9 min readConcept sheet2 or 3-mark question; often inside 4-mark case-based questions

Multiplication theorem on probability

The probability that E and F both happen is P(E ∩ F) = P(E) · P(F|E). For three events, keep multiplying: P(E) · P(F|E) · P(G|(E ∩ F)).

Two eventsP(E ∩ F) = P(E) · P(F|E), where P(E) ≠ 0

The same result the other wayP(E ∩ F) = P(F) · P(E|F), where P(F) ≠ 0

Three eventsP(E ∩ F ∩ G) = P(E) · P(F|E) · P(G|(E ∩ F))

Typical useDrawing items one after another without replacement

Where the theorem comes from

1
Definition: P(F|E) = P(E ∩ F)/P(E), with P(E) ≠ 0.
2
Multiply both sides by P(E): P(E ∩ F) = P(E) · P(F|E).
3
Swapping the roles of E and F: P(E ∩ F) = P(F) · P(E|F).
4★
For three events, apply the rule twice: P(E ∩ F ∩ G) = P(E ∩ F) · P(G|(E ∩ F)) = P(E) · P(F|E) · P(G|(E ∩ F)).

★ key step

Example 1: sweets without replacement

Problem

A sweet box has 5 laddoos and 3 jalebis. Two sweets are taken one after the other without replacement. Find the probability that both are laddoos.

Working

E: the first is a laddoo. P(E) = 5/8.
F: the second is a laddoo. After one laddoo is gone, 4 of the 7 left are laddoos, so P(F|E) = 4/7.
P(E ∩ F) = 5/8 × 4/7 = 20/56 = 5/14

Answer

5/14

Example 2: three events

Problem

A box of 12 bulbs has 4 defective ones. Three bulbs are taken one by one without replacement. Find the probability that all three are good.

Working

Good bulbs: 12 − 4 = 8.
P(1st good) = 8/12
P(2nd good | 1st good) = 7/11
P(3rd good | first two good) = 6/10
P = 8/12 × 7/11 × 6/10 = 336/1320 = 14/55

Answer

14/55

Example 3: order matters

Problem

From the same box, two bulbs are taken without replacement. Find the probability that the first is defective and the second is good.

Working

P(1st defective) = 4/12
P(2nd good | 1st defective) = 8/11
P = 4/12 × 8/11 = 32/132 = 8/33

Answer

8/33

With replacement vs without replacement

With replacementThe item goes back. The second draw has the same chances as the first, so the draws are independent: P = P(E) · P(F).

Without replacementThe item stays out. The chances for the second draw change, so use P(E) · P(F|E).

Two aces from a pack of 52: with replacement, (4/52)² = 1/169; without replacement, 4/52 × 3/51 = 1/221.

In real life

Quality inspectors in a factory test items one after another without putting them back. The chance that every sample passes comes from the multiplication theorem.

Exam angle

2 marks: “Two cards are drawn one after another without replacement from a pack of 52. Find the probability that both are hearts.” 13/52 × 12/51 = 1/4 × 12/51 = 12/204 = 1/17.

1 marks: “If P(A) = 0.4 and P(B|A) = 0.5, find P(A ∩ B).” P(A ∩ B) = P(A) · P(B|A) = 0.4 × 0.5 = 0.2.

3 marks: “A bag has 5 red and 4 blue balls. Three balls are drawn one by one without replacement. Find the probability that the first is red, the second blue and the third red.” 5/9 × 4/8 × 4/7 = 80/504 = 10/63.

Remember it

'And' means multiply, and each step uses what is left.