Question

A computer chip manufacturer finds that, historically, for ever 100 chips produced, 85 meet specifications, 10 need reworking, and 5 need to be discarded. Ten chips are chosen for inspection. A) What is the probability that all 10 meet specs? B) What is the probability that 2 or more need to be discarded? C) What is the probability that 8 meet specs,1 needs reworking, and 1 will be discarded?

Answer 1

Following is probabilities of disks based on historical data

Probability of chips meet specification , P(S) = 85/100 = 0.85

Probability of chips need rework, P(R) = 10/100 = 0.10

Probability of chips need to be discarded P(D) = 5/100 = 0.05

(A)

Sample size n = 10

From binomial distribution we know that probability of x success from n trials with p probability of getting success

Hence probability of finding 10 disks meet specs from sample of 10 disks = 0.1969

(B) Probability of 2 or more disks needs to be discarded = 1 - ( P( 0 discarded disk) + P( 1 discarded disk) )

Sample size n = 10

Lets assume discarded disk is success. So, p(D) = 0.05

probability of ( 0 discarted disk) :

n = 10 , x = 0 , p = 0.05

probability of ( 1 discarted disk) :  

n = 10 , x = 1 , p = 0.05

So

Probability of 2 or more disks needs to be discarded = 1 - (0.5987 + 0.3151) = 0.0862 = 8.62%

(C) Probability of 8 meet specs,1 needs reworking, and 1 will be discarded , from sample of 10 disks.

From multinomial distribution we know that

x = 3 events ( 8 disks meet specificatoin, 1 needs reworking , 1 will be discarded)

n = 10

So , Probability of 8 meet specs,1 needs reworking, and 1 will be discarded , from sample of 10 disks. would be 0.1226