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 binomial parameters

We have a binomial distribution with \( n = 12 \) (number of trials), \( k = 10 \) (number of successes), \( p = 0.6 \) (probability of success), and \( q=1 - p = 0.4 \) (probability of failure). The binomial probability formula is \( P(X = k)=\binom{n}{k}p^{k}q^{n - k} \), where \( \binom{n}{k}=\frac{n!}{k!(n - k)!} \).

Step2: Calculate the combination

First, calculate \( \binom{12}{10}=\frac{12!}{10!(12 - 10)!}=\frac{12!}{10!2!}=\frac{12\times11}{2\times1}=66 \).

Step3: Calculate \( p^{k} \) and \( q^{n - k} \)

\( p^{k}=(0.6)^{10}\approx0.0060466176 \), \( q^{n - k}=(0.4)^{2} = 0.16 \).

Step4: Multiply the three parts

Multiply the combination, \( p^{k} \), and \( q^{n - k} \): \( P(X = 10)=66\times0.0060466176\times0.16 \).
First, \( 66\times0.0060466176\approx0.3990767616 \), then \( 0.3990767616\times0.16\approx0.0638522819 \).

Answer:

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