Addition Rule for Probabilities

Formulas for mutually exclusive and overlapping events — with 8 solved examples

Overview

The addition rule of probability computes the probability that at least one of two events occurs. Read \( A \cup B \) as "\( A \) or \( B \)" and \( A \cap B \) as "\( A \) and \( B \)." Before proceeding, review the basic ideas of a sample space and events.

Throughout this lesson:

Summary of Formulas

Rule Formula
General Addition Rule
(non-mutually exclusive events)		 \( P(A \cup B) = P(A) + P(B) - P(A \cap B) \)
Mutually Exclusive Events
(\( P(A \cap B) = 0 \))		 \( P(A \cup B) = P(A) + P(B) \)

Why Subtract \( P(A \cap B) \)?

When you add \( P(A) + P(B) \), outcomes that lie in both events are counted twice — once inside \( A \) and once inside \( B \). Subtracting \( P(A \cap B) \) removes that double-count, so every outcome is counted exactly once. The example below demonstrates this in two ways.

Example 1 — Odd or ≤ 3 on a fair die

A fair die is rolled once. Find the probability of getting an odd number or a number less than or equal to 3.

View solution (two methods)

Method 1 — Direct sample-space counting.

\( S = \{1,2,3,4,5,6\} \), \( n(S) = 6 \).

The event "odd or ≤ 3" is:

\[ E = \{1, 2, 3, 5\} \implies n(E) = 4 \] \[ P(E) = \frac{4}{6} = \frac{2}{3} \]

Method 2 — Using the addition formula.

Let \( A \) = "odd" = \( \{1, 3, 5\} \) and \( B \) = "≤ 3" = \( \{1, 2, 3\} \).

Venn diagram for two overlapping events A = {1, 3, 5} and B = {1, 2, 3}, showing their intersection at {1, 3}
Figure 1 — The addition rule visualized.

The intersection is \( A \cap B = \{1, 3\} \), so \( n(A \cap B) = 2 \). Counts:

\[ n(A) = 3, \quad n(B) = 3, \quad n(A \cap B) = 2 \] \[ n(A \cup B) = 3 + 3 - 2 = 4 \]

Dividing by \( n(S) = 6 \):

\[ P(A \cup B) = \frac{3}{6} + \frac{3}{6} - \frac{2}{6} = \frac{4}{6} = \frac{2}{3} \]

The Mutually Exclusive Case

If \( A \) and \( B \) cannot occur on the same trial, their intersection is empty: \( A \cap B = \varnothing \), so \( P(A \cap B) = 0 \). The general rule simplifies to:

\[ P(A \text{ or } B) = P(A) + P(B) \]

Quick test: ask "can both events happen at once?" If not, they are mutually exclusive and the simplified rule applies.

More Solved Examples

Example 2 — Rolling a 1 or a 5

A fair die is rolled once. Find the probability of getting a "1" or a "5".

View solution

Let \( A \) = "rolling a 1" and \( B \) = "rolling a 5". A single roll cannot give both, so \( A \) and \( B \) are mutually exclusive:

\[ P(A) = \frac{1}{6}, \quad P(B) = \frac{1}{6} \] \[ P(A \cup B) = \frac{1}{6} + \frac{1}{6} = \frac{2}{6} = \frac{1}{3} \]

Example 3 — Drawing a green or blue ball

A box contains 3 red, 2 green, and 5 blue balls. One ball is drawn at random. Find the probability it is green or blue.

View solution

\( n(S) = 3 + 2 + 5 = 10 \).

Let \( A \) = "green" and \( B \) = "blue". A single ball cannot be both, so \( A \) and \( B \) are mutually exclusive:

\[ P(A) = \frac{2}{10} = \frac{1}{5}, \quad P(B) = \frac{5}{10} = \frac{1}{2} \] \[ P(A \cup B) = \frac{2}{10} + \frac{5}{10} = \frac{7}{10} \]

Example 4 — Football or basketball

In a school of 100 students, 50 play football, 20 play basketball, and 10 play both. A student is selected at random. What is the probability the student plays at least one sport?

View solution

Let \( A \) = "plays football" and \( B \) = "plays basketball". These events overlap because 10 students play both.

\[ P(A) = \frac{50}{100} = \frac{1}{2}, \quad P(B) = \frac{20}{100} = \frac{1}{5}, \quad P(A \cap B) = \frac{10}{100} = \frac{1}{10} \] \[ P(A \cup B) = \frac{50}{100} + \frac{20}{100} - \frac{10}{100} = \frac{60}{100} = \frac{3}{5} \]

Example 5 — Card problems from a 52-card deck

A single card is drawn from a standard deck. Find the probability of drawing:

  1. A "2" or a "5"
  2. An "8" or a "heart"
  3. A "Queen" or a "red card"
Standard deck of 52 playing cards arranged by suit and rank
Figure 2 — Standard 52-card deck: 4 suits × 13 ranks.
View solution
  1. A "2" or a "5": mutually exclusive (a card cannot be both a 2 and a 5). \[ P(\text{2 or 5}) = \frac{4}{52} + \frac{4}{52} = \frac{8}{52} = \frac{2}{13} \]
  2. An "8" or a "heart": overlapping (the 8 of hearts is both). \[ P(\text{8 or heart}) = \frac{4}{52} + \frac{13}{52} - \frac{1}{52} = \frac{16}{52} = \frac{4}{13} \]
  3. A "Queen" or a "red card": overlapping (2 red queens). \[ P(\text{Queen or red}) = \frac{4}{52} + \frac{26}{52} - \frac{2}{52} = \frac{28}{52} = \frac{7}{13} \]

Example 6 — Car inventory contingency table

A car dealer's inventory is classified in the table below. A car is chosen at random. Find the probability it is:

  1. A black or a white car
  2. A blue car or a coupe
  3. A black car or an SUV
SUV Sport Car Van Coupe Total
Black3510251585
White101520550
Blue151553065
Total60405050200
View solution

Total cars \( n(S) = 200 \).

  1. Black or White: mutually exclusive. \[ P(\text{Black or White}) = \frac{85}{200} + \frac{50}{200} = \frac{135}{200} = \frac{27}{40} \]
  2. Blue or Coupe: overlapping (30 blue coupes). \[ P(\text{Blue or Coupe}) = \frac{65}{200} + \frac{50}{200} - \frac{30}{200} = \frac{85}{200} = \frac{17}{40} \]
  3. Black or SUV: overlapping (35 black SUVs). \[ P(\text{Black or SUV}) = \frac{85}{200} + \frac{60}{200} - \frac{35}{200} = \frac{110}{200} = \frac{11}{20} \]

Example 7 — Homework hours distribution

The table below shows weekly hours spent on homework by high school students. A student is chosen at random. Find the probability the student spends at most 4 hours or at least 11 hours on homework.

Time (hours)Number of students
0–25
3–420
5–735
8–1050
11–1260
13+30
View solution

Total students: \( n(S) = 5 + 20 + 35 + 50 + 60 + 30 = 200 \).

Let \( A \) = "at most 4 hours" (0–2 or 3–4), so \( n(A) = 5 + 20 = 25 \).

Let \( B \) = "at least 11 hours" (11–12 or 13+), so \( n(B) = 60 + 30 = 90 \).

The two ranges do not overlap, so \( A \) and \( B \) are mutually exclusive:

\[ P(A \cup B) = \frac{25}{200} + \frac{90}{200} = \frac{115}{200} = \frac{23}{40} \]

Example 8 — Insurance coverage (solving for an unknown intersection)

An insurance company finds that 80% of policyholders have home insurance and 60% have auto insurance. Every policyholder has at least one of the two. A policyholder is selected at random. What is the probability that they hold both home and auto insurance?

View solution

Let \( A \) = "home insurance" and \( B \) = "auto insurance". We know:

\[ P(A) = 0.8, \quad P(B) = 0.6, \quad P(A \cup B) = 1 \;(\text{everyone has at least one}) \]

Apply the general addition rule and solve for \( P(A \cap B) \):

\[ 1 = 0.8 + 0.6 - P(A \cap B) \] \[ P(A \cap B) = 1.4 - 1 = 0.4 \quad (40\%) \]

Answer: 40% of policyholders have both types of insurance.

Practice Problems

Try each problem with the addition rule, then reveal the answer to check.

Practice 1 — Numbered balls

A bag contains balls numbered 1 through 20. One ball is drawn. What is the probability of drawing a ball that is a multiple of 3 or a multiple of 5?

Answer

Multiples of 3: {3, 6, 9, 12, 15, 18} → 6 balls.

Multiples of 5: {5, 10, 15, 20} → 4 balls.

Overlap: {15} → 1 ball.

\[ P = \frac{6}{20} + \frac{4}{20} - \frac{1}{20} = \frac{9}{20} \]

Practice 2 — Card draw

A single card is drawn from a 52-card deck. What is the probability of drawing a spade or a face card (J, Q, K)?

Answer

\( P(\text{spade}) = 13/52 \), \( P(\text{face}) = 12/52 \), overlap (face cards that are spades) \( = 3/52 \).

\[ P = \frac{13}{52} + \frac{12}{52} - \frac{3}{52} = \frac{22}{52} = \frac{11}{26} \]

Practice 3 — Commuting survey

In a survey of 200 commuters, 120 drive, 80 take public transit, and 40 use both. What is the probability a randomly chosen commuter drives or takes transit?

Answer
\[ P = \frac{120}{200} + \frac{80}{200} - \frac{40}{200} = \frac{160}{200} = \frac{4}{5} \]

Frequently Asked Questions

What is the addition rule for probability?

The general addition rule states \(P(A \cup B) = P(A) + P(B) - P(A \cap B)\). The term \(-P(A \cap B)\) corrects for double-counting outcomes in both \(A\) and \(B\).

What is the addition rule for mutually exclusive events?

If \(A\) and \(B\) cannot occur simultaneously, \(P(A \cap B) = 0\), so the rule simplifies to \(P(A \cup B) = P(A) + P(B)\).

Why do we subtract \(P(A \cap B)\)?

Outcomes in \(A \cap B\) are counted twice when you add \(P(A)\) and \(P(B)\) — once inside \(A\) and once inside \(B\). Subtracting \(P(A \cap B)\) removes the double-count.

How can you tell whether two events are mutually exclusive?

Two events are mutually exclusive if they share no outcomes — the intersection \(A \cap B\) is empty, so \(P(A \cap B) = 0\). In practice, ask: can both events happen on the same trial? If not, they are mutually exclusive.

Related Probability & Statistics Topics

Last updated: October 20, 2026