Question 938768: It is thought that, on average, 3% of light bulbs produced by a certain company last for less than 250 hours. This will be referred to as being defective.
In an inspection scheme, a sample of 25 light bulbs is selected at random from a large batch, they are tested for 250 hours and the number of defective bulbs is noted.
If the number is more than two, the whole batch is rejected; if it is less than two, the whole batch is accepted.
If there are exactly two defective bulbs in this batch, a further sample of size ten is taken.
The whole batch is rejected if there are any defective bulbs in this sample; otherwise the batch is accepted.
Find
(i)the probability that the batch is accepted after taking the first sample
(ii)the probability that the batch is accepted after taking the second sample
(iii)the probability that the batch is rejected.
Answer by ewatrrr(24785)   (Show Source):
You can put this solution on YOUR website! p(def) = .03, n = 25
i) P(accepted) = P(x < 2) = binomcdf(25, .03,1)
...
ii) P ( batch is accepted after taking the second sample)
P = [binompdf(25, .03,2)][10^(.97)]
...
iii) P(probability that the batch is rejected)
P = [1-binomcdf(25, .03, 2] + [1 - [binompdf(25, .03,2)][10^(.97)]
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