Maths Q9 – JEE-MAIN 2025

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

\begin{array}{cccc} Red & Blue & Green \\ Bag I & 3 & 2 & 5 \\ Bag II & 4 & 3 & 3 \\ Bag III & 5 & 1 & 4 \end{array}

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 :

Clue 1 of 3: Understand the problem setup

This problem involves conditional probability, where the probability of an event (choosing a specific bag) is calculated given that another event (drawing a specific color ball) has already occurred.

🧠 Full Solution Path

Step 1: Calculate Total Probabilities of Drawing Red and Green Balls✦ Active

Let $B_{1}, B_{2}, B_{3}$ be the events of choosing Bag I, Bag II, and Bag III respectively. Since a bag is chosen at random, $P (B_{1}) = P (B_{2}) = P (B_{3}) = \frac{1}{3}$ . Each bag contains 10 balls. Let $R$ be the event of drawing a Red ball and $G$ be the event of drawing a Green ball.

P (R | B_{1}) = \frac{3}{10}, P (R | B_{2}) = \frac{4}{10}, P (R | B_{3}) = \frac{5}{10}

P (G | B_{1}) = \frac{5}{10}, P (G | B_{2}) = \frac{3}{10}, P (G | B_{3}) = \frac{4}{10}

P (R) = P (R | B_{1}) P (B_{1}) + P (R | B_{2}) P (B_{2}) + P (R | B_{3}) P (B_{3})

P (R) = (\frac{3}{10}) (\frac{1}{3}) + (\frac{4}{10}) (\frac{1}{3}) + (\frac{5}{10}) (\frac{1}{3}) = \frac{1}{30} (3 + 4 + 5) = \frac{12}{30} = \frac{2}{5}

P (G) = P (G | B_{1}) P (B_{1}) + P (G | B_{2}) P (B_{2}) + P (G | B_{3}) P (B_{3})

P (G) = (\frac{5}{10}) (\frac{1}{3}) + (\frac{3}{10}) (\frac{1}{3}) + (\frac{4}{10}) (\frac{1}{3}) = \frac{1}{30} (5 + 3 + 4) = \frac{12}{30} = \frac{2}{5}

Step 2: Calculate Probabilities p and q using Bayes' Theorem○ Expand

The probability $p$ is $P (B_{1} | R)$ , and the probability $q$ is $P (B_{3} | G)$ . Using Bayes' Theorem:

p = P (B_{1} | R) = \frac{P (R | B_{1}) P (B_{1})}{P (R)} = \frac{(\frac{3}{10}) (\frac{1}{3})}{\frac{2}{5}} = \frac{\frac{3}{30}}{\frac{2}{5}} = \frac{1}{10} \times \frac{5}{2} = \frac{5}{20} = \frac{1}{4}

q = P (B_{3} | G) = \frac{P (G | B_{3}) P (B_{3})}{P (G)} = \frac{(\frac{4}{10}) (\frac{1}{3})}{\frac{2}{5}} = \frac{\frac{4}{30}}{\frac{2}{5}} = \frac{2}{15} \times \frac{5}{2} = \frac{10}{30} = \frac{1}{3}

💡 Teacher's Secret Hint

Ensure to use the correct conditional probabilities from the table for each specific bag and ball color.

Step 3: Calculate the Final Expression○ Expand

Now, substitute the values of $p$ and $q$ into the expression $(\frac{1}{p} + \frac{1}{q})$ :

\frac{1}{p} = \frac{1}{\frac{1}{4}} = 4

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

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

💡 Teacher's Secret Hint

Double-check the arithmetic for the reciprocals and the final sum.

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Maths Question 9 – JEE-MAIN 2025

Clue 1 of 3: Understand the problem setup

🧠 Full Solution Path