Question

A car hire firm has $2$ cars which it hires out day by day. If the number of demands for a car on each day follows poisson distribution with parameter $1.5$, then the probability that some demand is refused is

• A. $1.12\times e^{-1.5}$
• B. $\mathrm{1.2.5}\times e^{-1.5}$
• C. $1-3.625\times e^{-1.5}$
• D. $3.625\times e^{-1.5}$

Answer 1

Answer: (C)

$1-3.625\times e^{-1.5}$

let $X$ be a random variable denoting the number of demands for a car on any day.
So, $X$ is Poisson distributed with the parameter $\mu =1.5$
Therefore, the probability mass function is
$P(X=i)=\frac{e^{\mu }{\mu }^t}{i!}=e^{-1.5}\frac{{\left(1.5\right)}^2}{i!};i=0,1,2$
Now, the proportion of days in which neither car is used is actually the probability of there being no demand of cars which is given by
$P(X=0)=\frac{e^{-1.5}{\left(1.5\right)}^2}{0!}=e^{-1.5}$
and the proportion of days on which some demand is refused, is the probability that the number of demands become more than $2$ and is given by
$P(X>2)=1-P(X\le 2)$
$=1-P(X=0)+P(X=1)+P(X=2)$
$=1-\frac{e^{1.5}{\left(1.5\right)}^0}{0!}+\frac{e^{-1.5}{\left(1.5\right)}^1}{1!}+\frac{e^{-1.5}{\left(1.5\right)}^2}{2!}$
$=1-3.625\times e^{-1.5}$