Question

A restaurant has an electronic system that randomly selects customers when they pay for their meal to

Answer 1

Using the binomial distribution, it is found that there is a 0.81 = 81% probability that NEITHER customer is selected to receive a coupon.

For each customer, there are only two possible outcomes, either they receive the coupon, or they do not. The probability of a customer receiving the coupon is independent of any other customer, which means that the binomial distribution is used to solve this question.

Binomial probability distribution

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

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

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:

• For each customer, 10% probability of receiving a coupon, thus $p = 0.1$.
• 2 customers are selected, thus $n = 2$

The probability that neither receives a coupon is P(X = 0), thus:

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

$P(X = 0) = C_{2,0}.(0.1)^{0}.(0.9)^{2} = 0.81$

0.81 = 81% probability that NEITHER customer is selected to receive a coupon.

A similar problem is given at brainly.com/question/25326823