Question

suppose a basketball player makes 57% of shots and that the probabilty of making each shot is independent. if the basketball player attempts 5 shots, what is the probability of making at least one shot? enter your answer rounded to at least 4 decimals. question help: message instructor

Explanation:

Step1: Find probability of missing a shot

The probability of making a shot is $p = 0.57$, so the probability of missing a shot is $q=1 - p=1 - 0.57 = 0.43$.

Step2: Find probability of missing all 5 shots

Since the shots are independent, the probability of missing all $n = 5$ shots is given by the binomial - probability formula $P(X = k)=C(n,k)\times p^{k}\times q^{n - k}$. Here, $n = 5$, $k = 0$, $p = 0.57$, and $q = 0.43$. The binomial coefficient $C(5,0)=\frac{5!}{0!(5 - 0)!}=1$. So $P(X = 0)=1\times(0.57)^{0}\times(0.43)^{5}=(0.43)^{5}=0.0147008443$.

Step3: Find 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 $P(X\geq1)=1 - P(X = 0)=1-(0.43)^{5}=1 - 0.0147008443 = 0.9852991557\approx0.9853$.

Answer:

$0.9853$