Question

kuta software - infinite precalculus binomial probability find the probability of each event. 1) a basketball player has a 60% chance of making each free throw. what is the probability that the player makes exactly three out of six free throws?

Explanation:

Step1: Identify binomial - probability formula

The binomial - probability formula is $P(X = k)=C(n,k)\times p^{k}\times(1 - p)^{n - k}$, where $n$ is the number of trials, $k$ is the number of successful trials, $p$ is the probability of success on a single trial, and $C(n,k)=\frac{n!}{k!(n - k)!}$.

Step2: Determine the values of $n$, $k$, and $p$

Here, $n = 6$ (the number of free - throws), $k = 3$ (the number of made free - throws), and $p=0.6$ (the probability of making a single free - throw). Then $1 - p = 1-0.6 = 0.4$.

Step3: Calculate the combination $C(n,k)$

$C(6,3)=\frac{6!}{3!(6 - 3)!}=\frac{6!}{3!3!}=\frac{6\times5\times4\times3!}{3!\times3\times2\times1}=\frac{6\times5\times4}{3\times2\times1}=20$.

Step4: Calculate the probability $P(X = 3)$

$P(X = 3)=C(6,3)\times p^{3}\times(1 - p)^{6 - 3}=20\times(0.6)^{3}\times(0.4)^{3}$.
$(0.6)^{3}=0.6\times0.6\times0.6 = 0.216$, $(0.4)^{3}=0.4\times0.4\times0.4 = 0.064$.
$P(X = 3)=20\times0.216\times0.064 = 20\times0.013824=0.27648$.

Answer:

$0.27648$