Multiplication Rule: P(A and B) for Joint Events

The multiplication rule

For any two events $A$ and $B$: $P(A \cap B) = P(A) \cdot P(B \mid A)$

In words: the chance both happen is the chance of the first times the chance of the second given that the first happened.

If $A$ and $B$ are independent, $P(B \mid A) = P(B)$ and the rule simplifies to $P(A \cap B) = P(A) \cdot P(B).$

Step-by-step solution (without replacement)

Step 1 — First ace. Four aces in 52 cards: $P(A_1) = \dfrac{4}{52} = \dfrac{1}{13}$

Step 2 — Second ace, given the first was an ace. One ace already taken; three aces remain among 51 cards: $P(A_2 \mid A_1) = \dfrac{3}{51} = \dfrac{1}{17}$

Step 3 — Multiply: $P(A_1 \cap A_2) = \dfrac{4}{52} \cdot \dfrac{3}{51} = \dfrac{12}{2652} = \boxed{\dfrac{1}{221}}$

Numerically, $\approx 0.00452$, i.e. about 0.45% — roughly 1 in 221.

Comparison — with replacement

If we put the first card back and reshuffle, the two draws are independent, so $P(A_1 \cap A_2) = \dfrac{4}{52} \cdot \dfrac{4}{52} = \dfrac{1}{169} \approx 0.00592$

That's about 1 in 169 — slightly more likely than the without-replacement version, because you keep all 4 aces available for the second draw.

Verification by counting

Without replacement, the number of ordered 2-card hands that are both aces is $4 \cdot 3 = 12$. Total ordered 2-card hands: $52 \cdot 51 = 2652$. Ratio: $12/2652 = 1/221$. ✓

Generalizing to $n$ events

$P(A_1 \cap A_2 \cap \cdots \cap A_n) = P(A_1) \cdot P(A_2 \mid A_1) \cdot P(A_3 \mid A_1 \cap A_2) \cdots$

This "chain rule" is the backbone of many probability arguments, including Bayes updates and tree-diagram reasoning.

Common mistakes

• Treating draws without replacement as independent. The second denominator should shrink. Using $4/52 \cdot 4/52$ for without-replacement draws is a classic error.
• Mis-counting after removal. Once an ace is drawn, both the ace count (numerator) and the total card count (denominator) drop by 1.
• Mixing up "and" with "or." "And" multiplies (joint event); "or" adds (with overlap correction). See the addition-rule problem.

Try it in the visualization

Toggle between with- and without-replacement modes. A shrinking deck illustrates why the denominator drops when the drawn card is removed. The product $P(A_1) \cdot P(A_2 \mid A_1)$ is assembled factor by factor at the bottom of the panel.