Let B i i = 1 , 2 , 3 be three independent events in a sample space. The probability that only B 1 occur is…

given that $p_1\left(1-p_2\right)\left(1-p_3\right)=\alpha \dots \left(\mathrm{i}\right)$

$p_2\left(1-p_1\right)\left(1-p_3\right)=\beta \dots \left(\mathrm{ii}\right)$

$p_3\left(1-p_1\right)\left(1-p_2\right)=\gamma \dots \left(\mathrm{iii}\right)$

and $\left(1-p_1\right)\left(1-p_2\right)\left(1-p_3\right)=p \dots \left(\mathrm{iv}\right)$

$\Rightarrow \frac{p_1}{1-p_1}=\frac{\alpha }{p},\frac{p_2}{1-p_2}=\frac{\beta }{p}$ and $\frac{p_3}{1-p_3}=\frac{\gamma }{p}$

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

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

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

$\Rightarrow \frac{p_1}{1-p_1}-\frac{6p_3}{1-p_3}=\frac{5p_1{p}_3}{\left(1-p_1\right)\left(1-p_3\right)}$

$\Rightarrow p_1-6p_3=0$

$\Rightarrow \frac{p_1}{p_3}=6$