Question

Let there be three independent events $E_1,E_2$ and $E_3$. The probability that only $E_1$ occurs is $\alpha$, only $E_2$ occurs is $\beta$ and only $E_3$ occurs is $\gamma$. Let 'p' denote the probability of none of events occurs that satisfies the equations $(\alpha -2\beta )p=\alpha \beta$ and $(\beta -3\gamma )p=2\beta \gamma$. All the
given probabilities are assumed to lie in the interval $(0,1)$.
Then, $\frac{\text{ Probability of occurrence of }{E}_1}{\text{ Pr obability of occurrence of }{E}_3}$ is equal to _______.

Answer 1

Answer: 6

Let $P\left(E_1\right)=P_1;P\left(E_2\right)=P_2;P\left(E_3\right)=P_3$
$P\left(E_1\cap {\overline{E}}_2\cap {\overline{E}}_3\right)=\alpha =P_1\left(1-P_2\right)\left(1-P_3\right)\dots \dots (1)$
$P\left({\overline{E}}_1\cap E_2\cap {\overline{E}}_3\right)=\beta =\left(1-P_1\right)P_2\left(1-P_3\right)\dots \dots .(2)$
$P\left({\overline{E}}_1\cap {\overline{E}}_2\cap E_3\right)=\gamma =\left(1-P_1\right)\left(1-P_2\right)P_3\dots \dots .(3)$
$P\left({\overline{E}}_1\cap {\overline{E}}_2\cap {\overline{E}}_3\right)=P=\left(1-P_1\right)\left(1-P_2\right)\left(1-P_3\right)\dots \dots .(4)$
Given that, $(\alpha -2\beta )P=\alpha \beta$
$\Rightarrow \left(P_1\left(1-P_2\right)\left(1-P_3\right)-2\left(1-P_1\right)P_2\left(1-P_3\right)\right)P=P_1{P}_2$
$\left(1-P_1\right)\left(1-P_2\right){\left(1-P_3\right)}^2$
$\Rightarrow \left(P_1\left(1-P_2\right)-2\left(1-P_1\right)P_2\right)=P_1{P}_2$
$\Rightarrow \left(P_1-P_1{P}_2-2P_2+2P_1{P}_2\right)=P_1{P}_2$
$\Rightarrow P_1=2P_2\dots \dots (1)$
and similarly, $(\beta -3\gamma )P=2B\gamma$
$P_2=3P_3\dots \dots (2)$
So, $P_1=6P_3\Rightarrow \frac{P_1}{P_3}=6$