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:

99.24% probability that at least one client is dissatisfied

Step-by-step 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.

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

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

Answer 2

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

Step-by-step explanation: