Question

200 computers are built in a factory. They have a 0.5% defective rate.

What is the probability that the first defective item is within the first 10 items tested?

Answer 1

prob(found within first 10 tests)
= prob(found in 1st) + prob(found in 2nd) + .. + prob(found in 10th)
= (.05) + (.95)(.05) + (.95^2)(.05) + ... + (.95^9)(.05)
= .05(1 + .95 + .95^2 + ... + .95^9)
= .05(1)(1 - .95^10)/(1-.95) <----- using sum(n) = a (1 - r^n)/(1-r)
= appr .40126

Answer 2

the probability of any given piece being defective is .5% or .005
... not .05

there is about one defective piece expected in a lot of 200
... 200 * .5% = 1

does it seem reasonable that the probability of finding the defective piece
... in the 1st 5% of the lot would be almost 1/2 ?