suppose a basketball player makes 24% of shots and that the probability of making each shot is independent…

Explanation:

Step1: Find the probability of missing a shot

The probability of making a shot is $p = 0.24$. So the probability of missing a shot is $q=1 - p=1 - 0.24 = 0.76$.

Step2: Find the probability of missing all 12 shots

Since the shots are independent, the probability of missing all $n = 12$ shots is given by the binomial probability formula $P(X = k)=C(n,k)\times p^{k}\times q^{n - k}$, where $n = 12$, $k = 0$, $p = 0.24$ and $q = 0.76$. When $k = 0$, $C(12,0)=1$, and $P(X = 0)=1\times(0.24)^{0}\times(0.76)^{12}$. $P(X = 0)=(0.76)^{12}\approx0.03797$.

Step3: Find the probability of making at least one shot

The probability of making at least one shot is the complement of the probability of missing all shots. Let $P(X\geq1)$ be the probability of making at least one shot. Then $P(X\geq1)=1 - P(X = 0)$. $P(X\geq1)=1-(0.76)^{12}\approx1 - 0.03797 = 0.96203\approx0.9620$.

Answer:

$0.9620$