91影视

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

We have:

$\mathrm{鈩�}(A鈭�X=2)=\frac{\mathrm{鈩�}(X=2,A)}{\mathrm{鈩�}(X=2)}=\frac{\mathrm{鈩�}(X=2鈭�A)\mathrm{鈩�}\left(A\right)}{\mathrm{鈩�}(X=2)}$

$=\frac{\mathrm{鈩�}(X=2鈭�A)\mathrm{鈩�}\left(A\right)}{\mathrm{鈩�}(X=2鈭�A)\mathrm{鈩�}\left(A\right)+\mathrm{鈩�}\left(X=2鈭�A^c\right)\mathrm{鈩�}\left(A^c\right)}$

We calculate,

$\mathrm{鈩�}(X=2鈭�A)=\frac{e^{-3}脳3^2}{2!}=\frac{0.448}{2}=0.224$

$\mathrm{鈩�}\left(X=2鈭�A^c\right)=\frac{e^{鈭�5}鈰�5^2}{2!}=\frac{0.168}{2}=0.084$

Currently we plug in the results in previously accepted expressions:

$\mathrm{鈩�}(A鈭�X=2)=\frac{0.224路0.75}{0.224路0.75+0.084路0.25}=\frac{0.168}{0.189}=0.89$

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