Answer:
(a) 0.5282
(b) 0.4718
Step-by-step explanation:
It is provided that the shipment is accepted if not more than 1 of the 5 keyboards is defective. That is for the shipment to be acceptable there should be at most 1 defective .
It is also provided that the box selected has 6 defective keyboards.
So, the probability of selecting a defective keyboard is, p = 6/20.
Let X = number of defective keyboards.
The random variable X follows a binomial distribution with parameters n = 6 and p = 6/20.
(a)
Compute the probability that this shipment will be accepted as follows:
[tex]P(X\leq 1)=P(X=0)+P(X=1)[/tex]
[tex]=[{5\choose 0}(\frac{6}{20})^{0}(1-\frac{6}{20})^{5}]+[{5\choose 1}(\frac{6}{20})^{1}(1-\frac{6}{20})^{4}]\\\\=0.16807+0.36015\\\\=0.52822\\\\\approx 0.5282[/tex]
Thus, the probability that this shipment will be accepted is 0.5282.
(b)
Compute the probability that this shipment will not be accepted as follows:
P (shipment not accepted) = 1 - P(shipment accepted)
= 1 - 0.5282
= 0.4718
Thus, the probability that this shipment will not be accepted is 0.4718.