A car hire firm has two cars, which it hires out day by day. The number of demands for cars on each day is distributed as a Poisson distribution with mean 1.5. Calculate the probabilities of days on which neither car is used and the probabilities of days on which some demand is refused. (Use $e^{-1.5} = 0.2231$)

Probability (neither car used) = 0.2231; Probability (demand refused) = 0.1913

The correct answer is Option (1) → Probability (neither car used) = 0.2231; Probability (demand refused) = 0.1913

Here $λ$ = mean = 1.5

The probability of days when no car is required

$= P(0)=e^{-λ}=e^{-1.5} = 0.2231$

The probability of days when some demand is refused

= Prob. (more than 2 cars are demanded) = $1 - [P(0) + P(1) + P(2)]$

$=1-e^{-λ}\left(1+\frac{λ}{1!}+\frac{λ^2}{2!}\right)=1-0.2231 (1 + 1.5+ 1.125) = 0.1913$