Random variables of the discrete type?

A lot of 200 semiconductors is inpected by selecting 5 at random and without replacement. If at least one of the five is defective, the lot is rejected. Find the probability of rejecting the lot if in the 200,





a) zero are defective


b)10 are defective


c) 20 are defective





Please include an explantation. Thanks...

Random variables of the discrete type?
a) zero are defective so no 5 will have a defective one, so the probability of rejecting the lot is 0





Note that nCk = n!/k!(n-k)! which is the number of ways of choosing k objects from n.





For b and c it is easier to figure out the probability that we do not reject the lot and subtract that from 1. The probability we do not reject the lot is the probability that all 5 in the sample are not defective. For (b), there are 190C5 samples of 5 which have no defective semiconductors so the probability that we do not reject lot (b) is 190C5/200C5 (since 200C5 is the total number of samples of size 5. So the chance we do reject lot (b) is 1 - 190C5/200C5





Similarly the chance we reject lot (c) is 1 - 180C5/200C5
Reply:Because the lot is large, the binomial can approximate the hypergeometric distribution.


General equation:


P(x) = nCx[p^x(1-p^(n-x))


a) Probability of rejecting the lot is 0.


b) p ≡ fraction defective in the lot = 0.05


Probability of rejecting the lot is:


1 - P(x= 0) = 1 - 5C0[0.05^0(1 - 0.05)^(n-0) = 1 - 0.95^5 = 0.2262


c) p = 0.10; 1 - 0.90^5 = 0.4095
Reply:Let X be the number of defective units found. X has the binomial distribution with n = 5 trials and success probability of p. to make that true you should be sampling with replacement, but because of the size of the sample and the lower probabilities of success you can approximate by using the binomial distribution.





In general, if X has the binomial distribution with n trials and a success probability of p then


P[X = x] = n!/(x!(n-x)!) * p^x * (1-p)^(n-x)


for values of x = 0, 1, 2, ..., n





for part (a) p = 0, (b) p = 0.05, (c) p = 0.1





The lot is defective is P(X ≥ 1) = 1 - P(X=0)





a) P(rejected lot) = 1 - P(X= 0) = 0, prob the lot is reject is 1 -1 = 0





b) P(rejected lot) = 1 - P(X = 0) = 1 - .9^5 =1 - 0.59049 =0.40951





c) P(rejected lot) = 1- P(X = 0) = 0.67232