JEE PYQ: Probability Question 4
Question 4 - 2021 (17 Mar Shift 1)
Let there be 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 ‘p’ denote the probability that none of the events occurs and these 4 probabilities satisfy the equations $(\alpha - 2\beta)p = \alpha\beta$ and $(\beta - 3\gamma)p = 2\beta\gamma$. All the given probabilities are assumed to lie in the interval $(0,1)$. Then $\frac{P(\text{occurrence of } E_1)}{P(\text{occurrence of } E_3)}$ is equal to ______.
Show Answer
Answer: 6
Solution
Let $P_1, P_2, P_3$ be probabilities. $\alpha = P_1(1-P_2)(1-P_3)$, $\beta = (1-P_1)P_2(1-P_3)$, $\gamma = (1-P_1)(1-P_2)P_3$, $p = (1-P_1)(1-P_2)(1-P_3)$. From $(\alpha - 2\beta)p = \alpha\beta$: $P_1 = 2P_2$. From $(\beta - 3\gamma)p = 2\beta\gamma$: $P_2 = 3P_3$. So $P_1 = 6P_3$, hence $\frac{P_1}{P_3} = 6$.
BETA
