Q.Three cards are drawn successively, without replacement from a pack of 52 well shuffled cards. What is the probability that first two cards are kings and the third card drawn is an ace?

Concept: Conditional Probability — the probability of a sequence of dependent events is the product of the conditional probabilities at each step.

Step 1: Probability first card is a king:

$\frac{4}{52} = \frac{1}{13}$.

Step 2: Given first is a king, probability second is also a king (3 kings left in 51 cards):

$\frac{3}{51} = \frac{1}{17}$. …

The key here is that the draws are without replacement — each draw changes the deck. So the probability of the second event depends on what happened first, and the third depends on both previous draws. This is exactly what conditional probability handles: $P(A \cap B \cap C) = P(A) \cdot P(B|A) \cdot P(C|A \cap B)$.

Let’s walk through it.

1. First card is a king. There are 4 kings in a deck of 52 cards.

2. Second card is a king, given the first was a king. After removing one king, 3 kings remain in a deck of 51 cards.

3. Third card is an ace, given the first two were kings. Two kings are gone, but no aces have been drawn yet — all 4 aces remain. The deck now has 50 cards.

4. Multiply the chain.

Simplify step by step: …

Method: The chain multiplication rule for successive dependent draws

Use this for three or more cards/objects drawn without replacement, where you need a specific outcome at each position.

Steps

Step 1: Expand the joint probability as a full chain.

Each conditional is evaluated after the earlier draws have been removed.

Step 2: Track how each draw changes both numerator and denominator. …

Common Mistakes

Mistake 1: Reusing the same fractions as if the deck were replaced.

Why it's wrong: the draws are without replacement, so the deck shrinks from $52$ to $51$ to $50$ and counts change each time. Correct approach: use $\tfrac{4}{52}\times\tfrac{3}{51}\times\tfrac{4}{50}$.

Mistake 2: Reducing the ace count after drawing the kings. …