Q.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. Show that the probability of drawing a white ball now does not depend on $k$.

Concept: Conditional Probability — the probability of the second draw depends on the outcome of the first, so we condition on the colour of the first ball.

Let $W_1$ and $B_1$ be the events that the first ball is white or black, respectively. Let $W_2$ be the event that the second ball is white.

Step 1: Write the total probability for $W_2$:

Step 2: Substitute the probabilities. Initially there are $m$ white and $n$ black balls, so:

Step 3: After a white ball is drawn and $k$ white balls are added, the urn contains $m+k$ white and $n$ black balls. Thus:

After a black ball is drawn and $k$ black balls are added, the urn contains $m$ white and $n+k$ black balls. Thus:

$P(W_2 \mid B_1) = \frac{m}{m+n+k}.$ …

The key insight here is that the extra $k$ balls added after the first draw shift the urn’s composition, but the probability of the second draw being white ends up being exactly the same as the probability of the first draw being white. This is a classic example of exchangeability — the draws are not independent, but they are identically distributed.

Let’s see why.

1. Define the events clearly

Let:

• $W_1$ = event that the first ball drawn is white.
• $B_1$ = event that the first ball drawn is black.
• $W_2$ = event that the second ball drawn is white.

We want $P(W_2)$.

2. Use the law of total probability

The first draw determines which colour gets the $k$ extra balls. So:

We know:

• $P(W_1) = \frac{m}{m+n}$
• $P(B_1) = \frac{n}{m+n}$

3. Find the conditional probabilities

If the first ball was white, we add $k$ white balls. The urn then has:

• White: $m + k$
• Black: $n$
• Total: $m + n + k$

So:

If the first ball was black, we add $k$ black balls. The urn then has:

• White: $m$
• Black: $n + k$
• Total: $m + n + k$

So:

4. Put it together

Factor $\frac{m}{m+n}$ out of both terms:

The bracket simplifies: …

Method: Law of total probability by conditioning on the first outcome

Use this when a later event's probability depends on how an earlier random step turned out (a draw that changes the composition, a die that picks a box, etc.).

Steps

Step 1: Identify the conditioning event

Spot the earlier outcome that changes the later probabilities — here, the colour of the first ball drawn, since it decides which colour gets the $k$ extra balls.

Step 2: Split the target over the cases

$P(\text{target})=\sum_{\text{cases}} P(\text{case})\,P(\text{target}\mid\text{case}).$ …

Common Mistakes

Mistake 1: Treating the two draws as independent

Why it's wrong: the first draw changes the urn, so the second draw's probability must be conditioned on it, not read straight off the original composition. Correct approach: use total probability over the first-ball colour.

Mistake 2: Using the wrong updated total

Why it's wrong: after returning the ball plus $k$ of its colour, the urn holds $m+n+k$ balls, not $m+n$. Correct approach: recount the composition in each branch before writing the conditional. …