Q.A box has $5$ blue and $4$ red balls. One ball is drawn at random and not replaced. Its colour is also not noted. Then another ball is drawn at random. What is the probability of second ball being blue?

This problem uses the Law of Total Probability to account for the unknown outcome of the first draw.

Let $B_1$ be the event that the first ball drawn is blue, and $R_1$ be the event that the first ball drawn is red. Let $B_2$ be the event that the second ball drawn is blue.

1. The initial probabilities for the first draw are $P(B_1) = \frac{5}{9}$ and $P(R_1) = \frac{4}{9}$.
2. If the first ball was blue ($B_1$), the box now contains $4$ blue and $4$ red balls. The conditional probability of the second ball being blue is $P(B_2 | B_1) = \frac{4}{8} = \frac{1}{2}$.
3. If the first ball was red ($R_1$), the box now contains $5$ blue and $3$ red balls. The conditional probability of the second ball being blue is $P(B_2 | R_1) = \frac{5}{8}$. …

Why this works: Conditional probability and the law of total probability

When you draw without replacement, the second draw depends on what happened in the first draw. But here’s the twist: you don’t know the first ball’s colour. So you can’t simply say “if the first was blue, then…” — you have to consider both possibilities, weighted by how likely each was.

This is exactly the Law of Total Probability. It says: to find the probability of an event (second ball blue), break the situation into all the mutually exclusive ways it can happen, find the probability in each case, and add them up with the right weights.

Let’s walk through it.

1. Find the probability that the first ball is blue. There are 5 blue out of 9 total balls.

2. Find the probability that the first ball is red. There are 4 red out of 9.

3. If the first ball was blue, what’s the chance the second is blue? After removing one blue, 4 blue remain out of 8 total balls.

4. If the first ball was red, what’s the chance the second is blue? After removing one red, all 5 blue remain out of 8 total balls.

5. Now combine using the law of total probability:

Compute each term: …

Method: Total Probability over an Unobserved Draw (without replacement, colour not noted)

Use this when a first item is removed but its identity is unknown, and you need a probability about a later draw.

Steps

Step 1: Split on the unknown first outcome.

List the possibilities for the first (unseen) draw and their probabilities from the original collection.

Step 2: Condition the target on each first outcome.

For each case, update the remaining collection (one fewer item, total reduced by one) and compute the target's conditional probability.

Step 3: Combine. …

Common Mistakes

Mistake 1: Treating the first ball's colour as known.

Why it's wrong: the colour is not noted, so you cannot fix the second bag's composition — you must average over both possibilities. Correct approach: use the law of total probability across "first blue" and "first red".

Mistake 2: Adding the conditionals without their weights. …