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