Solve this following

Question:

Let $\mathrm{B}_{i}(i=1,2,3)$ be three independent events in a sample space. The probability that only $\mathrm{B}_{1}$ occur is $\alpha$, only $B_{2}$ occurs is $\beta$ and only $B_{3}$ occurs is $\gamma$. Let p be the probability that none of the events $\mathrm{B}_{i}$ occurs and these 4 probabilities satisfy the equations $(\alpha-2 \beta) \mathrm{p}=\alpha \beta$ and $(\beta-3 \gamma) p=2 \beta \gamma($ All the probabilities are assumed to lie in the interval $(0,1))$. Then

$\frac{\mathrm{P}\left(\mathrm{B}_{1}\right)}{\mathrm{P}\left(\mathrm{B}_{3}\right)}$ is equal to

 

Solution:

Let $\mathrm{P}\left(\mathrm{B}_{1}\right)=\mathrm{p}_{1}, \mathrm{P}\left(\mathrm{B}_{2}\right)=\mathrm{p}_{2}, \mathrm{P}\left(\mathrm{B}_{3}\right)=\mathrm{p}_{3}$

given that $p_{1}\left(1-p_{2}\right)\left(1-p_{3}\right)=\alpha$ ...................(I)

$\mathrm{p}_{2}\left(1-\mathrm{p}_{1}\right)\left(1-\mathrm{p}_{3}\right)=\beta$.................(II)

$\mathrm{p}_{3}\left(1-\mathrm{p}_{1}\right)\left(1-\mathrm{p}_{2}\right)=\gamma$.........................(III)

and    $\left(1-\mathrm{p}_{1}\right)\left(1-\mathrm{p}_{2}\right)\left(1-\mathrm{p}_{3}\right)=\mathrm{p} \ldots \ldots$ (iv)

$\Rightarrow \quad \frac{\mathrm{p}_{1}}{1-\mathrm{p}_{1}}=\frac{\alpha}{\mathrm{p}}, \frac{\mathrm{p}_{2}}{1-\mathrm{p}_{2}}=\frac{\beta}{\mathrm{p}} \quad \& \frac{\mathrm{p}_{3}}{1-\mathrm{p}_{3}}=\frac{\gamma}{\mathrm{p}}$

Also $\quad \beta=\frac{\alpha p}{\alpha+2 p}=\frac{3 \gamma p}{p-2 \gamma}$

$\Rightarrow \alpha p-2 \alpha \gamma=3 \alpha \gamma+6 p \gamma$

$\Rightarrow \alpha p-6 p \gamma=5 \alpha \gamma$

$\Rightarrow \frac{p_{1}}{1-p_{1}}-\frac{6 p_{3}}{1-p_{3}}=\frac{5 p_{1} p_{3}}{\left(1-p_{1}\right)\left(1-p_{3}\right)}$

$\Rightarrow \quad \mathrm{p}_{1}-6 \mathrm{p}_{3}=0$

$\Rightarrow \quad \frac{\mathrm{p}_{1}}{\mathrm{p}_{3}}=6$

 

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