Probability 3 Question 45
45. 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
Passage Type Questions
Passage
Football teams $T _1$ and $T _2$ have to play two games against each other. It is assumed that the outcomes of the two games are independent. The probabilities of $T _1$ winning, drawing and losing a game against $T _2$ are $\frac{1}{2}, \frac{1}{6}$ and $\frac{1}{3}$, respectively. Each team gets 3 points for a win, 1 point for a draw and 0 point for a loss in a game. Let $X$ and $Y$ denote the total points scored by teams $T _1$ and $T _2$, respectively, after two games.
(2016 Adv.)
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Answer:
Correct Answer: 45. (c)
Solution:
- 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 $y=3 z$
$$ \begin{array}{ll} \therefore & x=6 z \\ \Rightarrow & \frac{x}{z}=6 \end{array} $$
