Question

A certain drug, manufactured by a Company is tested chemically for its toxic nature. Let the event "THE DRUG IS TOXIC" be denoted by H & the event "THE CHEMICAL TEST REVEALS THAT THE DRUG IS TOXIC" be denoted by S. Let $P\left(H\right)=a,P\left(S/H\right)=P\left(\overline{S}/H\right)=1-a$. Then show that the probability that the drug is not toxic given that the chemical test reveals that it is toxic, is $\frac{k_1}{k_2}$ (HCF of k_1 and k_2 is 1) then find the value of $k_1+k_2$ for $a=3$, for

Answer 1

$P\left(\overline{H}/S\right)=\frac{P\left(\overline{H}S\right)}{P\left(S\right)}$ .....(1)
also $\frac{P\left(HS\right)}{P\left(H\right)}=1-a\Rightarrow P\left(HS\right)=a(1-a)$ .......(2)
also $\frac{P\left(\overline{S}\overline{H}\right)}{P\left(\overline{H}\right)}=1-aP\left(\overline{S}\overline{H}\right)=(1-a{)}^2$
$P(\overline{S}\cap \overline{H})=P\left(\overline{S\cup H}\right)=(1-a{)}^2$
$\Rightarrow 1\left[P\right(S)+P(H\left)P\right(SH\left)\right]=(1a{)}^2$
$\Rightarrow \left[P\right(S)+aa+a^2]=2a{a}^2$
$P\left(S\right)=2a\left(1a\right)$ ... (3)
also $P\left(\overline{H}S\right)=P\left(S\right)P\left(SH\right)$
$=2a\left(1a\right)a\left(1a\right)=a\left(1a\right)$ ... (4)
from (3) & (4)

$P\left(\overline{H}/S\right)=\frac{1}{2}$