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 ssumed 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 $P\left(E_1\right)=x,P\left(E_2\right)=y$ and $P\left(E_3\right)=z$

then $(1-x)(1-y)(1-z)=p$

$x(1-y)(1-z)=\alpha$

$(1-x)y(1-z)=\beta$

$(1-x)(1-y)z=\gamma$

So $\frac{1-x}{x}=\frac{p}{\alpha }$

So, $x=\frac{\alpha }{\alpha +p}$

Similarly $y=\frac{\beta }{\beta +p}$

and $z=\frac{\gamma }{\gamma +p}$

So, $\frac{P\left(E_1\right)}{P\left(E_3\right)}=\frac{\frac{\alpha }{\alpha +p}}{\frac{\gamma }{\gamma +p}}=\frac{\frac{\gamma +p}{\gamma }}{\frac{\alpha +p}{\alpha }}=\frac{1+\frac{p}{\gamma }}{1+\frac{p}{\alpha }}$

Also given $\frac{\alpha \beta }{\alpha -2\beta }=p=\frac{2\beta \gamma }{\beta -3\gamma }\Rightarrow \beta =\frac{5{\alpha }^{}\gamma }{\alpha +4\gamma }$

Substituting back $\left(\alpha -2\left(\frac{5\alpha \gamma }{\alpha +4\gamma }\right)\right)p=\frac{\alpha \cdot 5\alpha \gamma }{\alpha +4\gamma }$
$\Rightarrow \alpha p-6p\gamma =5\alpha \gamma$
$\Rightarrow \frac{p}{\gamma }+1=6(\frac{p}{\alpha }+1)\Rightarrow \frac{\frac{p}{\gamma }+1}{\frac{p}{\alpha }+1}=6$.