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

Let $P \left( E _{1}\right)= P _{1} ;\, P \left( E _{2}\right)= P _{2} ;\, P \left( E _{3}\right)= P _{3}$
$P \left( E _{1} \cap \overline{ E }_{2} \cap \overline{ E }_{3}\right)=\alpha= P _{1}\left(1- P _{2}\right)\left(1- P _{3}\right) \ldots \ldots(1)$
$P \left(\overline{ E }_{1} \cap E _{2} \cap \overline{ E }_{3}\right)=\beta=\left(1- P _{1}\right) P _{2}\left(1- P _{3}\right) \ldots \ldots .(2)$
$P \left(\overline{ E }_{1} \cap \overline{ E }_{2} \cap E _{3}\right)=\gamma=\left(1- P _{1}\right)\left(1- P _{2}\right) P _{3} \ldots \ldots .(3)$
$P \left(\overline{ E }_{1} \cap \overline{ E }_{2} \cap \overline{ E }_{3}\right)= P =\left(1- P _{1}\right)\left(1- P _{2}\right)\left(1- P _{3}\right) \ldots \ldots .(4)$
Given that, $(\alpha-2 \beta) P=\alpha \beta$
$\Rightarrow \left( P _{1}\left(1- P _{2}\right)\left(1- P _{3}\right)-2\left(1- P _{1}\right) P _{2}\left(1- P _{3}\right)\right) P = P _{1} P _{2}$
$\left(1- P _{1}\right)\left(1- P _{2}\right)\left(1- P _{3}\right)^{2}$
$\Rightarrow \left( P _{1}\left(1- P _{2}\right)-2\left(1- P _{1}\right) P _{2}\right)= P _{1} P _{2}$
$\Rightarrow \left( P _{1}- P _{1} P _{2}-2 P _{2}+2 P _{1} P _{2}\right)= P _{1} P _{2}$
$\Rightarrow P _{1}=2 P _{2} \ldots \ldots(1)$
and similarly, $(\beta-3 \gamma) P =2 B \gamma$
$P _{2}=3 P _{3} \ldots \ldots(2)$
So, $P _{1}=6 P _{3} \Rightarrow \frac{ P _{1}}{ P _{3}}=6$