Originally Posted by

**masimini25** A certain product is received in lots of 500 units. The quality control acceptance procedure is to choose 20 units at random from the lot and test them. If two or more units in the sample are found defective the entire lot is rejected.

What is the probability that a lot containing 40 defective units will be accepted? Can you see if I am on the right track Please help!

with n defectives by the total number of ways to choose any sample. There are

(40/n) ways to select the n defectrive units( 500/20-n) ways to select 920-n) good units and (500/20) ways to select any sample of 40 units

P(d =n) =(40/n) (500/20-n)/(500/20)

Since we want the probability that the sample will contain no more than two defective units, we simply sum

the probabilities of there being exactly zero, one, or two defectives in the sample, which gives us

P(d< 40) = P(d = 0) + P(d = 1) + P(d = 2) =

p(d<40) =P(d=0) + P(d=1)+ P(d=2)+ (400)(500/20) +(40/1) (500/3) =(40/2) =(500/2) all divided by 500/20