Question

The owner of a small deli is trying to decide whether to discontinue selling magazines. He suspects that only 9.8% 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 9.8% of his customers buy a magazine is correct, what is the probability that exactly 5 out of the first 10 customers buy a magazine? a) 0.00136008 b) 0.000796803 c) 0.002090 d) 0.00142299 e) 0.000882963 f) 0.000404963

Answer 1

Answer:

a) 0.00136008

Step-by-step explanation:

For each customer, there are only two possible outcomes. Either they buy a magazine, or they do not. The probability of a customer buying a magazine is independent of any other customer. Thus, the binomial probability distribution is used to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

9.8% of his customers buy a magazine

This means that $p = 0.098$

What is the probability that exactly 5 out of the first 10 customers buy a magazine?

This is $P(X = 5)$ when $n = 10$. So

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 5) = C_{10,5}.(0.098)^{5}.(0.902)^{5} = 0.00136008$

Thus, the correct answer is given by option A.