The number of times that a person contracts a cold in a given year is a Poisson random variable with parameter $\lambda =5$. Suppose that a new wonder drug (based on large quantities of vitamin $C$) has just been marketed that reduces the Poisson parameter to $\lambda =3$ for $75$ percent of the population. For the other $25$ percent of the population, the drug has no appreciable effect on colds. If an individual tries the drug for a year and has $2$ colds in that time, how likely is it that the drug is beneficial for him or her?

We assume that the number of times person contracts a cold is Poisson random variable where $\lambda =5$. a new drug reduces the Poisson parameter to $\lambda =3$ for $75\%$ of the population while for others nothing significant happens. If an individual tries the drug for a year and has precisely $2$ colds in that time we calculate the probability that it is beneficial for him.

Firstly, let $A$be an occasion such that $\mathrm{\mathbb{P}}\left(A\right)=0.75⟹\mathrm{\mathbb{P}}\left(A^c\right)=0.25$and it represents probability that drug is beneficial for the patient.

We have:

$\mathrm{\mathbb{P}}(A\mid X=2)=\frac{\mathrm{\mathbb{P}}(X=2,A)}{\mathrm{\mathbb{P}}(X=2)}=\frac{\mathrm{\mathbb{P}}(X=2\mid A)\mathrm{\mathbb{P}}\left(A\right)}{\mathrm{\mathbb{P}}(X=2)}$

$=\frac{\mathrm{\mathbb{P}}(X=2\mid A)\mathrm{\mathbb{P}}\left(A\right)}{\mathrm{\mathbb{P}}(X=2\mid A)\mathrm{\mathbb{P}}\left(A\right)+\mathrm{\mathbb{P}}\left(X=2\mid A^c\right)\mathrm{\mathbb{P}}\left(A^c\right)}$

We calculate,

$\mathrm{\mathbb{P}}(X=2\mid A)=\frac{e^{-3}\times 3^2}{2!}=\frac{0.448}{2}=0.224$

$\mathrm{\mathbb{P}}\left(X=2\mid A^c\right)=\frac{e^{-5}\cdot 5^2}{2!}=\frac{0.168}{2}=0.084$

Currently we plug in the results in previously accepted expressions:

$\mathrm{\mathbb{P}}(A\mid X=2)=\frac{0.224·0.75}{0.224·0.75+0.084·0.25}=\frac{0.168}{0.189}=0.89$

The probability of drug helping him is $0.89$. Hence, we are done.