Given three identical balls each containing 10 balls, whose colours are as follows: Red Blue Green Bag I 3 2 5 Bag II 4 3

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Solution

Q.
Given three identical balls each containing 10 balls, whose colours are as follows:

Red Blue Green

Bag I 3 2 5

Bag II 4 3 3

Bag III 5 1 4

A person chooses a bag at random and takes out a ball. If the ball is Red, the probability that it is from bag I is p and if the ball is Green, the probability that it is from bag III is q, then the value of $(\frac{1}{p} + \frac{1}{q})$ is : [2025]

1 6

2 9

3 7

4 8

Ans.
(3)

Let us define the Events

A : Ball drawn from bag I

B : Ball drawn from bag II

C : Ball drawn from bag III

R : Red ball is drawn

G : Green ball is drawn

$∴ P (R) = P (A) \cdot P (R / A) + P (B) \cdot P (R / B) + P (C) \cdot P (R / C)$

$= \frac{1}{3} \times \frac{3}{10} + \frac{1}{3} \times \frac{4}{10} + \frac{1}{3} \times \frac{5}{10} = \frac{12}{30}$

Now, $p = P (A / R) = \frac{P (A) P (R / A)}{P (R)} = \frac{\frac{1}{3} \times \frac{3}{10}}{\frac{12}{30}} = \frac{3}{12} = \frac{1}{4}$

Now, $P (G) = \frac{1}{3} \times \frac{5}{10} + \frac{1}{3} \times \frac{3}{10} + \frac{1}{3} \times \frac{4}{10} = \frac{12}{30}$

$∴ q = P (C / G) = \frac{P (C) \cdot P (G / C)}{P (G)} = \frac{\frac{1}{3} \times \frac{4}{10}}{\frac{12}{30}} = \frac{4}{12} = \frac{1}{3}$

$∴ \frac{1}{p} + \frac{1}{q} = 4 + 3 = 7$ .

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