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Mathematics · Probability

JEE Main 2025 — 2 April, Evening Shift — Question 29

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

RedBlueGreen
Bag I325
Bag II433
Bag III514

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 pp and if the ball is Green, the probability that it is from bag III is qq, then the value of (1p+1q)\left(\frac{1}{p}+\frac{1}{q}\right) is :

  1. Option A:

    8

  2. Option B:

    9

  3. Option C:

    7

    Correct
  4. Option D:

    6

Answer: C

Step-by-step solution

p(B1/R)=p(B1)⋅p(R/B1)p(R)p\left(B_{1} / R\right)=\frac{p\left(B_{1}\right) \cdot p\left(R / B_{1}\right)}{p(R)}

=13×31013×310+13×410+13×510=14=p\begin{aligned} & =\frac{\frac{1}{3} \times \frac{3}{10}}{\frac{1}{3} \times \frac{3}{10}+\frac{1}{3} \times \frac{4}{10}+\frac{1}{3} \times \frac{5}{10}} \\& =\frac{1}{4}=p \end{aligned}

p(B3/G)=p(B3)⋅p(G/B3)p(G)p\left(B_{3} / G\right)=\frac{p\left(B_{3}\right) \cdot p\left(G / B_{3}\right)}{p(G)}

=13×41013×510+13×310+13×410=\frac{\frac{1}{3} \times \frac{4}{10}}{\frac{1}{3} \times \frac{5}{10}+\frac{1}{3} \times \frac{3}{10}+\frac{1}{3} \times \frac{4}{10}}

=13=q=\frac{1}{3}=q

∴(1p+1q)=7\therefore\left(\frac{1}{p}+\frac{1}{q}\right)=7

Answer key and solution verified before publishing.

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Exam
JEE Main 2025
Paper
2 April, Evening Shift
Subject
Mathematics
Chapter
Probability
Topic
Total Probability and Baye's Theorem