Let B _ i (i = 1, 2, 3) be three independent events in a sample...

given that ${p}_{1}\left(1 - {p}_{2}\right)\left(1 - {p}_{3}\right) = \alpha \ldots \left(i\right)$

${p}_{2}\left(1 - {p}_{1}\right)\left(1 - {p}_{3}\right) = \beta \ldots \left(ii\right)$

${p}_{3}\left(1 - {p}_{1}\right)\left(1 - {p}_{2}\right) = \gamma \ldots \left(iii\right)$

and \left(1 - {p}_{1}\right)\left(1 - {p}_{2}\right)\left(1 - {p}_{3}\right) = p \ldots \left(iv\right)$</p><p>$⇒ \dfrac{{p}_{1}}{1 - {p}_{1}} = \dfrac{\alpha}{p}, \dfrac{{p}_{2}}{1 - {p}_{2}} = \dfrac{\beta}{p} and $\dfrac{{p}_{3}}{1 - {p}_{3}} = \dfrac{\gamma}{p}$

Also $\beta = \dfrac{\alphap}{\alpha + 2p} = \dfrac{3\gammap}{p - 2\gamma}$

$⇒ \alphap - 2\alpha\gamma = 3\alpha\gamma + 6p\gamma$

$⇒ \alphap - 6p\gamma = 5\alpha\gamma$

$⇒ \dfrac{{p}_{1}}{1 - {p}_{1}} - \dfrac{6{p}_{3}}{1 - {p}_{3}} = \dfrac{5{p}_{1}{p}_{3}}{\left(1 - {p}_{1}\right)\left(1 - {p}_{3}\right)}$

$⇒ {p}_{1} - 6{p}_{3} = 0$

$⇒ \dfrac{{p}_{1}}{{p}_{3}} = 6$