question on quality control

[funnybabe]

Hello,

My question is I have this box with 20 items in it. the odds there is no defectives is 60%. the odds of 1 defect is 30% and the odds of 2 defects is 10%. if 2 samples are randomly selected and are good, what is the probability this box has no defects?

the book I have has the formula:
P(A) = (k)(n-k)
(m)(r-m)
---------
(n)
(r)

k is the defective items, n is the # of items (20 in our case), r is sample size (2 in our case) and m is
m defects.

in this example, is k going to be 0 since it's supposed to be a non defective box? is this true for m?
also what do I do with the 60% probability that was given? does that play a role in this question?

thanks!

 

[funnybabe]

hello all,

I did another attempt at this because I don't think that formula would have helped. I calculated the total probability of good items so it would be 1/3 x 20/20 + 1/3 x 19/20 + 1/3 x 18/20 = 57/60= 0.95

the P(no defects) would be (1/3) / (57/60) = 0.3509
this method I feel isn't right because I'm not using the 60%

my other attempt was to do the probability that it's not defects so it's 0.60/0.95 = 0.6316

thanks!