Using the binomial distribution, we have that:
A. The salesperson should expect 1.54 customers until he finds a customer that makes a purchase.
B. There is a 0.0279 = 2.79% probability that a salesperson helps 3 customers until he finds the first person to make a purchase.
The formula is:
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]
The parameters are:
The expected number of trials until q successes are found is:
[tex]E(X) = \frac{q}{p}[/tex]
In this problem, the probability of a success is of p = 0.65, hence:
[tex]E(X) = \frac{1}{0.65} = 1.54[/tex]
The salesperson should expect 1.54 customers until he finds a customer that makes a purchase.
The probability is p = 0.65 multiplied by P(X = 0) when n = 3, hence:
0.65 x (0.35)³ = 0.0279.
There is a 0.0279 = 2.79% probability that a salesperson helps 3 customers until he finds the first person to make a purchase.
More can be learned about the binomial distribution at https://brainly.com/question/24863377
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