a) probability that on a given day no computers arrive for repair =e-3*30/0! =0.0498
b) probability that on a given day at least 3 computers arrive for repair =P(X>=3) =1-P(X<=2)
=1-(P(X=0)+P(X=1)+P(X=2)) =1-(e-3*30/0!+e-3*31/1!+e-3*32/2!) =0.8647
c)
distribution of the time between arrivals of computers to this facility is exponential with mean time between arrival =1/3 days
or (24/3) =8 hours
d)
best estimate for next computer will arrive for repair =* hour after noon or at 8.00 pm
e)
probability that the next one will not arrive for at least another 5 hours =P(T>5)=e-5/8 =0.5353
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