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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 ..??

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