Let there be three independent events E _ 1 , E _ 2 and E _ 3 . The...

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

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

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

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

$P\left(\overset{¯}{E}1 \cap \overset{¯}{E}2 \cap \overset{¯}{E}3\right) = p = \left(1 - {P}_{1}\right)\left(1 - {P}_{2}\right)\left(1 - {P}_{3}\right) \ldots \left(4\right)$

Given that, $( \alpha - 2\beta ) p = \alpha\beta$

$⇒ \left[{P}_{1}\left(1 - {P}_{2}\right)\left(1 - {P}_{3}\right) - 2\left(1 - {P}_{1}\right){P}_{2}\left(1 - {P}_{3}\right)\right]p = {P}_{1}{P}_{2}$

\left(1 - {P}_{1}\right)\left(1 - {P}_{2}\right){\left(1 - {P}_{3}\right)}^{2}$</p><p>$⇒ \left({P}_{1}\left(1 - {P}_{2}\right) - 2\left(1 - {P}_{1}\right){P}_{2}\right) = {P}_{1}{P}_{2}$</p><p>$⇒ \left({P}_{1} - {P}_{1}{P}_{2} - 2{P}_{2} + 2{P}_{1}{P}_{2}\right) = {P}_{1}{P}_{2}$</p><p>$⇒ {P}_{1} = 2{P}_{2} \ldots \left(1\right)$</p><p>and similarly,$( \beta - 3\gamma ) P = 2\beta\gamma$</p><p>${P}_{2} = 3{P}_{3} \ldots \left(2\right)$</p><p>So,${P}_{1} = 6{P}_{3} ⇒ \dfrac{{P}_{1}}{{P}_{3}} = 6$