Asked by Triple_A_Math
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?
What is the probability that the first defective item is within the first 10 items tested?
Answers
Answered by
mathhelper
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
= 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
Answered by
R_scott
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 ?
... 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 ?