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A restaurant has an electronic system that randomly selects customers when they pay for their meal to
receive a coupon for their next visit. Each customer has a 10% probability of being selected to receive a
coupon, and one customer being selected or not doesn't affect whether or not another customer will be
selected.
Suppose that 2 customers pay for their meals, one after the other.
What is the probability that NEITHER customer is selected to receive a coupon?
Round your answer to two decimal places.
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Answer :

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

[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:

  • 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 [tex]p = 0.1[/tex].
  • 2 customers are selected, thus [tex]n = 2[/tex]

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

[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]

[tex]P(X = 0) = C_{2,0}.(0.1)^{0}.(0.9)^{2} = 0.81[/tex]

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

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