Of the three independent events E_1, E_2 and | vedclass

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(B) Let $x, y, z$ be the probabilities of $E_1, E_2, E_3$ respectively. Since the events are independent,we have: $\alpha = x(1-y)(1-z)$,$\beta = y(1-

Vedclass · Accepted Answer

$6$

Vedclass · Answer

(B) Let $x, y, z$ be the probabilities of $E_1, E_2, E_3$ respectively. Since the events are independent,we have: $\alpha = x(1-y)(1-z)$,$\beta = y(1-x)(1-z)$,$\gamma = z(1-x)(1-y)$,and $p = (1-x)(1-y)(1-z)$. From these,we can write: $\frac{\alpha}{p} = \frac{x}{1-x}$,$\frac{\beta}{p} = \frac{y}{1-y}$,and $\frac{\gamma}{p} = \frac{z}{1-z}$. Given $(\alpha - 2\beta)p = \alpha\beta$,dividing by $p^2$ gives $\frac{\alpha}{p} - 2\frac{\beta}{p} = \frac{\alpha}{p} \cdot \frac{\beta}{p}$. Substituting $u = \frac{x}{1-x}, v = \frac{y}{1-y}, w = \frac{z}{1-z}$,we get $u - 2v = uv \Rightarrow u(1-v) = 2v \Rightarrow u = \frac{2v}{1-v}$. Similarly,$(\beta - 3\gamma)p = 2\beta\gamma \Rightarrow \frac{\beta}{p} - 3\frac{\gamma}{p} = 2\frac{\beta}{p} \cdot \frac{\gamma}{p} \Rightarrow v - 3w = 2vw \Rightarrow v = \frac{3w}{1-2w}$. Solving these relations leads to $x = 2y$ and $y = 3z$,hence $x = 6z$. Thus,the ratio $\frac{P(E_1)}{P(E_3)} = \frac{x}{z} = 6$.

List-$I$	List-$II$
$(A) P(\overline{A} \cup B)$	$(I) \frac{2}{3}$
$(B) P(\frac{A}{\overline{B}})$	$(II) \frac{11}{15}$
$(C) P(A \cup B)$	$(III) \frac{3}{5}$

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