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?
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.
1 answer
1 answer