The general manager of a pizza restaurant is running a surprise promotion in which every customer on a particular day has a 2% probability of winning free pizza for a % probability of win vear. (a) How many customer should the general manager expect until there is a winner?
(b) If there were 150 customers on a day of the promotion, what is the probability that (i) none of the customers won?) That exactly one customer won? and iii) That at least two customers won?

a) The general manager should expect 50 customers until there is a winner.

b - i) 0.0483 = 4.83% probability that none of the customers won.

b - ii) 0.1478 = 14.78% probability that exactly one customer won.

b - iii) 0.8039 = 80.39% probability that at least two customers won.

Step-by-step explanation:

For each customer, there are only two possible outcomes. Either they win the prize, or they do not. The probability of a customer winning the prize is independent of other customers. So we use the binomial probability distribution 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.

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

And p is the probability of X happening.

2% probability of winning free pizza

This means that

(a) How many customer should the general manager expect until there is a winner?

The expected number of trials until there is n sucesses, with p probability, is given by:

In this question, 1 winner(success). So

The general manager should expect 50 customers until there is a winner.

(b) If there were 150 customers on a day of the promotion

This means that

(i) none of the customers won?

This is . So

0.0483 = 4.83% probability that none of the customers won.

ii) That exactly one customer won?

This is . So

0.1478 = 14.78% probability that exactly one customer won.

iii) That at least two customers won?

This is , in which:

0.8039 = 80.39% probability that at least two customers won.