JEE Mathematics 2024 Question with Solution

Given: A bag has $8$ balls with some white and some black balls. $4$ balls are drawn without replacement and the observed outcome is $2$ white and $2$ black.

Find: The probability that the bag originally had equal numbers of white and black balls.

Use Bayes' theorem. Let $A$ be the event that the bag contains $4$ white and $4$ black balls, and let $E$ be the event of drawing $2$ white and $2$ black balls.

From the solution, the possible valid bag compositions are:

• $2W6B$
• $3W5B$
• $4W4B$
• $5W3B$
• $6W2B$

These are taken to be equally likely, so each has prior probability $\frac{1}{5}$.

Now,

In particular,

Using total probability,

So,

Evaluate the terms:

Hence,

Apply Bayes' theorem:

Cancelling common factors,

Therefore, the required probability is $\frac{2}{7}$. The correct option is B.

Given: The observed draw is $2$ white and $2$ black out of $4$ balls.

Find: The probability that the original bag had equal numbers of white and black balls.

Let $A$ denote the event $4W4B$, and let $E$ denote the event $2W2B$ in the draw. Then,

Since all five feasible compositions are equally likely,

Also,

Substitute these values:

Therefore, the probability that the bag contains equal numbers of white and black balls is $\frac{2}{7}$.