Probability Question (1 Viewer)

[juantheron]

 

[pikachu975]

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

 

[He-Mann]

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

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.

 

[He-Mann]

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.

 

[pikachu975]

Re: Probability

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.

 

[Blast1]

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.