Respuesta :
Answer:
0.762 = 76.2% probability that this shipment is accepted
Step-by-step explanation:
For each pen, there are only two possible outcomes. Either it is defective, or it is not. The probability of a pen being defective is independent from other pens. 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.
17 randomly selected pens
This means that [tex]n = 17[/tex]
(a) Find the probability that this shipment is accepted if 10% of the total shipment is defective. (Use 3 decimal places.)
This is [tex]P(X \leq 2)[/tex] when [tex]p = 0.1[/tex]. So
[tex]P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2)[/tex]
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 0) = C_{17,0}.(0.1)^{0}.(0.9)^{17} = 0.167[/tex]
[tex]P(X = 1) = C_{17,1}.(0.1)^{1}.(0.9)^{16} = 0.315[/tex]
[tex]P(X = 2) = C_{17,2}.(0.1)^{2}.(0.9)^{15} = 0.280[/tex]
[tex]P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2) = 0.167 + 0.315 + 0.280 = 0.762[/tex]
0.762 = 76.2% probability that this shipment is accepted
