Out of 4 , it's given that exactly two are faulty therefore two are non faulty
now, ways to permute them will be 4!/(2!*2!) = 6
so total sample space contain 6 items now probab. that in two test we detect the faulty machines = (probab. of detecting two faulty m/c in 1st two tests) OR (probab. of detecting two unfaulty m/c in 1st two tests)
= (1/6) + (1/6)
= (2/6)
=(1/3)