Question 1118652
Let X denote number of cars hired out per day
Poisson distribution mean = m = 1.5
P(X=x)=(((e^−m)(m^x))/(x!))= (((e^−1.5)(1.5^x))/(x!)) 
1) P(neither car is used):
P(X=0)=(e^−1.5)(1.5^0)/0.2231
2) P(Some demand is refused ) = P(Demand is more than 2 cars per days)
P(x>2)
=1−P(x≤2)
=1−[P(x=0)+P(x=1)+P(x=2)]
=1−[((e^1.5)(1.5^0)/0!)+ ((e^1.5)(1.5^1)/1!)+((e^1.5)(1.5^2)/2!)]
=1−e^1.5[1+1.5+(2.25/2)]=0.1912
Proportion of days on which neither car is used = 0.2231 = 22.31 %
Proportion of days on which some demand is refused = 0.1912 = 19.12 %