Given three identical bags each containing 10 balls, whose colour… — Solution | JEE Mathematics

MCQMediumJEE 2025Conditional Probability & Bayes Theorem

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
$9$
B
$7$
C
$8$
D
$AMBIGUOUS_OPTION_MISSING$

Answer

Correct answer:B

Step-by-step solution

Standard Method

Given:

Bag I: $3R, 2B, 5G$
Bag II: $4R, 3B, 3G$
Bag III: $5R, 1B, 4G$
One bag is chosen at random, then one ball is drawn.

Find: $\frac{1}{p} + \frac{1}{q}$ where $p = P(B_1\mid R)$ and $q = P(B_3\mid G)$ .

Using the probabilities from the bags,

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,

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

Therefore, the correct option is B.

Note: The listed options list appears incomplete, but the solution clearly concludes the value is $7$ .

Bayes' Theorem Interpretation

Given: A bag is chosen uniformly at random from the three bags.

Find: Conditional probabilities $P(B_1\mid R)$ and $P(B_3\mid G)$ .

For a red ball,

P(B_1\mid R) = \frac{P(B_1)P(R\mid B_1)}{P(B_1)P(R\mid B_1)+P(B_2)P(R\mid B_2)+P(B_3)P(R\mid B_3)}

Since each bag is chosen with probability $\frac{1}{3}$ ,

p = \frac{\frac{1}{3}\cdot \frac{3}{10}}{\frac{1}{3}\cdot \frac{3}{10}+\frac{1}{3}\cdot \frac{4}{10}+\frac{1}{3}\cdot \frac{5}{10}} = \frac{3}{3+4+5} = \frac{1}{4}

Similarly, for a green ball,

P(B_3\mid G) = \frac{P(B_3)P(G\mid B_3)}{P(B_1)P(G\mid B_1)+P(B_2)P(G\mid B_2)+P(B_3)P(G\mid B_3)}

So,

q = \frac{\frac{1}{3}\cdot \frac{4}{10}}{\frac{1}{3}\cdot \frac{5}{10}+\frac{1}{3}\cdot \frac{3}{10}+\frac{1}{3}\cdot \frac{4}{10}} = \frac{4}{5+3+4} = \frac{1}{3}

Hence,

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

Therefore, the required value is $7$ and the correct option is B.

Common mistakes

Ignoring that the bag is chosen at random first. This is wrong because the conditional probability must include the prior probability of choosing each bag. Use Bayes' theorem or note that equal priors cancel only after being included correctly.
Using $P(R\mid B_1) = \frac{3}{10}$ directly as $p$ . This is wrong because $p$ is the probability that the bag was Bag I given that a red ball was observed, not the probability of drawing red from Bag I.
Computing $q$ with red-ball counts instead of green-ball counts. This is wrong because the condition is that the observed ball is green. Restrict the calculation to the green counts $5, 3, 4$ .

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Step-by-step solution

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