Quantitative Aptitude · Probability

Addition Theorem of Probability for CA Foundation

Updated 1 October 2026 · Fact-checked

The addition theorem gives the probability that at least one of two events happens. For any events A and B, P(A ∪ B) = P(A) + P(B) − P(A ∩ B). If A and B are mutually exclusive, P(A ∩ B) = 0, so you simply add. Subtract the overlap so it is not counted twice.

Understand Addition Theorem of Probability

Probability measures how likely an event is, on a scale from 0 to 1. The addition theorem answers one kind of question: what is the chance that A happens, or B happens, or both? In set language this is the union, A ∪ B. The word "or" is your signal to think of addition.

Two events are mutually exclusive if they cannot happen together. On one throw of a die, getting 2 and getting 5 are mutually exclusive. Here nothing is shared, so P(A ∪ B) = P(A) + P(B).

If the events can happen together, they are not mutually exclusive. On one throw of a die, "an even number" and "a number greater than 3" can both happen when you get 4 or 6. If you add P(A) and P(B), you count those shared outcomes twice. So you subtract the overlap once: P(A ∪ B) = P(A) + P(B) − P(A ∩ B). This general formula also covers the mutually exclusive case, because the overlap is then zero.

The complement of A, written A' or Ac, means A does not happen. Since A and A' cover everything with no overlap, P(A) + P(A') = 1. This helps with "at least one" questions: P(at least one of A or B) = 1 − P(neither). Use it when counting the "neither" case is easier.

For three events, the same idea continues: add the singles, subtract the pairwise overlaps, and add back the triple overlap. Sets and Venn diagrams are the best way to see why.

Key formulas to remember

Addition theorem (general): P(A ∪ B) = P(A) + P(B) − P(A ∩ B); Works for any two events A and B.
Mutually exclusive events: P(A ∪ B) = P(A) + P(B); Use only when A and B cannot occur together, so P(A ∩ B) = 0.
Complement rule: P(A') = 1 − P(A); A' means A does not occur.
Neither A nor B: P(A' ∩ B') = 1 − P(A ∪ B); Probability that none of the two events occurs.
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); If the three are pairwise mutually exclusive, just add the three probabilities.
Only A (not B): P(A ∩ B') = P(A) − P(A ∩ B); Useful for "A but not B" questions.
Exactly one of A, B: P(exactly one) = P(A) + P(B) − 2P(A ∩ B); Counts A only and B only, excluding both.

How to solve Addition Theorem of Probability questions

Use this method for any addition-theorem question.

1Define the events clearly, such as A = card is a king, B = card is a heart.
2Mark the key word: "or" or "at least one" means union. "Neither" means the complement of the union.
3Decide whether the events can occur together. If not, they are mutually exclusive and P(A ∩ B) = 0.
4Find P(A), P(B) and, if needed, P(A ∩ B) by counting favourable outcomes ÷ total outcomes.
5Apply P(A ∪ B) = P(A) + P(B) − P(A ∩ B).
6If the question asks "neither" or "not", subtract your result from 1.
7Check that the answer lies between 0 and 1 and that it is at least as large as P(A) and P(B).

Quickest way: Count outcomes directly, then eliminate options

When to use it: Use it when the sample space is small, such as a die, a deck of cards or numbered tickets, and you have under a minute.

Count the favourable outcomes for "A or B" directly, listing shared ones only once, and divide by the total.
Alternatively, find P(A) + P(B) first. The answer must be this value or less. Remove any option above it.
The answer must also be at least the larger of P(A) and P(B). Remove options below it.
For "neither" questions, compute the union first and subtract from 1.
If the overlap is unclear and you have no time, skip. A wrong answer costs 0.25 marks.

Common mistakes in Addition Theorem of Probability

Adding P(A) and P(B) when the events overlap.
The word "or" triggers plain addition and the overlap is forgotten.
Fix: Ask: can both happen together? If yes, subtract P(A ∩ B).
Treating events as mutually exclusive when they are not.
Students confuse mutually exclusive with independent.
Fix: Mutually exclusive means P(A ∩ B) = 0. Test by looking for even one common outcome, such as the king of hearts.
Subtracting the overlap when the question gives events as mutually exclusive.
Students memorise the long formula and apply it blindly.
Fix: If the events are mutually exclusive, the overlap is 0 and you only add.
Getting "neither" wrong by using P(A') + P(B').
The complement is applied to each event separately.
Fix: Use P(neither) = 1 − P(A ∪ B).
Forgetting to add back P(A ∩ B ∩ C) in three-event questions.
The triple overlap is subtracted three times and then missed when adding back.
Fix: Remember the sign pattern: plus singles, minus pairs, plus triple.
Getting an answer above 1 and not noticing.
Wrong overlap or wrong counting.
Fix: Any probability must lie between 0 and 1. Recheck the counting if it does not.

Worked examples

Example 1

One card is drawn from a well-shuffled pack of 52 cards. What is the probability that it is a king or a heart? Options: (a) 4/13 (b) 17/52 (c) 1/2 (d) 3/13

Show the solution

Let A = king, B = heart.
P(A) = 4/52, P(B) = 13/52.
Both can happen: the king of hearts. P(A ∩ B) = 1/52.
P(A ∪ B) = 4/52 + 13/52 − 1/52 = 16/52 = 4/13.
Check by counting: 13 hearts plus 3 other kings = 16 cards, so 16/52.

Answer: (a) 4/13

Example 2

P(A) = 0.5, P(B) = 0.4 and P(A ∪ B) = 0.7. What is the probability that exactly one of A and B occurs? Options: (a) 0.2 (b) 0.3 (c) 0.5 (d) 0.6

Show the solution

P(A ∩ B) = P(A) + P(B) − P(A ∪ B) = 0.5 + 0.4 − 0.7 = 0.2.
Exactly one = P(A) + P(B) − 2P(A ∩ B) = 0.9 − 0.4 = 0.5.
Check: A only = 0.5 − 0.2 = 0.3, B only = 0.4 − 0.2 = 0.2, total = 0.5.

Answer: (c) 0.5

Example 3

A single die is thrown once. What is the probability of getting a number that is even or greater than 4? Options: (a) 1/2 (b) 2/3 (c) 5/6 (d) 1/3

Show the solution

Let A = even = {2, 4, 6}, so P(A) = 3/6.
Let B = greater than 4 = {5, 6}, so P(B) = 2/6.
Common outcome: 6. P(A ∩ B) = 1/6.
P(A ∪ B) = 3/6 + 2/6 − 1/6 = 4/6 = 2/3.
Check: the union is {2, 4, 5, 6}, which is 4 outcomes out of 6.

Answer: (b) 2/3

Exam tips

Look at the wording. "Or" and "at least one" point to the addition theorem. "And" and "both" point to the overlap.
With cards and dice, list or count the union directly. It is faster and safer than the formula.
If P(A ∪ B) is given, the overlap can be found as P(A) + P(B) − P(A ∪ B). Many questions work backwards like this.
Use the range check: the union is never less than the larger of P(A), P(B) and never more than 1.
Do not guess when unsure. Each wrong answer costs 0.25 marks, so skip a question you cannot finish quickly.

Practice questions from Probability

Addition Theorem of Probability: frequently asked questions

What is the addition theorem of probability?

It gives the probability that at least one of two events occurs. For any events, P(A ∪ B) = P(A) + P(B) − P(A ∩ B). For mutually exclusive events the last term is zero.

What is the difference between mutually exclusive and independent events?

Mutually exclusive events cannot occur together, so P(A ∩ B) = 0. Independent events do not affect each other, and P(A ∩ B) = P(A) × P(B). Two events with non-zero probabilities cannot be both.

How do I use the addition rule for three events?

Add the three single probabilities, subtract the three pairwise intersections, then add the triple intersection. If the events are pairwise mutually exclusive, just add the three probabilities.

When should I use the complement instead?

Use it for "at least one" or "neither" questions. Find the probability of the opposite case and subtract it from 1 when that is easier to count.