9 min readConcept sheetPart II (2 marks) and Part III (5 marks)

Addition theorem of probability

By the addition theorem, P(A ∪ B) = P(A) + P(B) − P(A ∩ B). Subtract the overlap so shared outcomes count once; for mutually exclusive events, simply add.

Addition theoremP(A ∪ B) = P(A) + P(B) − P(A ∩ B)

Mutually exclusive eventsP(A ∪ B) = P(A) + P(B)

Neither A nor B1 − P(A ∪ B)

Three eventsAdd the three single events, subtract the three pairs, add the triple

The addition theorem

For any two events A and B: P(A ∪ B) = P(A) + P(B) − P(A ∩ B).
We subtract P(A ∩ B) because outcomes in both A and B are counted twice in P(A) + P(B).
If A and B are mutually exclusive, P(A ∩ B) = 0, so P(A ∪ B) = P(A) + P(B).
For three events: P(A ∪ B ∪ C) = P(A) + P(B) + P(C) − P(A ∩ B) − P(B ∩ C) − P(A ∩ C) + P(A ∩ B ∩ C).
P(neither A nor B) = 1 − P(A ∪ B).

Why the theorem is true

1
A ∪ B = (A ∩ B̄) ∪ B, and A ∩ B̄ and B have no common outcome.
2
For mutually exclusive events, probabilities add: P(A ∪ B) = P(A ∩ B̄) + P(B).
3
From the algebra of events, P(A ∩ B̄) = P(A) − P(A ∩ B).
4
Substitute: P(A ∪ B) = P(A) − P(A ∩ B) + P(B) = P(A) + P(B) − P(A ∩ B).

Example 1: a king or a heart

Problem

A card is drawn from a well-shuffled pack of 52 cards. Find the probability that it is a king or a heart.

Working

A: a king, n(A) = 4, so P(A) = 4/52.
B: a heart, n(B) = 13, so P(B) = 13/52.
A ∩ B: the king of hearts, so P(A ∩ B) = 1/52.
P(A ∪ B) = P(A) + P(B) − P(A ∩ B) = 4/52 + 13/52 − 1/52
= 16/52 = 4/13

Answer

4/13

Example 2: cricket and kabaddi

Problem

In a class of 40 students, 22 play cricket, 15 play kabaddi and 6 play both. A student is chosen at random. Find the probability that the student plays cricket or kabaddi, and the probability that the student plays neither.

Working

P(C) = 22/40, P(K) = 15/40, P(C ∩ K) = 6/40.
P(C ∪ K) = 22/40 + 15/40 − 6/40 = 31/40
P(neither) = 1 − 31/40 = 9/40

Answer

P(cricket or kabaddi) = 31/40; P(neither) = 9/40

Example 3: three events

Problem

P(A) = 0.5, P(B) = 0.4, P(C) = 0.3, P(A ∩ B) = 0.2, P(B ∩ C) = 0.1, P(A ∩ C) = 0.15 and P(A ∩ B ∩ C) = 0.05. Find P(A ∪ B ∪ C).

Working

Formula: P(A ∪ B ∪ C) = P(A) + P(B) + P(C) − P(A ∩ B) − P(B ∩ C) − P(A ∩ C) + P(A ∩ B ∩ C)
= 0.5 + 0.4 + 0.3 − 0.2 − 0.1 − 0.15 + 0.05
= 1.2 − 0.45 + 0.05
= 0.8

Answer

P(A ∪ B ∪ C) = 0.8

In real life

Surveys of tea and filter coffee drinkers use the addition theorem, so that people who drink both are not counted twice.

Exam angle

1 marks: “A and B are mutually exclusive, with P(A) = 1/3 and P(B) = 1/4. Find P(A ∪ B).” P(A ∪ B) = 1/3 + 1/4 = 7/12.

2 marks: “If P(A) = 0.6, P(B) = 0.45 and P(A ∪ B) = 0.8, find P(A ∩ B).” P(A ∩ B) = P(A) + P(B) − P(A ∪ B) = 0.6 + 0.45 − 0.8 = 0.25.

5 marks: “Two dice are rolled. Find the probability that the sum is 8 or both dice show the same number.” A: sum 8, 5 outcomes. B: doublet, 6 outcomes. A ∩ B = {(4, 4)}. P(A ∪ B) = 5/36 + 6/36 − 1/36 = 10/36 = 5/18.

Remember it

Add, then take away the overlap: P(A) + P(B) − P(A ∩ B). No overlap? Just add.