LeBron scores $70\%$ of his free throws. If he shoots $30$ free throws, which of the following expressions gives the probability that he makes exactly $21$ of them?

• A. $\frac{30!}{21!9!}{0.3}^{21}{0.7}^9$
• B. $\frac{30!}{21!9!}{0.7}^{30}$
• C. $\frac{30!}{9!21!}{0.7}^9{0.3}^{21}$
• D. $\frac{30!}{21!9!}{0.7}^{21}{0.3}^9$

Answer: (D)

Verified step by step guidance

1. Identify the type of probability problem: This is a binomial probability problem because LeBron has a fixed number of independent free throw attempts (n = 30), each with two possible outcomes (make or miss), and a constant probability of success (p = 0.7).
2. Define the parameters for the binomial distribution: number of trials \(n = 30\), number of successes \(k = 21\), probability of success on each trial \(p = 0.7\), and probability of failure \(q = 1 - p = 0.3\).
3. Recall the binomial probability formula: the probability of exactly \(k\) successes in \(n\) trials is given by \[ P(X = k) = \binom{n}{k} p^{k} q^{n-k} \] where \(\binom{n}{k} = \frac{n!}{k!(n-k)!}\) is the binomial coefficient.
4. Substitute the values into the formula: \[ P(X = 21) = \frac{30!}{21! \times 9!} \times (0.7)^{21} \times (0.3)^{9} \] This expression calculates the probability that LeBron makes exactly 21 free throws out of 30 attempts.
5. Interpret the components: the binomial coefficient counts the number of ways to choose which 21 shots are made, \((0.7)^{21}\) is the probability of making those 21 shots, and \((0.3)^{9}\) is the probability of missing the remaining 9 shots.