Let B_{i} (i=1, 2, 3) be three independent ev | vedclass

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(B) Let $P(B_{1}) = p_{1}$,$P(B_{2}) = p_{2}$,and $P(B_{3}) = p_{3}$.Given that $p_{1}(1 - p_{2})(1 - p_{3}) = \alpha$ $...(i)$$p_{2}(1 - p_{1})(1 - p

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$6$

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(B) Let $P(B_{1}) = p_{1}$,$P(B_{2}) = p_{2}$,and $P(B_{3}) = p_{3}$.Given that $p_{1}(1 - p_{2})(1 - p_{3}) = \alpha$ $...(i)$$p_{2}(1 - p_{1})(1 - p_{3}) = \beta$ $...(ii)$$p_{3}(1 - p_{1})(1 - p_{2}) = \gamma$ $...(iii)$And $(1 - p_{1})(1 - p_{2})(1 - p_{3}) = p$ $...(iv)$From these,we get $\frac{p_{1}}{1 - p_{1}} = \frac{\alpha}{p}$,$\frac{p_{2}}{1 - p_{2}} = \frac{\beta}{p}$,and $\frac{p_{3}}{1 - p_{3}} = \frac{\gamma}{p}$.Given equations: $(\alpha - 2\beta)p = \alpha\beta \Rightarrow \frac{1}{p} = \frac{1}{\beta} - \frac{2}{\alpha} \Rightarrow \frac{1}{\beta} = \frac{1}{p} + \frac{2}{\alpha}$.Also,$(\beta - 3\gamma)p = 2\beta\gamma \Rightarrow \frac{1}{2\gamma} - \frac{3}{2\beta} = \frac{1}{p} \Rightarrow \frac{1}{2\gamma} = \frac{1}{p} + \frac{3}{2\beta}$.Substituting $\frac{1}{\beta} = \frac{1}{p} + \frac{2}{\alpha}$ into the second equation:$\frac{1}{2\gamma} = \frac{1}{p} + \frac{3}{2}(\frac{1}{p} + \frac{2}{\alpha}) = \frac{1}{p} + \frac{3}{2p} + \frac{3}{\alpha} = \frac{5}{2p} + \frac{3}{\alpha}$.Using $\frac{\alpha}{p} = \frac{p_{1}}{1 - p_{1}}$ and $\frac{\gamma}{p} = \frac{p_{3}}{1 - p_{3}}$,we have $\frac{1}{2} \cdot \frac{1 - p_{3}}{p_{3}} = \frac{5}{2} + 3 \cdot \frac{1 - p_{1}}{p_{1}}$.Simplifying leads to $\frac{p_{1}}{p_{3}} = 6$.

List-$I$	List-$II$
$(I)$ Probability of $(X_2 \geq Y_2)$ is	$(P)$ $\frac{3}{8}$
$(II)$ Probability of $(X_2 > Y_2)$ is	$(Q)$ $\frac{11}{16}$
$(III)$ Probability of $(X_3 = Y_3)$ is	$(R)$ $\frac{5}{16}$
$(IV)$ Probability of $(X_3 > Y_3)$ is	$(S)$ $\frac{355}{864}$
	$(T)$ $\frac{77}{432}$

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