Probability Ques 86
Of the three independent events $E _1, E _2$ and $E _3$, the probability that only $E _1$ occurs is $\alpha$, only $E _2$ occurs is $\beta$ and only $E _3$ occurs is $\gamma$. Let the probability $p$ that none of events $E _1, E _2$ or $E _3$ occurs satisfy 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, $\frac{\text { probability of occurrence of } E _1}{\text { probability of occurrence of } E _3}$ is equal to
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Answer:
Correct Answer: 86.$(6)$
Solution:
Formula:
- PLAN:
Forthe events to be independent,
$ \begin{aligned} P\left(E _1 \cap E _2 \cap E _3\right) & =P\left(E _1\right) \cdot P\left(E _2\right) \cdot P\left(E _3\right) \\ P\left(E _1 \cap \bar{E} _2 \cap \bar{E} _3\right) & =P\left(\text { only } E _1 \text { occurs }\right) \\ & =P\left(E _1\right) \cdot\left(1-P\left(E _2\right)\right)\left(1-P\left(E _3\right)\right) \end{aligned} $
Let $x, y$ and $z$ be probabilities of $E _1, E _2$ and $E _3$, respectively.
$ \begin{array}{lll} \therefore & & \alpha=x(1-y)(1-z) \\ & & \beta=(1-x) \cdot y(1-z) \\ \Rightarrow & & \gamma=(1-x)(1-y) z \\ & & p=(1-x)(1-y)(1-z) \end{array} $
Given, $(\alpha-2 \beta) p=\alpha \beta$ and $(\beta-3 \gamma) p=2 \beta \gamma$
From above equations, $x=2 y$ and $y=3 z$
$ \begin{array}{ll} \therefore & x=6 z \\ \Rightarrow & \frac{x}{z}=6 \end{array} $
BETA
