Thread: Discrete Probability Distributions

Ok I think this is a hypergeometric probabiltiy question but I can't break down the numbers.

Six Chevy vans are being tested by consumer Reports for quality control. In the past, 40% of the vans tested have had some safety problems on delivery such s headlight out or a seatbelt not working, If a sample of 6 vans is taken:
a) what is the probability of four of the vans having a safety problem?
b) what is the expected number of vans in the sample that will have a safety prolem?

I can them if I knew where to break it a part.
The formula I have is P(x)=aCx^x-aCr-x
But which numbers go where?
a=?
n=?
r=?
x=?

Originally Posted by risteek

Ok I think this is a hypergeometric probabiltiy question but I can't break down the numbers.

Six Chevy vans are being tested by consumer Reports for quality control. In the past, 40% of the vans tested have had some safety problems on delivery such s headlight out or a seatbelt not working, If a sample of 6 vans is taken:
a) what is the probability of four of the vans having a safety problem?
b) what is the expected number of vans in the sample that will have a safety prolem?

I can them if I knew where to break it a part.
The formula I have is P(x)=aCx^x-aCr-x
But which numbers go where?
a=?
n=?
r=?
x=?

You're probably meant to assume that there's a large number of vans you've chosen from. In which case, if X is the random variable 'number of bad vans' then X ~ Binomial(n = 6, p = 0.4).

Ok thanks. C: