Originally Posted by
jake.davis I was hoping someone could check my work as something seems off to me. Any help/suggestions would be greatly appreciated.
(a)From a lot of 12 flares, 4 are selected at random. If the lot contains 4 defective flares, what is the probability that all 4 will not work?
P(4 defective)= C(4,4) / C(12,4)
= 1/149
= 0.002 OR 0.2%
(b)What is the probability that at most 2 will not work?
P(at most 2 will not work) = P(0, 1 and 2 will not work)
P (0 will not work) = (C(4,0) * C(8,4)) / C(12,4) = 0.1414...
P (1 will not work) = (C(4,1) * C(8,3)) / C(12,4) = 0.1131
P (2 will not work) = (C(4,2) * C(8,2)) / C(12,4) = 0.3394
Total = 0.5924 or 59.24%
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