Why do I have to subtract P(A ∩ B) instead of just adding P(A) + P(B)?

When you add P(A) + P(B), you count every outcome in the overlap twice—once in P(A) and once in P(B). Subtracting P(A \cap B) removes that double-counting, so each outcome is counted exactly once. Think of it as a correction for the overlapping region in a Venn diagram.

What happens if the events are mutually exclusive?

If A and B cannot both happen, then P(A \cap B) = 0, and the Addition Rule becomes P(A \cup B) = P(A) + P(B). This is a special case of the full formula, not a different rule. Mutually exclusive events are simpler to work with, but the same formula applies.

Can I use Addition Rule for more than two events?

Yes, but it becomes more complex. For three events A, B, and C, you would add all three individual probabilities, subtract the three pairwise intersections, and add back the three-way intersection. A Venn diagram or careful case-by-case counting often helps avoid errors.

What is the difference between P(A ∪ B) and P(A ∩ B)?

P(A ∪ B) is the probability of A OR B (at least one event happens); find it with Addition Rule. P(A ∩ B) is the probability of A AND B (both events happen); find it with Multiplication Rule (for independent events) or Conditional Probability. These answer completely different questions.

Addition Rule

The Addition Rule calculates the probability that at least one of two events occurs by adding individual probabilities and subtracting their overlap.

P(A \cup B) = P(A) + P(B) - P(A \cap B)

Solve a problem with Addition Rule

Type the problem. The solver will use Addition Rule where Addition Rule is the right tool, and tell you when it is not.

What each symbol means

What Addition Rule takes

Symbol	Meaning
$A$	The first event of interest, representing any outcome or situation you care about. Could be rolling a 4, drawing a red card, or exceeding a data limit—the specific event depends on your problem.
$B$	The second event you're combining with A. If you're finding the probability of A or B, both events should apply to the same trial (like a single card draw or a single person), or your setup may be wrong.

When to use it

Use the Addition Rule when you need to find the probability of one event or the other occurring (or both).

Level

Usually taught in: Algebra II · Appears on: SAT

Worked examples

1. Find the probability of a number being even or greater than 6

Problem

A number is chosen randomly from 1 to 10. What is the probability that it is either even or greater than 6?

$P(A) = \frac{5}{10}, \quad P(B) = \frac{4}{10}, \quad P(A \cap B) = \frac{2}{10}$
Event A is even: {2, 4, 6, 8, 10}. Event B is greater than 6: {7, 8, 9, 10}. The overlap is {8, 10}.
$P(A \cup B) = P(A) + P(B) - P(A \cap B) = \frac{5}{10} + \frac{4}{10} - \frac{2}{10}$
Substitute the three probabilities into the Addition Rule formula.
$P(A \cup B) = \frac{7}{10}$
Combine the fractions: $5 + 4 - 2 = 7$ .

Answer: $\frac{7}{10}$

Simply adding $5/10 + 4/10$ would double-count the numbers 8 and 10 (which satisfy both conditions). Subtracting $P(A \cap B)$ corrects this overcounting, giving the true probability of at least one condition being met.

2. Apply Addition Rule with fractional probabilities and independence

Problem

Events A and B are independent, with

P(A) = \frac{1}{3}

and

P(B) = \frac{3}{5}

. Find

P(A \cup B)

$P(A \cap B) = P(A) \cdot P(B) = \frac{1}{3} \cdot \frac{3}{5} = \frac{3}{15} = \frac{1}{5}$
For independent events, multiply the individual probabilities to find the intersection.
$P(A \cup B) = \frac{1}{3} + \frac{3}{5} - \frac{1}{5}$
Apply the Addition Rule by substituting $P(A)$ , $P(B)$ , and $P(A \cap B)$ .
$P(A \cup B) = \frac{5}{15} + \frac{9}{15} - \frac{3}{15} = \frac{11}{15}$
Convert to a common denominator (15) and combine: $5 + 9 - 3 = 11$ .

Answer: $\frac{11}{15}$

Even though A and B are independent, they can still both occur in a single trial. We cannot just add $1/3 + 3/5$ because that counts the overlap twice; the subtraction removes the double-counting.

3. Calculate probability of exceeding at least one limit on a phone plan

Problem

A phone plan includes a monthly fee, 10 GB of data, and 1000 minutes. Data overages cost $5 per gigabyte, and minute overages cost $0.20 per minute. A phone company's data shows that 40% of customers exceed their data limit, 25% exceed their minute limit, and 10% exceed both. What is the probability that a randomly selected customer exceeds at least one limit?

$P(A) = 0.40, \quad P(B) = 0.25, \quad P(A \cap B) = 0.10$
Let A be the event "exceed data limit" and B be "exceed minute limit." The intersection is customers who incur both types of charges.
$P(A \cup B) = P(A) + P(B) - P(A \cap B) = 0.40 + 0.25 - 0.10$
The question asks for "at least one limit exceeded," which is the union of the two events.
$P(A \cup B) = 0.65 - 0.10 = 0.55$
Compute $0.40 + 0.25 = 0.65$ , then subtract $0.10$ to avoid double-counting.

Answer: $0.55$

About 55% of customers will incur overage charges. Without subtracting the 10% overlap, we would incorrectly calculate $0.40 + 0.25 = 0.65$ , which overcounts the 10% of customers in both groups. The Addition Rule gives the accurate count of customers paying for at least one type of overage.

Common mistakes

Where Addition Rule usually goes wrong

The mistake	Why it is wrong	The fix
Writing $P(A \cup B) = P(A) + P(B)$ and forgetting to subtract $P(A \cap B)$ .	This double-counts the outcomes where both A and B occur, inflating the probability.	Always include the full formula: $P(A \cup B) = P(A) + P(B) - P(A \cap B)$ . If events are mutually exclusive (cannot both happen), then $P(A \cap B) = 0$ and the formula simplifies, but do not assume this without checking.
Confusing the union symbol (∪) with the intersection symbol (∩), leading to applying Addition Rule to find $P(A \cap B)$ instead of $P(A \cup B)$ .	The symbols have opposite meanings: ∪ means OR (at least one event), and ∩ means AND (both events). Using the wrong symbol gives an answer to the wrong question.	Remember: Addition Rule uses ∪ (union, OR). For $P(A \cap B)$ (intersection, AND), use the Multiplication Rule instead (for independent events: $P(A \cap B) = P(A) \times P(B)$ ).
Assuming $P(A \cap B) = 0$ when it is actually nonzero, especially when events seem different or unrelated.	Two events can overlap even if they sound distinct. For example, drawing a face card overlaps with drawing a spade; rolling above 3 overlaps with rolling even.	Carefully verify whether the two events can occur together. Draw a Venn diagram if unsure. Calculate or look up $P(A \cap B)$ before substituting into the formula.

Tips and when to use something else

If events A and B are mutually exclusive (impossible to occur together), then $P(A \cap B) = 0$ , and the formula simplifies to $P(A \cup B) = P(A) + P(B)$ . Always check whether events can overlap before assuming zero intersection.
The minus sign in the formula is non-negotiable. It fixes the double-counting that happens when you add the two individual probabilities. If you forget it, your answer will be too large.
To find the probability of A AND B (not A OR B), use the Multiplication Rule instead: for independent events, $P(A \cap B) = P(A) \times P(B)$ . Addition Rule solves a different problem.
When unsure whether events overlap, draw a Venn diagram with two circles. Shade in A on the left, B on the right, and mark the overlap in the middle. This picture shows why you add the two regions and subtract the overlap.

Related formulas

Basic ProbabilityBasic Probability calculates the likelihood of an event by dividing favorable outcomes by the total number of possible outcomes.Mutually Exclusive EventsWhen two events cannot happen at the same time, use this formula to find the probability that at least one of them occurs.Multiplication RuleFind the probability that two events both occur when one might affect the chance of the other using multiplication of probabilities.Conditional ProbabilityConditional probability calculates the likelihood of one event given that another event has definitely already occurred, denoted P(A|B).Complement RuleFind the probability of the opposite event by subtracting from 1 — use this when it's easier to calculate what doesn't happen.

Frequently asked questions

Why do I have to subtract P(A ∩ B) instead of just adding P(A) + P(B)?: When you add $P(A) + P(B)$ , you count every outcome in the overlap twice—once in P(A) and once in P(B). Subtracting $P(A \cap B)$ removes that double-counting, so each outcome is counted exactly once. Think of it as a correction for the overlapping region in a Venn diagram.
What happens if the events are mutually exclusive?: If A and B cannot both happen, then $P(A \cap B) = 0$ , and the Addition Rule becomes $P(A \cup B) = P(A) + P(B)$ . This is a special case of the full formula, not a different rule. Mutually exclusive events are simpler to work with, but the same formula applies.
Can I use Addition Rule for more than two events?: Yes, but it becomes more complex. For three events A, B, and C, you would add all three individual probabilities, subtract the three pairwise intersections, and add back the three-way intersection. A Venn diagram or careful case-by-case counting often helps avoid errors.
What is the difference between P(A ∪ B) and P(A ∩ B)?: P(A ∪ B) is the probability of A OR B (at least one event happens); find it with Addition Rule. P(A ∩ B) is the probability of A AND B (both events happen); find it with Multiplication Rule (for independent events) or Conditional Probability. These answer completely different questions.

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Reviewed 2026-09-18