Step2: Calculate the combination

First, calculate $\binom{12}{10}$. The combination formula is $\binom{n}{k} = \frac{n!}{k!(n - k)!}$. So $\binom{12}{10} = \frac{12!}{10!(12 - 10)!} = \frac{12!}{10!2!} = \frac{12\times11}{2\times1} = 66$.

Step3: Calculate $p^k$ and $(1 - p)^{n - k}$

$p^k = 0.6^{10}$, and $(1 - p)^{n - k} = (0.4)^{2}$.

Step4: Multiply the three parts

Now, multiply these together: $P(X = 10) = 66\times0.6^{10}\times0.4^{2}$. Calculate $0.6^{10} \approx 0.0060466176$, $0.4^{2} = 0.16$. Then $66\times0.0060466176\times0.16 \approx 66\times0.0009674588 \approx 0.06385228$.

Answer:

The probability is approximately $0.0639$ (or $6.39\%$).