The Addition Rule for Overlapping Events

Two events that can happen together

Roll a fair dice once. Let A be "rolls an even number", the faces 2, 4 and 6, and let B be "rolls a multiple of 3", the faces 3 and 6. A roll of 6 is in both events, so A and B can happen together: they are not mutually exclusive.

Place the faces in a Venn diagram. The face 6 goes in the overlap. The faces 2 and 4 are even but not multiples of 3, so they go in A alone; the face 3 goes in B alone; and the faces 1 and 5 are in neither event. Check: 2 + 1 + 1 + 2 = 6.

The 6 faces of a dice: 2 even only, the face 6 in both, 1 multiple of 3 only, and 2 in neither.

Adding counts the overlap twice

What is the chance of A or B, a roll that is even or a multiple of 3? Adding the two chances gives $\frac{3}{6}+\frac{2}{6}=\frac{5}{6}$3/6 + 2/6 = 5/6. But count the faces inside either circle: 2, 3, 4 and 6. That is 4 faces, not 5.

The face 6 is one of the 3 even faces and also one of the 2 multiples of 3, so adding the two counts counted it twice.

Everything inside either circle: 2 + 1 + 1 = 4 faces, though the circle totals 3 and 2 add to 5.

Take the overlap off once

To count each face once, subtract the shared face once: 3 + 2 − 1 = 4. So the probability of A or B is $\frac{4}{6}=\frac{2}{3}$4/6 = 2/3.

In chances: $\frac{3}{6}+\frac{2}{6}-\frac{1}{6}=\frac{4}{6}$3/6 + 2/6 − 1/6 = 4/6. The $\frac{1}{6}$1/6 taken off is the chance of the overlap, a roll that is even and a multiple of 3.

The overlap holds the face 6 only. It is inside both circle totals, so it comes off once.

The addition rule

For any two events A and B, P(A or B) = P(A) + P(B) − P(A and B). Add both chances, then subtract the chance of both happening, because those outcomes were inside P(A) and inside P(B).

When A and B are mutually exclusive, P(A and B) = 0 and nothing is subtracted, so the rule becomes P(A or B) = P(A) + P(B), the rule for mutually exclusive events. The addition rule covers both cases.

A heart or a king

A card is drawn at random from a standard pack of 52. What is the probability that it is a heart or a king? There are 13 hearts and 4 kings, and the king of hearts is both.

So $P\left(\text{heart or king}\right)=\frac{13}{52}+\frac{4}{52}-\frac{1}{52}=\frac{16}{52}=\frac{4}{13}$P(heart or king) = 13/52 + 4/52 − 1/52 = 16/52 = 4/13. Count the cards to check: 12 hearts that are not kings, the king of hearts, and 3 other kings make 12 + 1 + 3 = 16 cards.

The 52 cards: 12 hearts that are not kings, the king of hearts in the overlap, 3 other kings, and 36 cards in neither. The 16 inside the circles are shaded.

From the chances alone

Often only the chances are given. If $P\left(A\right)=\frac{7}{20}$P(A) = 7/20 and $P\left(B\right)=\frac{5}{20}$P(B) = 5/20, and the chance of both is $\frac{2}{20}$2/20, then the chance of A or B is $\frac{7}{20}+\frac{5}{20}-\frac{2}{20}=\frac{10}{20}=\frac{1}{2}$7/20 + 5/20 − 2/20 = 10/20 = 1/2.

The rule also runs the other way. If P(A) = 0.6, P(B) = 0.5 and P(A or B) = 0.8, then 0.8 = 0.6 + 0.5 − P(A and B), so P(A and B) = 1.1 − 0.8 = 0.3. Two chances that add to more than 1, like 0.6 and 0.5, belong to events that must overlap.

The chance of neither event is everything outside both circles: P(neither) = 1 − P(A or B). On the dice that is $1-\frac{4}{6}=\frac{2}{6}$1 − 4/6 = 2/6, the faces 1 and 5.

Out of 20 equal parts: A holds 5 + 2 = 7, B holds 2 + 3 = 5, and the shaded union holds 5 + 2 + 3 = 10.

The usual mistakes

Adding the two chances when the events overlap: $\frac{7}{20}+\frac{5}{20}=\frac{12}{20}$7/20 + 5/20 = 12/20. The 2 shared parts are inside P(A) and inside P(B), so they are counted twice.

Subtracting the overlap twice, once from each event: 7 + 5 − 2 − 2 = 8, giving $\frac{8}{20}$8/20. The shared outcomes are still in A or B, so they belong in the answer once. On the dice, 3 + 2 − 2 = 3 would leave the face 6 out of "even or a multiple of 3".

Using P(A) + P(B) without checking whether the events can happen together. That shortcut is only for mutually exclusive events.

A raffle

In the application below, a ticket from 1 to 30 wins a prize for a multiple of 3 or a multiple of 5. The tickets 15 and 30 are in both lists, so the addition rule takes them off once.