Question

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

Bag I	3 Red	2 Blue	5 Green
Bag II	4 Red	3 Blue	3 Green
Bag III	5 Red	1 Blue	4 Green

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:

• A. 6
• B. 9
• C. 7
• D. 8

Hint:

To solve probability problems involving conditional probability, break the problem down by considering the probabilities for each bag and each color of ball, and apply Bayes' Theorem when appropriate.

Answer 1

The Correct Option is C

Bag I: \( 3R, 2B, 5G \) Bag II: \( 4R, 3B, 3G \) Bag III: \( 5R, 1B, 4G \) 
We need to calculate \( p \) and \( q \). \[ p = P\left( \frac{B_1}{R} \right) = \frac{\frac{3}{10}}{\frac{3}{10} + \frac{4}{10} + \frac{5}{10}} = \frac{1}{4} \] \[ q = P\left( \frac{B_3}{G} \right) = \frac{\frac{4}{10}}{\frac{5}{10} + \frac{3}{10} + \frac{4}{10}} = \frac{1}{3} \] Thus, we get: \[ \frac{1}{p} + \frac{1}{q} = 7 \]