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A piece of equipment contains six identical items and it is known that three of them are defective.
The items are tested one after the other until the three defective items are found.
(a) What is the probability that the testing process is stopped on the (i) third test (ii) fourth test.
(b) If the process is stopped on the fourth test, what is the probability that the first item is not defective?
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(a,i) If the testing process is stopped on the third item, it means that the first item was defective, second item
was defective and third item was defective. The probability of this event is
P = = . ANSWER
(a,ii) If the testing process is stopped on the fourth item, it may happen in these three cases, shown by sequences
of letters
nDDD, DnDD, DDnD
where "n" symbolizes non-defective item, "D" sympolises defective item. So,
P(process is stopped on the fourth item) = P(nDDD) + P(DnDD) + P(DDnD).
Next, P(nDDD) = = = = ;
P(DnDD) = = ;
P(DDnD) = = =
Thus final probability P(process is stopped on the fourth item) = + + = . ANSWER
(b) If the testing process is stopped on the fourth item, it means that one of the three cases nDDD, DnDD, or DDnD happens,
which has the probability of , as it is shown in part (a,ii) above.
If at the same time, the first item is not defective, it means that the event nDDD did not happens, but one of DnDD or DDnD happens,
which has the probability + = .
THEREFORE, the conditional probability under given question is P = = . ANSWER
Solved.
Really good probability problem.