MathsCBSE Class 11 · Probability

Addition Rule, Complement and Bounds on Probability (Class 11)

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) = 0. Also P(A') = 1 − P(A), P(neither) = 1 − P(A ∪ B) and P(exactly one) = P(A) + P(B) − 2P(A ∩ B). The smallest possible P(A ∩ B) is max(0, P(A) + P(B) − 1) and the largest is min(P(A), P(B)).

Adding P(A) and P(B) counts the orange overlap twice, so subtract it once; everything outside both circles is neither, 1 − P(A∪B).

Illustration drawn for QuestionBuddy.

Understand Addition Rule, Complement and Bounds on Probability

Algebra of events. $A \cup B$ is "A or B (or both)", $A \cap B$ is "both", $A^{'}$ is "not A". $A^{'} \cap B^{'}$ is "neither", and by De Morgan $A^{'} \cap B^{'} = (A \cup B)^{'}$ .

Addition rule. For any two events,
$ $P (A \cup B) = P (A) + P (B) - P (A \cap B) .$ $
The intersection is subtracted because it is counted twice when we add. If $A \cap B = \emptyset$ (mutually exclusive), $P (A \cup B) = P (A) + P (B)$ . Events $E_{1}, \dots, E_{n}$ are exhaustive if $E_{1} \cup \dots \cup E_{n} = S$ ; mutually exclusive and exhaustive events have probabilities adding to $1$ .

Useful derived results.

Complement: $P (A^{'}) = 1 - P (A)$ .
Neither: $P (A^{'} \cap B^{'}) = 1 - P (A \cup B)$ .
Only $A$ : $P (A \cap B^{'}) = P (A) - P (A \cap B)$ .
Exactly one of $A, B$ : $P (A) + P (B) - 2 P (A \cap B)$ .
At least one: $P (A \cup B)$ . At most one: $1 - P (A \cap B)$ .

Three-way partition. Draw a Venn diagram with regions "only A", "both", "only B", "neither"; fill them from the data and the total $1$ (or the total $n (S)$ ). Most word problems become simple subtractions.

Counting with the rule. For numbers $1$ to $N$ , "divisible by $a$ or $b$ " has $⌊ N / a ⌋ + ⌊ N / b ⌋ - ⌊ N / lcm (a, b)⌋$ elements; the overlap is the multiples of the lcm, not the product.

Bounds. Since $P (A \cup B) \leq 1$ , $P (A \cap B) \geq P (A) + P (B) - 1$ . Since $A \cap B \subseteq A$ and $\subseteq B$ , $P (A \cap B) \leq min {P (A), P (B)}$ . A data set violating either bound is impossible, which is a favourite assertion-reason test.

Axioms behind it. $P (E) \geq 0$ , $P (S) = 1$ , and for mutually exclusive events probabilities add. Every rule above follows from these three.

Addition Rule, Complement and Bounds on Probability: formulas to remember

Addition rule: $P (A \cup B) = P (A) + P (B) - P (A \cap B)$; Mutually exclusive: last term is 0.
Neither: $P (A^{'} \cap B^{'}) = 1 - P (A \cup B)$; De Morgan.
Exactly one: $P (A) + P (B) - 2 P (A \cap B)$; Only A plus only B.
Only A: $P (A \cap B^{'}) = P (A) - P (A \cap B)$; Draw the Venn diagram.
Bounds on the intersection: $ma x (0, P (A) + P (B) - 1) \leq P (A \cap B) \leq min (P (A), P (B))$; Limits for impossible data.

How to solve Addition Rule, Complement and Bounds on Probability questions

Use these steps for problems on P(A or B), neither, exactly one and bounds.

1Translate each phrase into set language: or, and, not, neither, exactly one.
2Draw a Venn diagram with the four regions and fill in what the data gives.
3Use $P (A \cup B) = P (A) + P (B) - P (A \cap B)$ to find the missing value.
4Find the other regions by subtraction and check that all four add to $1$ .
5For counting problems, subtract the overlap (multiples of the lcm).
6For bounds, apply $max (0, P (A) + P (B) - 1) \leq P (A \cap B) \leq min (P (A), P (B))$ .

Addition Rule, Complement and Bounds on Probability 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.

Union of two groups

Suppose in a school of 200 students, 120 play football, 90 play basketball and 40 play both. One student is chosen at random.

Find the probability that the student plays at least one of the two games, and that the student plays neither.

Try it first, then show the solutionSolution

$P (F) = \frac{120}{200} = 0.6$ , $P (B) = \frac{90}{200} = 0.45$ , $P (F \cap B) = \frac{40}{200} = 0.2$ .
$P (F \cup B) = 0.6 + 0.45 - 0.2 = 0.85$ .
Neither: $1 - 0.85 = 0.15$ .

Answer: 0.85 for at least one; 0.15 for neither.

Memory hook: Subtract the overlap because the 40 students are in both lists.

Weather over a weekend

Suppose the chance of rain on Saturday is 0.4, on Sunday 0.5, and on both days 0.2.

Find the probability of rain on at least one day, and of both days being dry.

Try it first, then show the solutionSolution

At least one: $0.4 + 0.5 - 0.2 = 0.7$ .
Both dry: $1 - 0.7 = 0.3$ .

Answer: 0.7 and 0.3.

Memory hook: Both dry is the complement of at least one wet.

Tricks to remember Addition Rule, Complement and Bounds on Probability

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

Subtract the overlap
Adding P(A) and P(B) counts "both" twice, so take it out once.
Use it when: Addition rule.
Neither is the complement of the union
P(neither) = 1 − P(A or B).
Use it when: Neither and none questions.
Use the lcm for the overlap
For "divisible by 4 or 6" subtract multiples of 12, not 24.
Use it when: Counting problems.
Four regions add to one
Only A, both, only B and neither always total 1.
Use it when: Checking answers.

Addition Rule, Complement and Bounds on Probability: solved examples

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

Example 1: Basic union · 2 marks · Very short answerBoard level

If $P (A) = 0.4$ , $P (B) = 0.35$ and $P (A \cap B) = 0.15$ , find $P (A \cup B)$ and the probability of neither event.

Try it first, then show the solutionSolution

$P (A \cup B) = 0.4 + 0.35 - 0.15 = 0.6$ . (1 mark)
$P (neither) = 1 - 0.6 = 0.4$ . (1 mark)

Answer: 0.6; 0.4

Example 2: Missing intersection · 3 marks · Short answerBoard level

If $P (A) = 0.6$ , $P (B) = 0.5$ and $P (A \cup B) = 0.8$ , find $P (A \cap B)$ , $P (A \cap B^{'})$ and the probability that exactly one of the events occurs.

Try it first, then show the solutionSolution

$P (A \cap B) = 0.6 + 0.5 - 0.8 = 0.3$ . (1 mark)
$P (A \cap B^{'}) = 0.6 - 0.3 = 0.3$ ; $P (A^{'} \cap B) = 0.5 - 0.3 = 0.2$ . (1 mark)
Exactly one $= 0.3 + 0.2 = 0.5$ (check: $0.6 + 0.5 - 2 \times 0.3 = 0.5$ ). (1 mark)

Answer: 0.3; 0.3; 0.5

Example 3: Dice and the addition rule · 3 marks · Short answerHarder board level

Two dice are thrown. Find the probability that the sum is 7 or the product is 12.

Try it first, then show the solutionSolution

Sum 7: $(1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1)$ : 6 outcomes. (1 mark)
Product 12: $(2, 6), (3, 4), (4, 3), (6, 2)$ : 4 outcomes. Common to both: $(3, 4), (4, 3)$ : 2 outcomes. (1 mark)
$n (A \cup B) = 6 + 4 - 2 = 8$ , so $P = \frac{8}{36} = \frac{2}{9}$ . (1 mark)

Answer: $\frac{2}{9}$

Example 4: Divisibility union · 4 marks · Short answerHarder board level

A number is chosen at random from 1, 2, ..., 100. Find the probability that it is divisible by 4 or by 6, and the probability that it is divisible by neither.

Try it first, then show the solutionSolution

Multiples of 4: $⌊ 100/4 ⌋ = 25$ ; of 6: $⌊ 100/6 ⌋ = 16$ . (1 mark)
Divisible by both means by $lcm (4, 6) = 12$ : $⌊ 100/12 ⌋ = 8$ . (1 mark)
$n (A \cup B) = 25 + 16 - 8 = 33$ , so $P = \frac{33}{100}$ . (1 mark)
Neither: $1 - \frac{33}{100} = \frac{67}{100}$ . (1 mark)

Answer: $\frac{33}{100}$ ; $\frac{67}{100}$

Example 5: Bounds · 5 marks · Long answerHOTS / competency

If $P (A) = 0.7$ and $P (B) = 0.8$ , find the least and the greatest possible value of $P (A \cap B)$ , and the corresponding range of $P (neither)$ .

Try it first, then show the solutionSolution

Since $P (A \cup B) \leq 1$ : $P (A \cap B) = P (A) + P (B) - P (A \cup B) \geq 1.5 - 1 = 0.5$ . (1 mark)
Since $A \cap B \subseteq A$ : $P (A \cap B) \leq 0.7$ . (1 mark)
So $0.5 \leq P (A \cap B) \leq 0.7$ . (1 mark)
$P (neither) = 1 - P (A \cup B) = 1 - (1.5 - P (A \cap B)) = P (A \cap B) - 0.5$ . (1 mark)
Hence $P (neither)$ ranges from $0$ (when $P (A \cap B) = 0.5$ ) to $0.2$ (when $P (A \cap B) = 0.7$ ). (1 mark)

Answer: $0.5 \leq P (A \cap B) \leq 0.7$ ; $0 \leq P (neither) \leq 0.2$

Where students lose marks in Addition Rule, Complement and Bounds on Probability

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

Mistake: Not subtracting the overlap
P(A) + P(B) counts the common part twice.
Wrong working
$P (A \cup B) = 0.4 + 0.35 = 0.75$ .
Right working
$0.4 + 0.35 - 0.15 = 0.6$ .
Do this: Write the full addition rule as the first line.
Usually costs: The answer.
Mistake: Using 24 as the overlap for "4 or 6"
The common multiples are multiples of the lcm, 12.
Wrong working
Multiples of 24: 4.
Right working
Multiples of 12: 8.
Do this: Compute the lcm first.
Usually costs: The answer.
Mistake: Confusing "exactly one" with "at least one"
"Exactly one" excludes the intersection.
Wrong working
Exactly one $= P (A \cup B) = 0.8$ .
Right working
$0.8 - 0.3 = 0.5$ .
Do this: Subtract P(A ∩ B) from the union, or use P(A) + P(B) − 2P(A ∩ B).
Usually costs: The answer.
Mistake: Accepting impossible data
A probability above 1 or a break of the bounds is not checked.
Wrong working
P(A) = 0.3, P(B) = 0.4, P(A ∪ B) = 0.8 accepted.
Right working
Impossible, since $0.8 > 0.3 + 0.4$ .
Do this: Test the data against the bounds.
Usually costs: The reasoning mark.

Exam tips

Draw a Venn diagram with four regions; it turns phrases like "exactly one" and "neither" into subtractions.
Write the addition rule in full before substituting numbers.
For counting unions use the lcm for the overlap.
Say which bound you use when finding the least or greatest value of $P (A \cap B)$ .
Check that the four regions add up to 1.

Practice questions on Addition Rule, Complement and Bounds on Probability

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
If $A$ and $B$ are mutually exclusive with $P (A) = \frac{2}{5}$ and $P (B) = \frac{1}{3}$ , then $P (A \cup B)$ is:
(a) $\frac{2}{15}$
(b) $\frac{11}{15}$
(c) $\frac{1}{15}$
(d) $\frac{4}{5}$
Hint
Mutually exclusive: add the probabilities.
Answer
(b). $\frac{2}{5} + \frac{1}{3} = \frac{6 + 5}{15} = \frac{11}{15}$ .
Q2 · 1 mark · MCQHarder board level
If $P (A \cup B) = 0.9$ and $P (A \cap B) = 0.3$ , then $P (A^{'}) + P (B^{'})$ equals:
(a) 0.8
(b) 1.2
(c) 0.6
(d) 1.0
Hint
$P (A) + P (B) = P (A \cup B) + P (A \cap B)$ .
Answer
(a). $P (A) + P (B) = 1.2$ , so $P (A^{'}) + P (B^{'}) = 2 - 1.2 = 0.8$ .
Q3 · 3 marks · Short answerHarder board level
A card is drawn at random from a pack of 52. Find the probability that it is neither a spade nor a face card (J, Q, K).
Hint
Count spades or face cards by the addition rule.
Answer
Spades 13, face cards 12, spade face cards 3. Spade or face: $13 + 12 - 3 = 22$ . Neither: $52 - 22 = 30$ . $P = \frac{30}{52} = \frac{15}{26}$ .
Q4 · 1 mark · Assertion-reasonHarder board level
Assertion (A): There are events $A$ and $B$ with $P (A) = 0.3$ , $P (B) = 0.4$ and $P (A \cup B) = 0.8$ .
Reason (R): For any events, $P (A \cup B) \leq P (A) + P (B)$ .
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
Hint
Compare $0.8$ with $0.3 + 0.4$ .
Answer
(d). $0.8 > 0.3 + 0.4 = 0.7$ , which breaks $P (A \cup B) \leq P (A) + P (B)$ , so such events cannot exist (A is false); R is a true result.
Q5 · 4 marks · Short answerHOTS / competency
For events $A$ and $B$ , $P (A) = \frac{1}{2}$ , $P (B) = \frac{1}{3}$ and $P (A^{'} \cap B^{'}) = \frac{1}{4}$ . Find $P (A \cap B)$ , $P (A \cap B^{'})$ and the probability that exactly one of $A, B$ occurs.
Hint
$P (A^{'} \cap B^{'}) = 1 - P (A \cup B)$ .
Answer
$P (A \cup B) = 1 - \frac{1}{4} = \frac{3}{4}$ . $P (A \cap B) = \frac{1}{2} + \frac{1}{3} - \frac{3}{4} = \frac{1}{12}$ . $P (A \cap B^{'}) = \frac{1}{2} - \frac{1}{12} = \frac{5}{12}$ . Exactly one $= \frac{5}{12} + (\frac{1}{3} - \frac{1}{12}) = \frac{5}{12} + \frac{3}{12} = \frac{2}{3}$ .

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.What is P(A ∪ B)?Show answer
P(A) + P(B) − P(A ∩ B).
2.What is P(neither)?Show answer
1 − P(A ∪ B).
3.What is P(exactly one)?Show answer
P(A) + P(B) − 2P(A ∩ B).
4.What is P(A ∪ B) for mutually exclusive events?Show answer
P(A) + P(B).
5.What is the smallest possible P(A ∩ B)?Show answer
max(0, P(A) + P(B) − 1).

Addition Rule, Complement and Bounds on Probability: frequently asked questions

What is the difference between mutually exclusive and exhaustive?

Mutually exclusive events cannot happen together. Exhaustive events cover the whole sample space between them.

Why does the lcm appear when counting multiples?

A number is divisible by both a and b exactly when it is divisible by their lcm.

Can P(A ∩ B) be larger than P(A)?

No. The intersection is part of A, so its probability cannot exceed P(A) or P(B).

Is the axiomatic approach examined?

The three axioms are the base for these rules. The chapter page notes that the introductory sample-space section is formative only in 2026-27.