My approach: Given EMAT = 4
4 = (1-p)(m) + p(page fault service + m) ……………. p = page fault rate
Page fault service = (0.6) * 10 + (0.4)* (3) ==> 7.2
4 = (1-p)(1) + p(7.2 + 1)
They have taken 4 = (1-p)(1) + p(7.2)
In page fault also we should consider Memory access time right ..??