Homework Answers
a)
expected number of crash in 2 hour =2*6/24=0.5
P(computer will not crash in 2 hour) =P(X=0) =e-0.5*0.50/0! =0.6065
b)
expected number of crash in 5 hour =5*6/24=1.25
P(crash exactly twice )=P(X=2)=e-1.2522/2! =0.2238
c)
P(2 crashes in 5 hours )=P(X>=2) =1-P(X<=1)
=1-P(X=0)-P(X=1)
=1-e-1.2520/0!-e-1.2521/1! =1-0.6446 =0.3554
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