Question

Let $B_i(i=1,2,3)$ be three independent events in a sample space. The probability that only $B_1$ occur is $\alpha$, only $B_2$ occurs is $\beta$ and only $B_3$ occurs is $\gamma$. Let $p$ be the probability that none of the events $B_i$ occurs and these 4 probabilities satisfy the equations $(\alpha -2\beta )p=\alpha \beta$ and $(\beta -3\gamma )p=2\beta \gamma$ (All the probabilities are assumed to lie in the interval $(0,1))$. Then $\frac{P\left(B_1\right)}{P\left(B_3\right)}$ is equal to

Answer 1

Let $x,y,z$ be probability of $B_1,B_2,B_3$ respectively.
$\Rightarrow x(1-y)(1-z)=\alpha$
$\Rightarrow y(1-x)(1-z)=\beta$
$\Rightarrow z(1-x)(1-y)=\gamma$
$\Rightarrow (1-x)(1-y)(1-z)=p$
Now $(\alpha -2\beta )P=\alpha \beta$
$\Rightarrow \left[x\right(1-y\left)\right(1-z)-2y(1-x\left)\right(1-z\left)\right](1-x)(1-y)(1-z)=xy(1-x)(1-y)(1-z{)}^2$
$\Rightarrow x+xy-2y=xy$
$\therefore x=2y\dots \left(1\right)$
similarly, $(\beta -3\gamma )p=2\beta \gamma$
$\Rightarrow y=3z\cdots \left(2\right)$
From $\left(1\right)$ & $\left(2\right)$
$\Rightarrow x=6z$
Hence $\frac{x}{z}=\frac{P\left(B_1\right)}{P\left(B_3\right)}=6$