From the above formula
$p=$miss rate$,1-p=$hit rate$,ma=$memory access time$,ps=$page fault time $(or)$ service time
$EMAT=p(ps+ma)+(1-p)\times ma$
$EMAT=p(ps)+p(ma)+ma-p(ma)$
$EMAT=p(ps)+ma$
Put the values $p=\frac{0.001}{100}=0.0001=1\times 10^{-4}=10^{-4},ps=10\times 10^{-3}sec,ma=10^{-6}sec$
we get $EMAT=10^{-4}(10\times10^{-3}sec)+10^{-6}sec$
$EMAT=10^{-4}\times(10^{-2}sec)+10^{-6}sec$
$EMAT=10^{-6}sec+10^{-6}sec$
$EMAT=1\mu sec+1\mu sec$
$EMAT=2\mu sec$
please correct me if I'm wrong$?$