There are 4 machines $M_{1},M_{2},M_{3},M_{4}$
Here say $M_{3},M_{4}$ are faulty
So, we can select it either by $M_{3},M_{4}$ or $M_{4},M_{3}$ =$2$ ways
Now among 4 machines we can select 2 in $\binom{4}{1}\times \binom{3}{1}$ ways=$12$ ways
So, total probability that only $2$ test cases required to get both mchines are faulty is $\frac{2}{12}=\frac{1}{6}$