JEE Mathematics 2023 Question with Solution

Given: A bag contains $6$ white and $4$ black balls, so total balls are $10$. A fair die is rolled once.

Find: The probability that all the balls drawn are white.

Condition on the die outcome. If the die shows $k$, then $k$ balls are drawn without replacement.

The conditional probability that all $k$ balls are white is

$$P(\text{all white}\mid k)=\frac{\binom{6}{k}}{\binom{10}{k}}, \quad k\le 6$$

Now evaluate for each possible die outcome:

$$\begin{aligned} k=1:&\ \frac{\binom{6}{1}}{\binom{10}{1}}=\frac{6}{10}=\frac{3}{5} \\ k=2:&\ \frac{\binom{6}{2}}{\binom{10}{2}}=\frac{15}{45}=\frac{1}{3} \\ k=3:&\ \frac{\binom{6}{3}}{\binom{10}{3}}=\frac{20}{120}=\frac{1}{6} \\ k=4:&\ \frac{\binom{6}{4}}{\binom{10}{4}}=\frac{15}{210}=\frac{1}{14} \\ k=5:&\ \frac{\binom{6}{5}}{\binom{10}{5}}=\frac{6}{252}=\frac{1}{42} \\ k=6:&\ \frac{\binom{6}{6}}{\binom{10}{6}}=\frac{1}{210} \end{aligned}$$

Using the law of total probability,

$$P=\sum_{k=1}^{6}\frac{1}{6}\,P(\text{all white}\mid k)$$

So,

$$P=\frac{1}{6}\left(\frac{3}{5}+\frac{1}{3}+\frac{1}{6}+\frac{1}{14}+\frac{1}{42}+\frac{1}{210}\right)$$

From the given working,

$$\frac{3}{5}+\frac{1}{3}+\frac{1}{6}+\frac{1}{14}+\frac{1}{42}+\frac{1}{210}=\frac{6}{5}$$

Therefore,

$$P=\frac{1}{6}\cdot\frac{6}{5}=\frac{1}{5}$$

Thus, the required probability is $\frac{1}{5}$ and the correct option is A.

Given: The number of balls drawn depends on the die outcome.

Find: The probability that every drawn ball is white.

The quickest approach is to split the experiment into $6$ cases according to the die result and average the corresponding probabilities, because each die outcome has probability $\frac{1}{6}$.

For outcome $k$, use

$$\frac{\binom{6}{k}}{\binom{10}{k}}$$

for the probability that all drawn balls are white. Then take the average over $k=1,2,3,4,5,6$:

$$\frac{1}{6}\left(\frac{3}{5}+\frac{1}{3}+\frac{1}{6}+\frac{1}{14}+\frac{1}{42}+\frac{1}{210}\right)=\frac{1}{5}$$

This works because the die first chooses how many balls are drawn, and then the bag experiment occurs under that condition. Therefore, the correct option is A.