JEE Main 2025 · 2 April · evening · Single correct · +4 / −1
Given three indentical bags 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 (1/p+1/q) is:
- (a) 6
- (b) 9
- (c) 7
- (d) 8
Answer and solution
Answer (c) 7
Solution
Maths sir · 4 steps
Same chance for each bag and 10 balls in each, so Bayes reduces to counting balls of that colour.
- Bayes setup. Each bag is chosen with probability 1/3, and each bag holds 10 balls. So in Bayes' formula the factor 1/3 × 1/10 is the same in every term and cancels.
P(Bag I | Red) = (1/3·3/10)/(1/3·3/10+1/3·4/10+1/3·5/10) = 3/(3+4+5)
- Find p. There are 3 + 4 + 5 = 12 red balls in total, and 3 of them are in Bag I. So p = 3/12 = 1/4.
- Find q. There are 5 + 3 + 4 = 12 green balls in total, and 4 of them are in Bag III. So q = 4/12 = 1/3.
- So. 1/p + 1/q = 4 + 3 = 7, which is option (c).
The idea to keep. When the bags are equally likely and all hold the same number of balls, P(bag | colour) is that bag's count of the colour divided by the total count of that colour across all bags.
Answer from the NTA final answer key. Solution by PrepWiser.