EQN-05-09

Compound & Conditional Probability

Multiply along a sequence of events and adjust the pool when draws are made without replacement.

~5 min read•Difficulty 3/5

The probability that several events all happen is the product of their probabilities, taking each in turn.

P (A then B) = P (A) \times P (B given A)

If the item is returned, the second probability is unchanged.

Two draws of a red from a bag with 3 red of 10, replaced :

3/10 \times 3/10 = 9/100

If the item is NOT returned, both the favourable count and the total drop for the next draw.

Two reds from 3 red of 10, not replaced :

3/10 \times 2/9 = 6/90 = 1/15

P (at least one) = 1 - P (none)

Computing "none" is usually a single short product.

Diagram

Without replacement, the second draw is out of 9, not 10.

Worked examples

Example 1 (difficulty 3/5)

A box has 4 red and 6 blue balls. Two are drawn without replacement. What is the probability both are red?

◎ Predict the answer first:

Example 2 (difficulty 3/5)

A fair die is rolled twice. What is the probability of at least one 6?

◎ Predict the answer first:

Keeping the same denominator on the second draw without replacement

Why: The first fraction is reused.

Fix: Removing an item lowers the total (and the matching count) for the next draw.

Adding probabilities for a sequence of events

Why: "And then" feels additive.

Fix: Sequential "and" events multiply.