Question

the owner of a small deli is trying to decide whether to discontinue selling magazines. he suspects that only 8.6% of his customers buy a magazine and he thinks that he might be able to use the display space to sell something more profitable. before making a final decision, he decides that for one day he will keep track of the number of customers that buy a magazine. assuming his suspicion that 8.6% of his customers buy a magazine is correct, what is the probability that exactly 4 out of the first 11 customers buy a magazine?
0.00961903
0.0103850
0.0143026
0.0133644
0.00525630
0.0184515

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 successes, $p$ is the probability of success on a single trial, and $C(n,k)=\frac{n!}{k!(n - k)!}$.

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

Here, $n = 11$ (the number of customers), $k = 4$ (the number of customers who buy a magazine), and $p=0.086$ (the probability that a single customer buys a magazine), and $1 - p = 1-0.086 = 0.914$.

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

$C(11,4)=\frac{11!}{4!(11 - 4)!}=\frac{11!}{4!7!}=\frac{11\times10\times9\times8}{4\times3\times2\times1}=330$.

Step4: Calculate the probability

$P(X = 4)=C(11,4)\times(0.086)^{4}\times(0.914)^{7}$
$P(X = 4)=330\times(0.086)^{4}\times(0.914)^{7}$
$(0.086)^{4}=0.086\times0.086\times0.086\times0.086\approx0.0000547$
$(0.914)^{7}=0.914\times0.914\times0.914\times0.914\times0.914\times0.914\times0.914\approx0.577$
$P(X = 4)=330\times0.0000547\times0.577$
$P(X = 4)\approx0.0103850$.

Answer:

$0.0103850$