Let there be three independent events E₁, E₂ and E₃. The probability that only E₁ occurs is $\alpha$, only E₂ occurs is $\beta$ and only E₃ occurs is $\gamma$. Let 'p' denote the probability of none of events occurs that satisfies the equations
($\alpha$ $-$ 2$\beta$)p = $\alpha$$\beta$ and ($\beta$ $-$ 3$\gamma$)p = 2$\beta$$\gamma$. All the given probabilities are assumed to lie in the interval (0, 1).

Then, $$\frac{Probability\ of\ occurrence\ of\ E_{1}}{Probability\ of\ occurrence\ of\ E_{3}} $$ is equal to _____________.

Let P(E₁) = x, P(E₂) = y and P(E₃) = z <br><br>$\alpha$ = P$\left( {{E_1} \cap {{\overline E }_2} \cap {{\overline E }_3}} \right)$ = $$P\left( {{E_1}} \right).P\left( {{{\overline E }_2}} \right).P\left( {{{\overline E }_3}} \right)$$ <br><br>$\Rightarrow$ $\alpha$ = x(1 $-$ y) (1 $-$ z) ......(i) <br><br>Similarly <br><br>β = (1 – x).y(1 – z) ...(ii) <br><br>$\gamma$ = (1 – x)(1 – y).z ...(iii) <br><br>p = (1 – x)(1 – y)(1 – z) ...(iv) <br><br>From (i) and (iv) <br><br>${x \over {1 - x}} = {\alpha \over p}$ <br><br>$\Rightarrow$ x = ${\alpha \over {\alpha + p}}$ <br><br>From (iii) and (iv) <br><br>${z \over {1 - z}} = {\gamma \over p}$ <br><br>$\Rightarrow$ z = ${\gamma \over {\gamma + p}}$ <br><br>$\therefore$ $${{P\left( {{E_1}} \right)} \over {P\left( {{E_3}} \right)}} = {x \over z} = {{{\alpha \over {\alpha + p}}} \over {{\gamma \over {\gamma + p}}}}$$ $$ = {{{{\gamma + p} \over \gamma }} \over {{{\alpha + p} \over \alpha }}} = {{1 + {p \over \gamma }} \over {1 + {p \over \alpha }}}$$ ..(v) <br><br>Also given, <br><br>($\alpha$ $-$ 2$\beta$)p = $\alpha$$\beta$ $\Rightarrow$ $\alpha$p = ($\alpha$ + 2p)$\beta$ ....(vi) <br><br>$\beta$ $-$ 3$\gamma$)p = 2$\beta$$\gamma$ $\Rightarrow$ 3$\gamma$p = (p - 2$\gamma$)$\beta$ .....(vii) <br><br>From (vi) and (vii), <br><br>${\alpha \over {3\gamma }} = {{\alpha + 2p} \over {p - 2\gamma }}$ <br><br>$\Rightarrow$ p$\alpha$ - 6p$\gamma$ = 5$\gamma$$\alpha$ <br><br>$\Rightarrow$ ${p \over \gamma } - {{6p} \over \alpha } = 5$ <br><br>$\Rightarrow$ ${p \over \gamma } + 1 = 6\left( {{p \over \alpha } + 1} \right)$ ....(viii) <br><br>Now from (v) and (viii), <br><br>${{P\left( {{E_1}} \right)} \over {P\left( {{E_3}} \right)}}$ = 6