91Ó°ÊÓ

We assume that the number of times person contracts a cold is Poisson random variable where $\mathrm{λ}=5$. a new drug reduces the Poisson parameter to $\mathrm{λ}=3$ 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 $\mathrm{ℙ}\left(A\right)=0.75⟹\mathrm{ℙ}\left(A^c\right)=0.25$and it represents probability that drug is beneficial for the patient.

We have:

$\mathrm{â„™}(Aâˆ\pounds X=2)=\frac{\mathrm{â„™}(X=2,A)}{\mathrm{â„™}(X=2)}=\frac{\mathrm{â„™}(X=2âˆ\pounds A)\mathrm{â„™}\left(A\right)}{\mathrm{â„™}(X=2)}$

$=\frac{\mathrm{â„™}(X=2âˆ\pounds A)\mathrm{â„™}\left(A\right)}{\mathrm{â„™}(X=2âˆ\pounds A)\mathrm{â„™}\left(A\right)+\mathrm{â„™}\left(X=2âˆ\pounds A^c\right)\mathrm{â„™}\left(A^c\right)}$

We calculate,

$\mathrm{â„™}(X=2âˆ\pounds A)=\frac{e^{-3}×3^2}{2!}=\frac{0.448}{2}=0.224$

$\mathrm{â„™}\left(X=2âˆ\pounds A^c\right)=\frac{e^{−5}â‹\dots 5^2}{2!}=\frac{0.168}{2}=0.084$

Currently we plug in the results in previously accepted expressions:

$\mathrm{â„™}(Aâˆ\pounds X=2)=\frac{0.224·0.75}{0.224·0.75+0.084·0.25}=\frac{0.168}{0.189}=0.89$

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