@srestha I think this approach is not correct.
We have considered ordering of machines M1,M2,M3 and M4. So even in denominator we will have to consider the orderings.
No. of ways of arranging M1,M2,M3,M4 = 4! = 24
No. of ways in which the faulty machines(M3 and M4) appear as the first 2 machines
or non-faulty machines(M1 and M2) appear as the first 2 machines
= 2 * {2!} + 2 * {2!} .........(cases: M3,M4,_,_ M4,M3,_,_ or M1,M2,_,_ M2,M1,_,_)
= 4 + 4 = 8
Hence the required probability = 8/24 = 1/3
Let me know if anything is wrong here. :)