Let there be three independent events E_{1}, | vedclass

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(B) Let $P(E_{1}) = P_{1}$,$P(E_{2}) = P_{2}$,and $P(E_{3}) = P_{3}$.Since the events are independent,the probability that only $E_{1}$ occurs is $\al

Vedclass · Accepted Answer

$6$

Vedclass · Answer

(B) Let $P(E_{1}) = P_{1}$,$P(E_{2}) = P_{2}$,and $P(E_{3}) = P_{3}$.Since the events are independent,the probability that only $E_{1}$ occurs is $\alpha = P_{1}(1 - P_{2})(1 - P_{3})$.The probability that only $E_{2}$ occurs is $\beta = (1 - P_{1})P_{2}(1 - P_{3})$.The probability that only $E_{3}$ occurs is $\gamma = (1 - P_{1})(1 - P_{2})P_{3}$.The probability that none of the events occur is $p = (1 - P_{1})(1 - P_{2})(1 - P_{3})$.Given $(\alpha - 2\beta)p = \alpha\beta$,we substitute the expressions:$\left(P_{1}(1 - P_{2})(1 - P_{3}) - 2(1 - P_{1})P_{2}(1 - P_{3})\right)p = P_{1}(1 - P_{2})(1 - P_{3}) \cdot (1 - P_{1})P_{2}(1 - P_{3})$.Dividing both sides by $(1 - P_{1})(1 - P_{2})(1 - P_{3})^2$,we get:$\frac{P_{1}}{1 - P_{1}} - \frac{2P_{2}}{1 - P_{2}} = \frac{P_{1}P_{2}}{(1 - P_{1})(1 - P_{2})}$.This simplifies to $P_{1} = 2P_{2}$.Similarly,from $(\beta - 3\gamma)p = 2\beta\gamma$,we get $P_{2} = 3P_{3}$.Therefore,$P_{1} = 2(3P_{3}) = 6P_{3}$.Thus,$\frac{P_{1}}{P_{3}} = 6$.

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