Question

The probability a car salesman sells a car to a customer is 0.05

Assuming the salesmen sees 16 customers in a week, what is the probability he sells less than 3 cars?

Write answer using three decimal places

Answer 1

Using the binomial distribution, it is found that there is a 0.957 = 95.7% probability that he sells less than 3 cars.

What is the binomial distribution formula?

The formula is:

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

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

The parameters are:

• x is the number of successes.

• n is the number of trials.

• p is the probability of a success on a single trial.

In this problem:

• The probability a car salesman sells a car to a customer is 0.05, hence p = 0.05.

• The salesmen sees 16 customers in a week, hence n = 16.

The probability he sells less than 3 cars is:

P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2).

In which:

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

$P(X = 0) = C_{16,0}.(0.05)^{0}.(0.95)^{16} = 0.4401$

$P(X = 1) = C_{16,1}.(0.05)^{1}.(0.95)^{15} = 0.3706$

$P(X = 2) = C_{16,2}.(0.05)^{2}.(0.95)^{14} = 0.1463$

Then:

P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2) = 0.4401 + 0.3706 + 0.1463 = 0.957.

0.957 = 95.7% probability that he sells less than 3 cars.

More can be learned about the binomial distribution at https://brainly.com/question/24863377