Question

Assume 10% of the engines manufactured on an assembly line are defective. If engines are randomly selected one at a time and tested. What is the probability that the third nondefective engine will be found
a)on the fifth trial?
b) on or before the fifth trial?

MY answers:
a) P(X=5)=.0437 (Used the negative bin. dist)
b)P(X<=5)=P(X=3)+P(X=4)+P(X=5)=.99117
I think b) is definitely wrong. I used the negative binomial distribution to get all of the probabilities, I don't know what else to do for b.
Probability and Statistics - Reiny, Sunday, September 29, 2013 at 8:14am
If I understand correctly, there are 5 trials and you are expecting 3 S's and 2 F's, (S = success, F = failure)

defective - F -----> prob(F) = .1
non-defective - S --> prob(S) = .9

a) you want an S in the 5th spot
e.g. SFFSS is one of these

number of ways for your specified event (having an S at the end) = C(4,2) x C(2,2) = 6
or 4!/(2!2!) x 1 = 6

prob( a specific 2F,2S, with S at end) = (.9^2)(.1^2)(.1) = .00081

prob of your event = 6(.00081) = .0486

mmmh, close to your answer.
Did you interpret this differently?


I used [4!/(2!2!)](.9^3)(.1^2) for a)

Do I have the right formula for part b)? a probability of .99117 does not seem to make sense.

Answer 1

Katie, you were right, there should have been
3 successes and 2 failures, which would be
6( .9^3)(.1^2)
Don't know why I miscounted.

in my "prob( a specific 2F,2S, with S at end) = (.9^2)(.1^2)(.1) = .00081 "
I had the first part correct, with 2F's and 3S's, but somehow put in .1 instead of .9

b) your formula seems right
could be 3 S's OR 3 S's 1 F OR 3 S's 2 F , in each case S at the end
= .9^3 + 3 (.9^2) (.1) + 6(.9^3)(.1^2)
= .729 + .192 + .04374
= .96474

you probably just made a calculation error