Answer:it doesnt matter what you choose for this one a or b
Step-by-step explanation:
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.
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
In which [tex]C_{n,x}[/tex] is the number of different combinations of x objects from a set of n elements, given by the following formula.
[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]
And p is the probability of X happening.
85% of the clients of your company are highly satisfied
So 15% are dissatisfied, so [tex]p = 0.15[/tex]
If you select a random sample of 30 clients, what is the probability that at least one client is dissatisfied?
This is [tex]P(X \geq 1)[/tex] when [tex]n = 30[/tex]
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
[tex]P(X = 0) + P(X \geq 1) = 1[/tex]
[tex]P(X \geq 1) = 1 - P(X = 0)[/tex]
In which
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 0) = C_{30,0}.(0.15)^{0}.(0.85)^{30} = 0.0076[/tex]
[tex]P(X \geq 1) = 1 - P(X = 0) = 1 - 0.0076 = 0.9924[/tex]
99.24% probability that at least one client is dissatisfied
