An urn contains m white and n black balls. A ball is drawn at random and is put back into the urn al

Q: Wait, why does the parameter $k$ completely vanish from the final answer? It feels like adding balls should change the probability!

That is the beautiful, counter-intuitive magic of Polya's Urn! While $k$ does change the composition of the urn, it does so in a way that perfectly balances the probabilities. When you draw a white ball first, you increase the chance of drawing white again, but you also increase the total number of balls. When you draw a black ball first, you increase the chance of drawing black again. These two effects cancel each other out perfectly in the summation. The initial ratio of white to black balls is the only thing that dictates the long-term probability.

Q: Can I just assume the probability is $\frac{m}{m+n}$ without doing the calculation? It seems too simple.

In an exam, never assume! While your intuition is correct here, the JEE Advanced examiners want to see the rigorous application of the Law of Total Probability. The 'why' is just as important as the 'what'. Showing the step-by-step derivation proves you understand the underlying mechanism of conditional probability, which is a core skill for higher-level physics and statistics.

Q: What if we had to draw a third ball? Would the probability still be $\frac{m}{m+n}$?

Yes, it would! This is a fascinating property of Polya's Urn model. The probability of drawing a white ball on any draw $n$ remains $\frac{m}{m+n}$. It is a stationary process in terms of marginal probability. It is a great exercise to try and prove this for the third draw using the same method—you will see the algebra simplifies just as elegantly.

Question

An urn contains $m$ white and $n$ black balls. A ball is drawn at random and is put back into the urn along with $k$ additional balls of the same colour as that of the ball drawn. A ball is again drawn at random. What is the probability that the ball drawn now is white?

Sigma Percentile · Accepted Answer

### The Urn of Mystery: A Journey into Polya's Urn

Imagine you are standing before an ancient urn. Inside, there are $m$ white balls and $n$ black balls, waiting for your hand to reach in.

This is not just a probability problem; it is a story of dependency and balance. When we draw a ball and replace it along with $k$ additional balls of the same color, we are engaging in a classic experiment known as Polya's Urn.

It feels like the urn is 'learning' from our choices, doesn't it? Let's unravel this mystery together.

### Phase 1: The Branching Reality

We want to know the probability that the second ball drawn is white, which we denote as $P(W_2)$. But here is the catch: the second draw is not independent.

It is shackled to the outcome of the first draw. If we draw a white ball first, the urn becomes 'whiter'. If we draw a black ball first, it becomes 'blacker'.

To handle this, we use the **Law of Total Probability**. Think of this as a bridge between two possible realities. We partition our sample space into two mutually exclusive events: the first ball is white ($W_1$) or the first ball is black ($B_1$).

The total probability is the sum of these two paths:

$$P(W_2) = P(W_1) \cdot P(W_2|W_1) + P(B_1) \cdot P(W_2|B_1)$$

### Phase 2: Walking the Paths

Let's walk down the first path. The probability of drawing a white ball initially is simply the number of white balls over the total: $P(W_1) = \frac{m}{m+n}$.

Now, we add $k$ white balls. Our new urn composition is $(m+k)$ white balls and $n$ black balls, with a new total of $(m+n+k)$.

Thus, the conditional probability of drawing white again is:

$$P(W_2|W_1) = \frac{m+k}{m+n+k}$$

Now, let's walk the second path. The probability of drawing a black ball initially is $P(B_1) = \frac{n}{m+n}$.

We add $k$ black balls. The urn now holds $m$ white balls and $(n+k)$ black balls, with the same total $(m+n+k)$.

The conditional probability of drawing a white ball here is:

$$P(W_2|B_1) = \frac{m}{m+n+k}$$

### Phase 3: The Algebraic Revelation

Now, we bring these pieces together into our master equation:

$$P(W_2) = \left( \frac{m}{m+n} \cdot \frac{m+k}{m+n+k} \right) + \left( \frac{n}{m+n} \cdot \frac{m}{m+n+k} \right)$$

Take a deep breath. I know this looks like a messy fraction, but look closely at the denominators. They are identical!

We can combine the numerators over the common denominator $(m+n)(m+n+k)$:

$$P(W_2) = \frac{m(m+k) + nm}{(m+n)(m+n+k)}$$

Here is where most students panic and start expanding everything. Don't! Look at the numerator: $m(m+k) + nm$.

Let's pull out the common factor of $m$:

$$P(W_2) = \frac{m(m+k+n)}{(m+n)(m+n+k)}$$

### The Final Symmetry

Do you see it? The term $(m+n+k)$ is sitting in both the numerator and the denominator, waiting to be cancelled.

When we strike it out, we are left with the most elegant result imaginable:

$$P(W_2) = \frac{m}{m+n}$$

It is the same as the initial probability! The parameter $k$, which seemed so crucial, has vanished.

This is the beauty of Polya's Urn. It teaches us that while individual events are influenced by the past, the underlying probability structure maintains a beautiful, persistent symmetry.

You have just mastered a fundamental concept in stochastic processes. Keep this intuition with you—it will serve you well in your journey through **JEE Advanced** physics and mathematics.

Visualized Solution

Initial State of the Urn

Total Probability Theorem

Case 1: Drawing a White Ball First

Updating the Urn (Case 1)

Second Draw (Case 1)

Case 2: Drawing a Black Ball First

Updating the Urn (Case 2)

Second Draw (Case 2)

Substituting the Probabilities

Combining the Terms

Factoring the Numerator

Canceling Common Terms

Final Conclusion

The Sigma Insight: Total Probability Theorem

The Urn of Mystery

Phase 1

Phase 2

Phase 3

The Final Symmetry

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