Machine probability

1/3 is the right answer. The number of ways to pick the machines is 6 (g=good, b=bad): {(ggbb), (gbbg), (gbgb), (bbgg),(bgbg), (bggb)}

To find the good machines in two tries, you must pick (ggbb) or (bbgg) . So two out of 6 is 1/3.

He knows that 2 machines are incorrect, and he can find out which two if he tests either two faulty ones or two good ones (if he test two and it turns out they are good, he will automatically know that the other two are faulty). The probability that he will test two faulty machines is 1/2 * 1/3 = 1/6 . The same is the probability of testing two good ones, hence the probability that exactly two tests are needed is 1/6 + 1/6 = 1/3 .

Let us say that the two faulty machines are, $F_1$ and $F_2$ and the two working ones are, $W_1$ and $W_2$ .Now,we see that we can identify the faulty ones in these two ways: $1)$ The two faulty ones are the first two to get checked;and $2)$ The two working ones are the first two to get checked.Thus,for $1)$ we have $4$ cases in total(two ways of arranging the faulty ones and two ways of arranging the working ones) and similarly for $2)$ we have four cases.Hence total number of favourable cases is equal to $8$ ,and the total number of ways of arranging the four machines is $4\times3\times2\times1=24$ . $\Longrightarrow P(E)=\dfrac{8}{24}=\dfrac{1}{3}$ .

The result of the first test doesn't matter. If he gets the same result in the second test, he knows which ones work. The probability for this is obviously 1/3 cause after the first test 3 machines are left from which 1 is in the same condition as the machine from the first test.

When there are exactly two tests, that means that the first machine tested was faulty, but the second one works. The probability of finding a faulty machine out of the four machines is 1/2. The probability of finding a working machine on the second try out of the remaining three machines is 2/3. (1/2) * (2/3) = 1/3

First test will identify either valid or invalid machine. Since we can determine 2 valid machines in 2 trials only if we are lucky enough to pick 2 valid or 2 invalid machines in first 2 trials (because there are 4 machines in total and those which are not valid are invalid), it is enough for test to show the same validity of the machines, i.e. the second machine should have the same test result as the first machine. Without loss of generality, we can assume that the first machine is tested valid - which means we're left with 3 machines, 2 invalid and 1 valid, and the question reduces to probability of picking a valid machine between 3 machines, p= $\frac{1}{3}$ .

A probability tree shows that there are only two ways in which he will only need two tries:

• take two times one of the the faulty ( $\frac{2}{4}$ x $\frac{1}{3}$ )

• take two times one of the perfectly fine ( $\frac{2}{4}$ x $\frac{1}{3}$ )

=> Solution is 2x(2/12) = 1/3

That's because if he would have chosen one faulty and than one fine machine (+the other way around) he has left one faulty and one fine and has therefore to do another try with probability 1/2.

In total there are ${4 \choose 2}$ ways to test two out of four machines. We can identify the faulty machines by either checking the two faulty ones or the two working ones. This gives us 2 favourable cases out of the ${4 \choose 2}$ in total.

This means that the probability of identifying the two faulty machines is $\frac {2}{4 \choose 2}$ = $\frac {2}{6}$ = $\frac {1}{3}$ .

As both the faulty machines are found in the first two trials, the answer is (2/4)(1/3) = 1/6.