Question

suppose a basketball player makes 24% of shots and that the probability of making each shot is independent. if the basketball player attempts 12 shots, what is the probability of making at least one shot? enter your answer rounded to at least 4 decimals.

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:

$0.9620$