Question

Let there be three independent events $E_1,E_2\&E_3$. The probability that only $E_1$ occurs is $\alpha$, only $E_2$ occurs is $\beta$ and only $E_3$ occurs is $\gamma$ . Let ‘$p$’ denote the probability of none of the events that occur that satisfies the equations $(ɑ–2\beta )p=ɑ\beta$ 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 ___

Answer 1

Step 1: Finding the ratio

Considering $P\left(E_1\right)=x,P(E_2)=y\&P(E_3)=z$

Then,

By given condition

$(1-x)(1-y)(1-z)=p...\left(i\right)$

$x(1-y)(1-z)=\alpha ...\left(ii\right)$

$(1-x)y(1-z)=\beta ...\left(iii\right)$

$(1-x)(1-y)z=\gamma ...\left(iv\right)$

From $\left(i\right)and\left(ii\right)$

So, $x_1-x=\alpha p$

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

​Similarly, $y=\frac{\beta }{\beta +p}\&z=\frac{\gamma }{\gamma +p}$

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

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

Step 2: Substituting

$$\begin{gathered} \Rightarrow \left(\alpha -2\left(\frac{5\alpha \gamma }{\alpha +4\gamma }\right)\right)p=\frac{5\alpha \gamma }{\alpha +4\gamma \alpha } \\ \Rightarrow \alpha p-6p\gamma =5\alpha \gamma \\ \Rightarrow \frac{p}{\gamma }+1=6\left(\frac{p}{\alpha }+1\right) \\ \Rightarrow \frac{\frac{p}{\gamma }+1}{{\displaystyle \frac{p}{\alpha }}+1}=6. \end{gathered}$$

Hence,$\frac{\mathbf{}\mathbf{[}\mathbf{probability}\mathbf{}\mathbf{of}\mathbf{}\mathbf{occurrence}\mathbf{}\mathbf{of}\mathbf{}\mathbf{}\mathbf{}{\mathbf{E}}_{\mathbf{1}}\mathbf{}\mathbf{]}\mathbf{}}{\mathbf{}\mathbf{[}\mathbf{probability}\mathbf{}\mathbf{of}\mathbf{}\mathbf{occurrence}\mathbf{}\mathbf{of}\mathbf{}\mathbf{}\mathbf{}{\mathbf{E}}_{\mathbf{3}}\mathbf{}\mathbf{}\mathbf{]}\mathbf{}}$ $=6$