using the binomial distribution with n = 9 and p = 0.2, find the following probability. round answer to four…

Explanation:

Step1: Recall 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 successes, $p$ is the probability of success on a single - trial, and $C(n,k)=\frac{n!}{k!(n - k)!}$. Here, $n = 9$, $k = 5$, and $p=0.2$, so $1 - p = 0.8$.

Step2: Calculate the combination $C(n,k)$

$C(9,5)=\frac{9!}{5!(9 - 5)!}=\frac{9!}{5!4!}=\frac{9\times8\times7\times6\times5!}{5!\times4\times3\times2\times1}=\frac{9\times8\times7\times6}{4\times3\times2\times1}=126$.

Step3: Calculate the binomial probability

$P(X = 5)=C(9,5)\times(0.2)^{5}\times(0.8)^{9 - 5}$ $=126\times(0.2)^{5}\times(0.8)^{4}$ $=126\times0.00032\times0.4096$ $=126\times0.000131072$ $=0.016515072\approx0.0165$.

Answer:

$0.0165$