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
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
2016 HSC (Accelerated)
// 2 Unit Maths // Studies of Religion 1 //
2017 HSC
// Biology // Physics // Maths Extension 1 // Maths Extension 2 // English Advanced //
@juantheron, your logic checks out but strange notation.
Maybe the given answer is for three or four tests? Too lazy to do the computation.
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.
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.
There are currently 1 users browsing this thread. (0 members and 1 guests)
Bookmarks