1. ## Probability Question

$There are four machines and it is known that exactly two of them are$

$faulty. They are tested, one by one, in a random order till both the faulty$

$machines are identified. Then the probability that only two tests are$

$need is$

$What i have Try:$

$Let A be the event in whic first machine is fault.$

$And B be the machine in which second machine is fault.$

$So \displaystyle P(A \cap B) = P(A)\cdot P\left(\frac{B}{A}\right) = \frac{2}{4}\cdot \frac{1}{3} = \frac{1}{6}$

$But anser given as \displaystyle \frac{1}{3}$

$What,s wrong with my solution, plz explain me , Thanks$

2. ## Re: Probability

The desired event can be achieved through getting 1 machine in each test, or 0 machines first test and 2 machines in second test.

P(only 2 tests) = P(1 machine then 1 machine) + P(0 machines then 2 machines)
= 1/4 x 1/3 + 2/4 x 2/4
= 1/3

3. ## Re: Probability

Originally Posted by pikachu975
The desired event can be achieved through getting 1 machine in each test, or 0 machines first test and 2 machines in second test.

P(only 2 tests) = P(1 machine then 1 machine) + P(0 machines then 2 machines)
= 1/4 x 1/3 + 2/4 x 2/4
= 1/3
P(1 machine then 1 machine) = 2/4 x 1/3, because there is a 2/4 chance of identifying the faulty machine with 1 test.

The event "0 machines then 2 machines" is 3 tests.

4. ## Re: Probability

@juantheron, your logic checks out but strange notation.

Maybe the given answer is for three or four tests? Too lazy to do the computation.

5. ## Re: Probability

Originally Posted by He-Mann
P(1 machine then 1 machine) = 2/4 x 1/3, because there is a 2/4 chance of identifying the faulty machine with 1 test.

The event "0 machines then 2 machines" is 3 tests.
Nevermind misinterpreted the question, 1/6 seems right after re-reading.

6. ## Re: Probability

If the question was supposed to be for three machines (instead of four), then 1/3 would be the answer. Answer would then be (2/3)*(1/2) = 1/3, using similar logic to juantheron.

7. ## Re: Probability

The answer should be 1/3 for the version with 4 machines. P (both first two tests reveal faulty machines) = 1/6 as established by earlier posts. Note that we can also identify which two machines are faulty by also having the two tests reveal non faulty machines, meaning that the untested machines must both be faulty. The probability of this occurring is, by symmetry, 1/6. So the total probability that only 2 tests are needed is 1/6 + 1/6 = 1/3.

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