Question

From recent polls about customer satisfaction, you know that 85% of the clients of your company are highly satisfied and want to renew their contracts. If you select a random sample of 30 clients, what is the probability that at least one client is dissatisfied?

Answer 1

Answer:it doesnt matter what you choose for this one a or b

Explanation:

Answer 2

99.24% probability that at least one client is dissatisfied

Explanation:

For each client, there are only two possible outcomes. Either they are dissatisfied, or they are not. The probability of a client being dissatisfied is independent from other clients. 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.

`P(X = x) = C_(n,x).p^(x).(1-p)^(n-x)`

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

`C_(n,x) = (n!)/(x!(n-x)!)`

And p is the probability of X happening.

85% of the clients of your company are highly satisfied

So 15% are dissatisfied, so
`p = 0.15`

If you select a random sample of 30 clients, what is the probability that at least one client is dissatisfied?

This is
`P(X \geq 1)` when
`n = 30`

We know that either no clients are dissatisfied, or at least one is. The sum of the probabilities of these events is decimal 1. So

`P(X = 0) + P(X \geq 1) = 1`

`P(X \geq 1) = 1 - P(X = 0)`

In which

`P(X = x) = C_(n,x).p^(x).(1-p)^(n-x)`

`P(X = 0) = C_(30,0).(0.15)^(0).(0.85)^(30) = 0.0076`

`P(X \geq 1) = 1 - P(X = 0) = 1 - 0.0076 = 0.9924`

99.24% probability that at least one client is dissatisfied