There are two factories that produce computers. Each computer produced at factor
ID: 3065095 • Letter: T
Question
There are two factories that produce computers. Each computer produced at factory A is defective with probability .05 and each one produced at factory B is defective with probability .02. Suppose you purchase 5 computers that were produced at the same factory. Suppose also that there is a .75 chance the computers were purchased at factory A and a .25 chance the computers were purchased at factory B. If the first computer you check is not defective, what is the probability that the other 4 are also not defective?
Explanation / Answer
Define the event
A : The computer purchased at factory A
B : The computer purchased at factory B
P(A) = 0.75 and P(B) = 0.25
A and B are mutually exclusive events and forms a partition of sample space.
Define another event C
C : The computer purchased is not defective
From the information
P(C/A) = 0.95 and P(C/B) = 0.98
By using Baye's Theorem
P(C) = P(A) * P(C/A) + P(B) * P(C/B)
= 0.75 * 0.95 + 0.25 *0.98
= 0.9575
p = P ( purchased computer is not defective) = 0.9575
n= number of computers purchased = 5
X : number of purchased computer which are not defective.
X ~ Bin ( n=5, p=0.9575)
If first computer selected from 5 is not defective then probability of remaining 4 which is also not defective
= P ( All the computers are not defective )
i.e. P(X=5)
P(X=5) = 5C5 (0.9575) 5 * ( 0.0425)0
= 0.8048