Let there be three independent events
Question:

Let there be three independent events $\mathrm{E}_{1}, \mathrm{E}_{2}$ 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 ‘ $\mathrm{p}$ ‘ 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, $\frac{\text { Pr obability of oceurrence of } E_{1}}{\text { Probability of occurrence of } E_{3}}$ is equal to___________.

Solution:

Let $\mathrm{P}\left(\mathrm{E}_{1}\right)=\mathrm{P}_{1} ; \mathrm{P}\left(\mathrm{E}_{2}\right)=\mathrm{P}_{2} ; \mathrm{P}\left(\mathrm{E}_{3}\right)=\mathrm{P}_{3}$

$\mathrm{P}\left(\mathrm{E}_{1} \cap \overline{\mathrm{E}}_{2} \cap \overline{\mathrm{E}}_{3}\right)=\alpha=\mathrm{P}_{1}\left(1-\mathrm{P}_{2}\right)\left(1-\mathrm{P}_{3}\right) \ldots \ldots(1)$

$\mathrm{P}\left(\overline{\mathrm{E}}_{1} \cap \mathrm{E}_{2} \cap \overline{\mathrm{E}}_{3}\right)=\beta=\left(1-\mathrm{P}_{1}\right) \mathrm{P}_{2}\left(1-\mathrm{P}_{3}\right) \ldots \ldots(2)$

$\mathrm{P}\left(\overline{\mathrm{E}}_{1} \cap \overline{\mathrm{E}}_{2} \cap \mathrm{E}_{3}\right)=\gamma=\left(1-\mathrm{P}_{1}\right)\left(1-\mathrm{P}_{2}\right) \mathrm{P}_{3} \ldots \ldots(3)$

$\mathrm{P}\left(\overline{\mathrm{E}}_{1} \cap \overline{\mathrm{E}}_{2} \cap \overline{\mathrm{E}}_{3}\right)=\mathrm{P}=\left(1-\mathrm{P}_{1}\right)\left(1-\mathrm{P}_{2}\right)\left(1-\mathrm{P}_{3}\right) \ldots \ldots(4)$

Given that, $(\alpha-2 \beta) \mathrm{P}=\alpha \beta$

$\left(P_{1}\left(1-P_{2}\right)\left(1-P_{3}\right)-2\left(1-P_{1}\right) P_{2}\left(1-P_{3}\right)\right) P=P_{1} P_{2}$

$\left(1-P_{1}\right)\left(1-P_{2}\right)\left(1-P_{3}\right)^{2}$

$\Rightarrow\left(\mathrm{P}_{1}\left(1-\mathrm{P}_{2}\right)-2\left(1-\mathrm{P}_{1}\right) \mathrm{P}_{2}\right)=\mathrm{P}_{1} \mathrm{P}_{2}$

$\Rightarrow\left(\mathrm{P}_{1}-\mathrm{P}_{1} \mathrm{P}_{2}-2 \mathrm{P}_{2}+2 \mathrm{P}_{1} \mathrm{P}_{2}\right)=\mathrm{P}_{1} \mathrm{P}_{2}$

$\Rightarrow \mathrm{P}_{1}=2 \mathrm{P}_{2} \ldots \ldots .(1)$

and similarly, $(\beta-3 \gamma) \mathrm{P}=2 \mathrm{~B} \gamma$

$\mathrm{P}_{2}=3 \mathrm{P}_{3} \ldots .(2)$

So, $\mathrm{P}_{1}=6 \mathrm{P}_{3} \Rightarrow \frac{\mathrm{P}_{1}}{\mathrm{P}_{3}}=6$

 

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