JEE Mathematics 2025 Question with Solution

Given: Bag $1$ contains $4$ white and $5$ black balls. Bag $2$ contains $n$ white and $3$ black balls. One ball is transferred from Bag $1$ to Bag $2$, then one ball is drawn from Bag $2$.

Find: The value of $n$ given that the probability of drawing a white ball from Bag $2$ is $\frac{29}{45}$.

Use total probability.

Probability of transferring a white ball from Bag $1$:

$$P(E_1) = \frac{4}{9}$$

Probability of transferring a black ball from Bag $1$:

$$P(E_2) = \frac{5}{9}$$

If a white ball is transferred, Bag $2$ has $n+1$ white balls and total $n+4$ balls, so

$$P(\text{White from Bag 2} \mid E_1) = \frac{n+1}{n+4}$$

If a black ball is transferred, Bag $2$ has $n$ white balls and total $n+4$ balls, so

$$P(\text{White from Bag 2} \mid E_2) = \frac{n}{n+4}$$

Therefore,

$$P(\text{White}) = P(E_1) \times P(\text{White} \mid E_1) + P(E_2) \times P(\text{White} \mid E_2)$$ $$\frac{29}{45} = \frac{4}{9} \times \frac{n+1}{n+4} + \frac{5}{9} \times \frac{n}{n+4}$$ $$\frac{29}{45} = \frac{1}{9(n+4)}\left[4(n+1) + 5n\right] = \frac{9n+4}{9(n+4)}$$

Multiply both sides by $9(n+4)$:

$$\frac{29}{45} \times 9(n+4) = 9n+4$$ $$\frac{29}{5}(n+4) = 9n+4$$

Multiply by $5$:

$$29(n+4) = 45n+20$$ $$29n+116 = 45n+20$$ $$96 = 16n$$ $$n = 6$$

Therefore, the correct option is D.

Given: The transferred ball from Bag $1$ can be either white or black.

Find: The value of $n$.

Break the experiment into two cases.

1. White transferred from Bag $1$

• Probability of this case:

$$\frac{4}{9}$$

• Then Bag $2$ contains $n+1$ white and $3$ black balls.
• Probability of drawing white from Bag $2$ in this case:

$$\frac{n+1}{n+4}$$

2. Black transferred from Bag $1$

• Probability of this case:

$$\frac{5}{9}$$

• Then Bag $2$ contains $n$ white and $4$ black balls.
• Probability of drawing white from Bag $2$ in this case:

$$\frac{n}{n+4}$$

Now add the two weighted probabilities:

$$\frac{29}{45} = \frac{4}{9}\cdot\frac{n+1}{n+4} + \frac{5}{9}\cdot\frac{n}{n+4}$$ $$\frac{29}{45} = \frac{4n+4+5n}{9(n+4)}$$ $$\frac{29}{45} = \frac{9n+4}{9(n+4)}$$

Cross-multiplying:

$$29 \cdot 9(n+4) = 45(9n+4)$$

This simplifies to the same linear equation shown in the solution working:

$$29(n+4) = 45n+20$$ $$29n+116 = 45n+20$$ $$96 = 16n$$ $$n = 6$$

Hence, the required value is $6$, so the correct option is D.