Probability – Bayes’ Theorem – JEE Main 02 April 2025 Shift 2

Solution:

Let $E_1, E_2, E_3$ be the events of choosing Bag I, Bag II, and Bag III respectively.
$$P(E_1) = P(E_2) = P(E_3) = \frac{1}{3}$$

Let $R$ be the event of drawing a Red ball and $G$ be the event of drawing a Green ball.
From the given table, the probability of drawing the respective balls from each bag is:
$$P(R|E_1) = \frac{3}{10}, \quad P(R|E_2) = \frac{4}{10}, \quad P(R|E_3) = \frac{5}{10}$$
$$P(G|E_1) = \frac{5}{10}, \quad P(G|E_2) = \frac{3}{10}, \quad P(G|E_3) = \frac{4}{10}$$

Using Bayes’ Theorem, the probability $p$ that the drawn Red ball is from Bag I is:
$$p = P(E_1|R) = \frac{P(E_1)P(R|E_1)}{P(E_1)P(R|E_1) + P(E_2)P(R|E_2) + P(E_3)P(R|E_3)}$$
$$p = \frac{\frac{1}{3} \cdot \frac{3}{10}}{\frac{1}{3} \cdot \frac{3}{10} + \frac{1}{3} \cdot \frac{4}{10} + \frac{1}{3} \cdot \frac{5}{10}}$$
$$p = \frac{3}{3 + 4 + 5} = \frac{3}{12} = \frac{1}{4}$$

Similarly, the probability $q$ that the drawn Green ball is from Bag III is:
$$q = P(E_3|G) = \frac{P(E_3)P(G|E_3)}{P(E_1)P(G|E_1) + P(E_2)P(G|E_2) + P(E_3)P(G|E_3)}$$
$$q = \frac{\frac{1}{3} \cdot \frac{4}{10}}{\frac{1}{3} \cdot \frac{5}{10} + \frac{1}{3} \cdot \frac{3}{10} + \frac{1}{3} \cdot \frac{4}{10}}$$
$$q = \frac{4}{5 + 3 + 4} = \frac{4}{12} = \frac{1}{3}$$

Calculate the value of $\left(\frac{1}{p} + \frac{1}{q}\right)$:
$$\frac{1}{p} + \frac{1}{q} = \frac{1}{1/4} + \frac{1}{1/3}$$
$$\frac{1}{p} + \frac{1}{q} = 4 + 3 = 7$$

Ans. (3)