JEE Mathematics 2024 Question with Solution

Given: Urn A has $7$ red and $5$ black balls, urn B has $5$ red and $7$ black balls, and urn C has $6$ red and $6$ black balls. One urn is selected at random and the drawn ball is black.

Find: The probability that the black ball was drawn from urn A.

The solution is inconsistent with the question because it discusses only two bags and white balls. So, using the question data and Bayes' theorem,

$$P(A)=P(B)=P(C)=\frac{1}{3}$$

Also,

$$P(\text{black}\mid A)=\frac{5}{12}, \quad P(\text{black}\mid B)=\frac{7}{12}, \quad P(\text{black}\mid C)=\frac{6}{12}=\frac{1}{2}$$

Hence,

$$P(\text{black})=P(A)P(\text{black}\mid A)+P(B)P(\text{black}\mid B)+P(C)P(\text{black}\mid C)$$ $$=\frac{1}{3}\cdot\frac{5}{12}+\frac{1}{3}\cdot\frac{7}{12}+\frac{1}{3}\cdot\frac{1}{2}$$ $$=\frac{5+7+6}{36}=\frac{18}{36}=\frac{1}{2}$$

Now by Bayes' theorem,

$$P(A\mid \text{black})=\frac{P(A)P(\text{black}\mid A)}{P(\text{black})}$$ $$=\frac{\frac{1}{3}\cdot\frac{5}{12}}{\frac{1}{2}}=\frac{5}{36}\cdot 2=\frac{5}{18}$$

Therefore, the required probability is $\frac{5}{18}$, so the correct option is C. The listed option labels and the answer key string disagree because the value $\frac{5}{18}$ appears under option B, while the solution states C.

Given: Each urn is chosen with equal probability.

Find: $P(A\mid \text{black})$.

For equally likely urn selection, compare the weighted black-ball chances from each urn:

$$A: \frac{1}{3}\cdot\frac{5}{12}=\frac{5}{36}, \qquad B: \frac{1}{3}\cdot\frac{7}{12}=\frac{7}{36}, \qquad C: \frac{1}{3}\cdot\frac{1}{2}=\frac{6}{36}$$

So the total probability of getting a black ball is

$$\frac{5}{36}+\frac{7}{36}+\frac{6}{36}=\frac{18}{36}=\frac{1}{2}$$

Hence,

$$P(A\mid \text{black})=\frac{\frac{5}{36}}{\frac{1}{2}}=\frac{5}{18}$$

Therefore, the probability that the black ball came from urn A is $\frac{5}{18}$.