MathsISC Class 11 · Probability

The Addition Theorem: P(A or B), Neither, Exactly One and Combined Events

Updated 17 October 2026 · Fact-checked

For any two events, P(A ∪ B) = P(A) + P(B) − P(A ∩ B). If A and B are mutually exclusive, P(A ∪ B) = P(A) + P(B). The probability of neither is 1 − P(A ∪ B), of exactly one is P(A) + P(B) − 2P(A ∩ B), and of A but not B is P(A) − P(A ∩ B).

The overlap A ∩ B is inside both circles, so adding P(A) and P(B) counts it twice; subtract it once to get P(A ∪ B).

Illustration drawn for QuestionBuddy.

Understand Addition Theorem and Probability of Combined Events

Addition theorem. $P (A \cup B) = P (A) + P (B) - P (A \cap B)$ . The overlap is subtracted once because it is counted twice in P(A) + P(B). For mutually exclusive events $P (A \cap B) = 0$ , so $P (A \cup B) = P (A) + P (B)$ . For three events, $P (A \cup B \cup C) = P (A) + P (B) + P (C) - P (A \cap B) - P (B \cap C) - P (A \cap C) + P (A \cap B \cap C)$ .

Derived events. *Neither A nor B:* $P (A' \cap B') = 1 - P (A \cup B)$ (De Morgan). *A but not B:* $P (A \cap B') = P (A) - P (A \cap B)$ . *Exactly one of A, B:* $P (A) + P (B) - 2 P (A \cap B)$ . *At least one:* $P (A \cup B)$ . *Not both:* $1 - P (A \cap B)$ .

Strategy. Identify the events A and B in words; find P(A), P(B) and P(A ∩ B) (the outcomes in both); use the theorem. For problems with percentages, treat the percentage as the probability of a randomly chosen person.

Cards. Typical events: heart, king, face card, red card, queen. The overlap is the cards satisfying both (for example the king of hearts). For dice: write the favourable ordered pairs for each event and count the overlap directly.

Venn diagram. Draw two overlapping circles and put the probabilities of the regions: only A, both, only B, neither; they total 1.

Addition Theorem and Probability of Combined Events: formulas to remember

Addition theorem: $$P(A \cup B) = P(A) + P(B) - P(A \cap B)$$; Subtract the overlap.
Mutually exclusive: $$P(A \cup B) = P(A) + P(B)$$; Since $A \cap B = \emptyset$ .
Neither, exactly one: $Neither: $1 - P(A \cup B)$ | Exactly one: $P(A) + P(B) - 2P(A \cap B)$$; A only: $P (A) - P (A \cap B)$ .
Complement: $$P(A') = 1 - P(A)$$; Useful for "at least one".

Addition Theorem and Probability of Combined Events in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Pass percentages

Suppose in a school 40% of the students pass maths (M), 30% pass physics (P), and 15% pass both.

Find the probability that a randomly chosen student passes at least one, exactly one, and neither.

Try it first, then show the solutionSolution

At least one: $P (M \cup P) = 0.40 + 0.30 - 0.15 = 0.55$ .
Exactly one: $0.40 + 0.30 - 2 (0.15) = 0.40$ .
Neither: $1 - 0.55 = 0.45$ .

Answer: 0.55; 0.40; 0.45.

Memory hook: Exactly one subtracts the overlap twice.

Tricks to remember Addition Theorem and Probability of Combined Events

Short memory aids. Say each one aloud once, then use it on a question the same day.

Add, then subtract the overlap
P(A) + P(B) − P(A ∩ B).
Use it when: Any "A or B".
Draw the Venn diagram
Fill the overlap first, then the "only" regions.
Use it when: Percentage problems.

Addition Theorem and Probability of Combined Events: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Mutually exclusive events · 1 mark · MCQBoard level

If A and B are mutually exclusive with P(A) = 0.4 and P(B) = 0.3, then P(A ∪ B) is:
(a) 0.12
(b) 0.7
(c) 0.1
(d) 0.3

Try it first, then show the solutionSolution

Add the probabilities: $0.4 + 0.3 = 0.7$ .

Answer: (b) 0.7

Example 2: Using the theorem · 3 marks · Short answerBoard level

If P(A) = 0.5, P(B) = 0.4 and P(A ∩ B) = 0.2, find P(A ∪ B), P(neither) and P(A only).

Try it first, then show the solutionSolution

$P (A \cup B) = 0.5 + 0.4 - 0.2 = 0.7$ . (1 mark)
Neither $= 1 - 0.7 = 0.3$ . (1 mark)
A only $= 0.5 - 0.2 = 0.3$ . (1 mark)

Answer: 0.7; 0.3; 0.3.

Example 3: Cards · 3 marks · Short answerHarder board level

A card is drawn from a pack of 52. Find the probability that it is (i) a heart or a king, (ii) a red card or a queen.

Try it first, then show the solutionSolution

(i) $P (H) = \frac{13}{52}$ , $P (K) = \frac{4}{52}$ , $P (H \cap K) = \frac{1}{52}$ (king of hearts). (1 mark)
$P (H \cup K) = \frac{13 + 4 - 1}{52} = \frac{16}{52} = \frac{4}{13}$ . (1 mark)
(ii) Red cards 26; queens 4 of which 2 are red: $\frac{26 + 4 - 2}{52} = \frac{28}{52} = \frac{7}{13}$ . (1 mark)

Answer: 4/13; 7/13.

Example 4: Finding P(B) · 4 marks · Short answerHarder board level

If P(A ∪ B) = 0.7, P(A) = 0.4 and P(A ∩ B) = 0.15, find P(B), P(A − B) and P(exactly one of A, B).

Try it first, then show the solutionSolution

$P (B) = P (A \cup B) - P (A) + P (A \cap B) = 0.7 - 0.4 + 0.15 = 0.45$ . (1.5 marks)
$P (A - B) = P (A) - P (A \cap B) = 0.25$ . (1 mark)
Exactly one $= 0.4 + 0.45 - 2 (0.15) = 0.55$ (check: $P (A \cup B) - P (A \cap B) = 0.55$ ). (1.5 marks)

Answer: P(B) = 0.45; P(A − B) = 0.25; exactly one = 0.55.

Example 5: Dice and a survey · 5 marks · Long answerHOTS / competency

Two dice are thrown: find the probability that the sum is 7 or 11. In another sample, 60% of students like tea, 50% like coffee and 20% like neither; find P(both) and P(only tea).

Try it first, then show the solutionSolution

Sum 7: 6 outcomes; sum 11: 2 outcomes; mutually exclusive: $P = \frac{6 + 2}{36} = \frac{2}{9}$ . (1.5 marks)
At least one drink: $1 - 0.20 = 0.80$ . (1 mark)
Both: $0.60 + 0.50 - 0.80 = 0.30$ . (1.5 marks)
Only tea: $0.60 - 0.30 = 0.30$ . (1 mark)

Answer: 2/9; 0.30; 0.30.

Where students lose marks in Addition Theorem and Probability of Combined Events

The slips examiners see most often, and what to write instead.

Mistake: Forgetting the overlap
Adding P(A) and P(B) only.
Do this: Subtract P(A ∩ B) unless the events are exclusive.
Usually costs: 2 marks.
Mistake: Misreading "exactly one"
Giving P(A ∪ B).
Do this: Exactly one = P(A ∪ B) − P(A ∩ B).
Usually costs: 1 to 2 marks.
Mistake: Not checking totals
Probabilities summing to more than 1.
Do this: Check that all regions add up to 1.
Usually costs: 1 mark.

Exam tips

Name the events and write what each represents.
Draw a Venn diagram for percentage problems.
Check that the answer is between 0 and 1.
Use the complement for "neither".

Practice questions on Addition Theorem and Probability of Combined Events

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
For any two events, $P (A \cup B)$ equals:
(a) $P (A) + P (B)$
(b) $P (A) + P (B) - P (A \cap B)$
(c) $P (A) P (B)$
(d) $P (A) - P (B)$
Hint
Subtract the overlap.
Answer
(b)
Q2 · 2 marks · Short answerBoard level
If P(A) = 0.6, P(B) = 0.3 and A, B are mutually exclusive, find P(neither).
Hint
1 − P(A ∪ B).
Answer
$1 - 0.9 = 0.1$ .
Q3 · 3 marks · Short answerHarder board level
A die is thrown once. Find the probability of getting an even number or a number greater than 4.
Hint
Even: 2, 4, 6; >4: 5, 6.
Answer
Union: {2, 4, 5, 6}; $P = \frac{4}{6} = \frac{2}{3}$ . Check: $\frac{3}{6} + \frac{2}{6} - \frac{1}{6}$ .
Q4 · 3 marks · Short answerHarder board level
A card is drawn from a pack. Find the probability that it is neither a spade nor a king.
Hint
Union of spades and kings.
Answer
Spades 13, kings 4, overlap 1: union 16; neither $= \frac{36}{52} = \frac{9}{13}$ .
Q5 · 3 marks · Short answerHarder board level
P(A) = 0.35, P(B) = 0.45 and P(A ∪ B) = 0.65. Find P(A ∩ B) and P(only B).
Hint
Rearrange the theorem.
Answer
$P (A \cap B) = 0.35 + 0.45 - 0.65 = 0.15$ ; $P (only B) = 0.45 - 0.15 = 0.30$ .
Q6 · 4 marks · Case-basedHOTS / competency
Case study: Suppose in a town 70% of the people read newspaper X and 50% read newspaper Y; 30% read both.
(i) Find P(at least one). (1 mark)
(ii) Find P(neither). (1 mark)
(iii) Find P(only X). (1 mark)
(iv) Find P(exactly one). (1 mark)
Hint
Draw the Venn diagram.
Answer
(i) $0.70 + 0.50 - 0.30 = 0.90$ . (ii) $0.10$ . (iii) $0.70 - 0.30 = 0.40$ . (iv) $0.40 + 0.20 = 0.60$ (only Y is $0.50 - 0.30 = 0.20$ ).

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.P(A ∪ B)?Show answer
P(A) + P(B) − P(A ∩ B).
2.Exactly one?Show answer
P(A) + P(B) − 2P(A ∩ B).
3.Neither?Show answer
1 − P(A ∪ B).
4.A only?Show answer
P(A) − P(A ∩ B).

Addition Theorem and Probability of Combined Events: frequently asked questions

What is the difference between "mutually exclusive" and "independent"?

Mutually exclusive events cannot occur together; independent events do not affect each other. This unit uses only exclusive events and the general addition theorem.

Why do we subtract P(A ∩ B)?

The outcomes common to both events are counted once in P(A) and once in P(B).

Can P(A ∪ B) exceed 1?

No. If it appears to, the overlap has been missed.