Q.There are three urns containing $2$ white and $3$ black balls, $3$ white and $2$ black balls, and $4$ white and $1$ black balls, respectively. There is an equal probability of each urn being chosen. A ball is drawn at random from the chosen urn and it is found to be white. Find the probability that the ball drawn was from the second urn.

Concept: Conditional Probability (Bayes' Theorem)

Let $U_1, U_2, U_3$ be the events of choosing urn 1, 2, or 3.

$P(U_1) = P(U_2) = P(U_3) = \frac{1}{3}$.

Let $W$ be the event of drawing a white ball.

Step 1 – Likelihoods

• Urn 1: $P(W|U_1) = \frac{2}{5}$
• Urn 2: $P(W|U_2) = \frac{3}{5}$
• Urn 3: $P(W|U_3) = \frac{4}{5}$

Step 2 – Total probability of white

$P(W) = \frac{1}{3}\cdot\frac{2}{5} + \frac{1}{3}\cdot\frac{3}{5} + \frac{1}{3}\cdot\frac{4}{5} = \frac{1}{3}\cdot\frac{9}{5} = \frac{3}{5}$ …

Why Bayes’ theorem works here

We have three urns, each equally likely to be chosen. A white ball is observed. The question is: given that white happened, what’s the chance it came from urn 2? This is a classic inverse probability problem — we know the probability of white given each urn, but we need the probability of the urn given white. That’s exactly what Bayes’ theorem does: it flips the conditional probability using the prior probabilities and the likelihoods.

The intuition: each urn has a different fraction of white balls. Urn 3 has the most white (4 out of 5), so if we see white, it’s more likely to have come from urn 3 than from urn 1 (which has only 2 white out of 5). Urn 2 sits in the middle. Bayes’ theorem lets us compute the exact posterior probability.

Step-by-step solution

1. Define the events

Let $U_1$, $U_2$, $U_3$ be the events that urn 1, urn 2, or urn 3 is chosen.

Let $W$ be the event that a white ball is drawn.

2. Write the prior probabilities

Since each urn is equally likely:

3. Write the likelihoods (probability of white given each urn)

• Urn 1: 2 white, 3 black → $P(W \mid U_1) = \frac{2}{5}$
• Urn 2: 3 white, 2 black → $P(W \mid U_2) = \frac{3}{5}$
• Urn 3: 4 white, 1 black → $P(W \mid U_3) = \frac{4}{5}$

4. Apply Bayes’ theorem

We want $P(U_2 \mid W)$. Bayes’ theorem says:

The denominator $P(W)$ is the total probability of drawing a white ball, found by the law of total probability:

Substitute the values:

Factor out $\frac{1}{3}$: …

Method: Bayes' theorem for the source of an observation

Use this "a white ball was drawn — which urn did it come from?" pattern whenever you must infer the hidden source from an observed result.

Steps

Step 1: Write priors and likelihoods

List $P(U_i)$ (chance of choosing each urn) and $P(W\mid U_i)$ (chance of white from that urn's composition).

Step 2: Total probability of the observation

Step 3: Posterior via Bayes …

Common Mistakes

Mistake 1: Comparing likelihoods and picking one

Why it's wrong: choosing the "middle" urn because its likelihood $\tfrac35$ is between the others ignores normalisation. Correct approach: divide by $P(W)$ to get an actual probability.

Mistake 2: Skipping the denominator $P(W)$

Why it's wrong: without it the posteriors will not sum to $1$. Correct approach: compute $P(W)$ by total probability first. …