 # Solve the following Question:

Let $\mathrm{B}_{\mathrm{i}}(\mathrm{i}=1,2,3)$ be three independent events in a sample space. The probability that only $\mathrm{B}_{1}$ occur is $\alpha$, only $\mathrm{B}_{2}$ occurs is $\beta$ and only $\mathrm{B}_{3}$ occurs is $\gamma$. Let $\mathrm{p}$ be the probability that none of the events $B_{i}$ occurs and these 4 probabilities satisfy the equations $(\alpha-2 \beta) \mathrm{p}=\alpha \beta$ and $(\beta-3 \gamma) \mathrm{p}=2 \beta \gamma($ All the probabilities are assumed to lie in the interval $(0,1)$ ). Then $\frac{\mathrm{P}\left(\mathrm{B}_{1}\right)}{\mathrm{P}\left(\mathrm{B}_{3}\right)}$ is equal to

Solution:

Let $x, y, z$ be probability of $B_{1}, B_{2}, B_{3}$ respectively

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

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

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

$\Rightarrow(1-x)(1-y)(1-z)=p$

$(\alpha-2 \beta) p=\alpha \beta$

$(x(1-y)(1-z)-2 y(1-x)(1-z))(1-x)(1-y)(1-z)=x y(1-x)(1-y)(1-z)$

$x-x y-2 y+2 x y=x y$

$x=2 y \ldots$ (1)

Similarly $(\beta-3 r) \mathrm{p}=2 \beta r$

$\Rightarrow y=3 z \ldots(2)$

From (1) and (2)

$x=6 z$

Now

$\frac{x}{z}=6$