The multiplication rule

A bag contains 5 red and 3 blue marbles. You draw two without replacement. What's the probability both are red?

What do you think?

You draw one red marble, then another. Is the second draw's probability affected by the first?

From definition to formula

Rearranging the definition of conditional probability gives the multiplication rule.

Deriving the Multiplication Rule

P(A|B) = \frac{P(A \cap B)}{P(B)}

Start with the definition of conditional probability.

Step 1 of 3

Multiplication Rule

$P(A \cap B) = P(B) \cdot P(A|B) = P(A) \cdot P(B|A)$

The probability that both events occur equals one event's probability times the other's conditional probability.

See the area

The multiplication rule has a clean geometric interpretation. $P(B)$ is a width, $P(A|B)$ is a height, and $P(A \cap B)$ is the area of the rectangle they form.

Drag the sliders and watch how the shaded region — the joint probability — depends on both factors.

Multiplication Rule — Area Model

The shaded area is P(A∩B) = P(B) × P(A|B)

P(A∩B) = P(B) × P(A|B) = 0.40 × 0.60 = 0.240

P(B): 0.40

P(A|B): 0.60

What do you think?

If P(B) = 0.5 and P(A|B) = 0.5, what is P(A∩B)?

decimal, e.g. 0.42

The chain rule

The multiplication rule extends to three or more events. Each factor conditions on everything that came before.

The Chain Rule

P(A \cap B \cap C) = P(A) \cdot P(B|A) \cdot P(C|A \cap B)

Three events: start with A, condition B on A, condition C on both A and B.

Step 1 of 2

Try building a chain yourself. Draw aces one at a time without replacement and watch each conditional probability shrink.

Chain Rule Builder

Draw aces one at a time, without replacement. Watch each conditional probability shrink.

Solving the marble problem

Back to the original puzzle: 5 red and 3 blue marbles, draw two without replacement.

P(both red)

P(R_1) = \frac{5}{8}

5 red out of 8 total marbles.

Step 1 of 3

What do you think?

Three cards are drawn without replacement from a 52-card deck. What is P(all three are hearts)?

decimal to 3 places, e.g. 0.042

Intersection vs independence

The multiplication rule is general. It works whether events are independent or not. When they are independent, it simplifies.

Relationship	Formula
General	$P(A \cap B) = P(A) \cdot P(B\\|A)$
Independent	$P(A \cap B) = P(A) \cdot P(B)$

Independence is the special case where conditioning on A doesn't change the probability of B — so $P(B|A)$ collapses to $P(B)$ .

P(A) = 0.6, P(B|A) = 0.3. What is P(A ∩ B)? (decimal, e.g. 0.42)

P(Rain) = 0.3, P(Traffic | Rain) = 0.8, P(Late | Rain ∩ Traffic) = 0.9. What's P(all three)? (decimal, e.g. 0.042)

A bag has 4 green, 6 yellow balls. Draw 2 without replacement. P(both yellow)? (decimal, e.g. 0.042)

Test your understanding

P(A ∩ B) = 0.12 and P(B) = 0.4. What is P(A|B)? (decimal, e.g. 0.42)

1/2

What's next

The multiplication rule tells you how to compute $P(A \cap B)$ . But what if you need $P(A)$ itself, and A is entangled with several different scenarios?

The Law of Total Probability partitions the sample space into cases and adds up the contributions. It's the key to computing the denominator of Bayes' Rule.