JEE Main 2021 (Online) 24th February Morning ShiftQuestion 1
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 $\frac{P(B_1)}{P(B_3)}$ is equal to $\_\_\_\_\_\_$ .
Probability
Answer$6$
Solution
Let $a=\frac{P(B_1)}{1-P(B_1)}$ etc.; then $\alpha=ap,\ \beta=bp,\ \gamma=cp$. Equations give $a-2b=ab\Rightarrow\frac1b-\frac2a=1$ and $b-3c=2bc\Rightarrow\frac1c-\frac3b=2$. Let $\frac1b=t$: $\frac1a=\frac{t-1}2$, $\frac1c=2+3t$. $P(B_1)=\frac{1}{1+1/a}=\frac2{t+1}$, $P(B_3)=\frac1{1+1/c}=\frac1{3(t+1)}$. Ratio $=6$.