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FSMQ Binomial Distribution

Hi I'm doing FSMQ at the moment and I am stuck on one question which includes Binomial Distribution:

?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 this number is more than two, the whole batch is rejected; if it is less than two, the whole bacth 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.?

I understand the questions which are shorter but this is a bit too far for me.

Thanks
Reply 1
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Asklepios
19
I would start by drawing a probability tree to show all the possible outcomes. Then use Binomial distribution to work out the probability of each branch
Probability of choosing a good light bulb (P) is 0.97, whereas the probability of choosing a defective light bulb (Q) is 0.03.

i) The 1st sample can only be accepted if there are either 1 or 0 defective light blubs. Probability of 1st sample accepted is (25C24 x P^24 x Q^1) + (25C25 x P^25 x Q^0) = 0.83

ii) The 2nd sample can only be accpeted if there are exactly 2 defective light bulbs from the 1st sample and 0 defective light bulbs from the 2nd sample. Probability of 2nd sample accepted is (25C23 x P^23 x Q^2) x (10C10 x P^10 x Q^0) = 0.099

iii) Probability that the batch is rejected is 1 minus the answers to i and ii because they cover all the possibilities for the batch to be accepted = 0.071
(edited 11 years ago)

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