Answer :
Answer:
14.78% probability that a sample of 10 covers will contain exactly 2 defectives
Step-by-step explanation:
For each cover, there are only two possible outcomes. Either it is defective, or it not. The probability of a cover being defective is independent from other covers. 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.
If the fraction of defective covers produced on the USB Mouse Factory production line is known to be 8%, what is the probability that a sample of 10 covers will contain exactly 2 defectives?
This is [tex]P(X = 2)[/tex] when [tex]p = 0.08, n = 10[/tex]. So
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 2) = C_{10,2}.(0.08)^{2}.(0.92)^{8} = 0.1478[/tex]
14.78% probability that a sample of 10 covers will contain exactly 2 defectives