Question

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 $\frac{\text{Probability of occurrence of}{E}_1}{\text{Probability of occurrence of}{E}_3}=$

Answer 1

Let the probabilities of $E_1$, $E_2$ and $E_3$ be $p_1$, $p_2$ and $p_3$ respectively.

Given, $p_1(1-p_2)(1-p_3)=\alpha \cdots \left(1\right)$
$p_2(1-p_1)(1-p_3)=\beta$,
$p_3(1-p_1)(1-p_2)=\gamma$,
and $(1-p_1)(1-p_2)(1-p_3)=p\cdots \left(2\right)$

From $\left(1\right)$ and $\left(2\right)$,
$\frac{p}{\alpha }=\frac{1-p_1}{p_1}$
Similarly, $\frac{p}{\beta }=\frac{1-p_2}{p_2}$
and $\frac{p}{\gamma }=\frac{1-p_3}{p_3}$

$(\alpha -2\beta )p=\alpha \beta$
$\Rightarrow \frac{p}{\beta }-\frac{2p}{\alpha }=1\cdots \left(3\right)$

$(\beta -3\gamma )p=2\beta \gamma$
$\Rightarrow \frac{p}{\gamma }-\frac{3p}{\beta }=2\cdots \left(4\right)$

From $\left(3\right)$ and $\left(4\right)$, we get
$\frac{p}{\gamma }-\frac{6p}{\alpha }=5$
$\Rightarrow \frac{1}{p_3}-1-\frac{6}{p_1}+6=5$
$\Rightarrow \frac{p_1}{p_3}=6$