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:
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:
Show Hint
- 6
- 9
- 7
- 8
The Correct Option is C
Solution and Explanation
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: \[ \frac{1}{p} + \frac{1}{q} = 7 \]
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