Addition Rule for Probability

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The addition rule for probability tells you how to find the probability that at least one of two events happens: event A, event B, or both. In probability, "A or B" always includes the case where both happen.

P(A or B) = P(A) + P(B) − P(A and B)

In set notation: P(A ∪ B) = P(A) + P(B) − P(A ∩ B). The symbol ∪ means "or" (union) and ∩ means "and" (intersection).

Why do we subtract P(A and B)?

When you add P(A) and P(B), the outcomes that belong to both events get counted twice: once as part of A and once as part of B. Subtracting P(A and B) removes the extra copy, so every outcome is counted exactly once.

Here is a quick example with a standard deck of 52 cards. What is the probability of drawing a king or a heart?

  • There are 4 kings, so P(king) = 4/52.
  • There are 13 hearts, so P(heart) = 13/52.
  • The king of hearts is both a king and a heart, so it was counted twice: P(king and heart) = 1/52.

P(king or heart) = 4/52 + 13/52 − 1/52 = 16/52 = 4/13, or about 0.31.

Without the subtraction you would get 17/52, which counts the king of hearts twice.

Worked example: gifted students

The table below shows 27,000 students, sorted by gender and by whether they are gifted.

Addition rule for probability: table of 27,000 students, males 150 gifted and 14,850 not gifted (15,000 total); females 240 gifted and 11,760 not gifted (12,000 total); totals 390 gifted and 26,610 not gifted

Question: What is the probability that a randomly chosen student is gifted or male?

Let G = gifted and M = male. We want P(G ∪ M).

  1. Gifted students: 150 + 240 = 390, so P(G) = 390/27,000.
  2. Male students: 150 + 14,850 = 15,000, so P(M) = 15,000/27,000.
  3. Students who are both gifted and male (where the Males row meets the Gifted column): 150, so P(G ∩ M) = 150/27,000.
  4. Apply the addition rule:
    P(G ∪ M) = 390/27,000 + 15,000/27,000 − 150/27,000 = 15,240/27,000
  5. Simplify: 15,240/27,000 ≈ 0.564, or about 56.4%.

So there is about a 56.4% chance that a randomly chosen student is gifted, male, or both.

The picture below shows where each part of the count comes from:

Addition rule for probability: G ∪ M = (150 + 240) + (150 + 14850) − 150, where 150 + 240 is G, 150 + 14850 is M, and 150 is G ∩ M

Common mistake: adding P(G) + P(M) and stopping there. That gives 15,390/27,000 ≈ 0.570, which is too high because the 150 gifted males were counted twice.

Addition rule for mutually exclusive events

Two events are mutually exclusive when they cannot happen at the same time. Then A ∩ B = { }, so P(A and B) = 0, and the rule becomes simpler:

P(A or B) = P(A) + P(B)

Use this shortcut only when A and B have no outcomes in common.

Example: You roll one die. What is the probability of rolling a 2 or a 5?

A single roll can't be both 2 and 5, so the events are mutually exclusive.
P(2 or 5) = 1/6 + 1/6 = 2/6 = 1/3.

Practice problems

1. Using the table above, find P(female or gifted).

P(F) = 12,000/27,000, P(G) = 390/27,000 and P(F ∩ G) = 240/27,000.
P(F ∪ G) = (12,000 + 390 − 240)/27,000 = 12,150/27,000 = 0.45

2. A card is drawn from a standard deck. Find P(queen or face card).

Every queen is a face card, so P(queen and face card) = P(queen) = 4/52.
P = 4/52 + 12/52 − 4/52 = 12/52 = 3/13. The answer is just P(face card).

3. A die is rolled. Find P(even number or a number greater than 3).

Even: {2, 4, 6}, greater than 3: {4, 5, 6}, both: {4, 6}.
P = 3/6 + 3/6 − 2/6 = 4/6 = 2/3

4. A die is rolled. Find P(1 or 6). Are the events mutually exclusive?

Yes. A roll can't be both 1 and 6.
P(1 or 6) = 1/6 + 1/6 = 1/3

Joint probability