probability

Question

There are four machines and it is known that exactly two of them are faulty .They are tested one by one in a random order till both the faulty machines are identified ,then the probability that only two tests are needed ?

Answer 1

We will require only two test only if the first two test gives both faulty or both non-faulty machines otherwise 3 test will be required.

So, probability that in first two test we get faulty is machine is, (2/4) × (1/3) also the probability of getting two non-faulty machine is same.

So, final answer is 2 * (2/4) * (1/3) = 1/3

Answer 2

Total machines: 4

Total faulty machines: 2

Total Non faulty machines : 2

Only two test are needed: Both faulty machines come up in the first two checks OR Both non faulty machines come up in the first two checks

$Prob(\text{Only two checks are needed}) =  \frac{2}{4}*\frac{1}{3}+\frac{2}{4}*\frac{1}{3}$

$ = \frac{1}{6}+\frac{1}{6}$

$ = \frac{1}{3}$

Comments

how can u make sure that both the faulty machine will come in 2 checksor both non faulty. It may be possible that in frist check u r getting faulty one and in second case u are getting good one ?

how to remove this ambiguity please explian in detail

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@Shashank we need to identify faulty in only 2 trails. Lets say we get one faulty and other non-faulty in second trail in this case we need 3 trials to identify 2 faulty machines. So in order to complete our task in 2 trails if we get two faulty machine then our task is completed and if we get two non-faulty than it implies other 2 are faulty bcoz it is known 2 are faulty. And in other case only we need 3 trails.