Let there be three independent events $E_{1}, E_{2}$ and $\mathrm{E}_{3}$. The probability that only $\mathrm{E}_{1}$ occurs is $\alpha$, only $\mathrm{E}_{2}$ occurs is $\beta$ and only $\mathrm{E}_{3}$ occurs is $\gamma$. Let ' $p$ ' denote the probability of none of events occurs that satisfies the equations $(\alpha-2 \beta) \mathrm{p}=\alpha \beta$ and $(\beta-3 \gamma) \mathrm{p}=2 \beta \gamma .$ All the given probabilities are assumed to lie in the interval $(0,1)$.
Then, $\frac{\text { Probability of occurrence of } E_{1}}{\text { Probability of occurrence of } E_{3}}$ is equal to
Let $\mathrm{P}\left(\mathrm{E}_{1}\right)=\mathrm{P}_{1} ; \mathrm{P}\left(\mathrm{E}_{2}\right)=\mathrm{P}_{2} ; \mathrm{P}\left(\mathrm{E}_{3}\right)=\mathrm{P}_{3}$
$\mathrm{P}\left(\mathrm{E}_{1} \cap \overline{\mathrm{E}}_{2} \cap \overline{\mathrm{E}}_{3}\right)=\alpha=\mathrm{P}_{1}\left(1-\mathrm{P}_{2}\right)\left(1-\mathrm{P}_{3}\right) \ldots \ldots(1)$..(1)
$\mathrm{P}\left(\overline{\mathrm{E}}_{1} \cap \mathrm{E}_{2} \cap \overline{\mathrm{E}}_{3}\right)=\beta=\left(1-\mathrm{P}_{1}\right) \mathrm{P}_{2}\left(1-\mathrm{P}_{3}\right) \ldots \ldots(2)$..(2)
$\mathrm{P}\left(\overline{\mathrm{E}}_{1} \cap \overline{\mathrm{E}}_{2} \cap \mathrm{E}_{3}\right)=\gamma=\left(1-\mathrm{P}_{1}\right)\left(1-\mathrm{P}_{2}\right) \mathrm{P}_{3} \ldots \ldots(3)$..(3)
$\mathrm{P}\left(\overline{\mathrm{E}}_{1} \cap \overline{\mathrm{E}}_{2} \cap \overline{\mathrm{E}}_{3}\right)=\mathrm{P}=\left(1-\mathrm{P}_{1}\right)\left(1-\mathrm{P}_{2}\right)\left(1-\mathrm{P}_{3}\right) \ldots \ldots(4)$...(4)
Given that, $(\alpha-2 \beta) \mathrm{P}=\alpha \beta$
$\begin{aligned} \Rightarrow &\left(\mathrm{P}_{1}\left(1-\mathrm{P}_{2}\right)\left(1-\mathrm{P}_{3}\right)-2\left(1-\mathrm{P}_{1}\right) \mathrm{P}_{2}\left(1-\mathrm{P}_{3}\right)\right) \mathrm{P}=\mathrm{P}_{1} \mathrm{P}_{2} \\ &\left(1-\mathrm{P}_{1}\right)\left(1-\mathrm{P}_{2}\right)\left(1-\mathrm{P}_{3}\right)^{2} \end{aligned}$
$\Rightarrow\left(P_{1}\left(1-P_{2}\right)-2\left(1-P_{1}\right) P_{2}\right)=P_{1} P_{2}$
$\Rightarrow\left(\mathrm{P}_{1}-\mathrm{P}_{1} \mathrm{P}_{2}-2 \mathrm{P}_{2}+2 \mathrm{P}_{1} \mathrm{P}_{2}\right)=\mathrm{P}_{1} \mathrm{P}_{2}$
$\Rightarrow \mathrm{P}_{1}=2 \mathrm{P}_{2}.....(1)$
and similarly, $(\beta-3 \gamma) \mathrm{P}=2 \mathrm{~B} \gamma$
$\mathrm{P}_{2}=3 \mathrm{P}_{3}$..(2)
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