Probability · Lesson 2 of 9

The rules of probability

Counting only works when outcomes are equally likely. The rules in this lesson work always. There are just three axioms; every formula here, including the ones examiners like to trap you with, is a consequence of them.

← Probability moduleLesson 2 of 9 · about 30 minutes

A question to hold on to

P(A) = 0.4 and P(B) = 0.7. What are the largest and the smallest values P(A and B) can possibly take?

Delhi School of Economics entrance, 2009Solved at the end ↓

Language

Events are sets

An event is a set of outcomes, so the everyday words in a question translate directly into set operations. Get the translation right and half the problem is solved.

The question says	Set	Read as
A or B (or both)	$A \cup B$	union
A and B	$A \cap B$	intersection
not A	$A^{c}$	complement
A but not B	$A \cap B^{c}$	difference, A − B
neither A nor B	$(A \cup B)^{c} = A^{c} \cap B^{c}$	De Morgan
not both	$(A \cap B)^{c} = A^{c} \cup B^{c}$	De Morgan
exactly one of them	$(A \cap B^{c}) \cup (A^{c} \cap B)$	symmetric difference

De Morgan in one sentence

To negate “or”, negate both parts and switch to “and”; to negate “and”, negate both and switch to “or”. “Not (rain or wind)” is “no rain and no wind”.

Foundations

Three axioms, and everything that follows

In 1933 Andrey Kolmogorov showed that all of probability follows from three rules. A probability P assigns each event a number such that:

Kolmogorov's axioms

(1) P (A) \geq 0 (2) P (S) = 1 (3) P (A \cup B) = P (A) + P (B) if A \cap B = \emptyset

Everything else is a consequence. Four you will use constantly:

Complement. A and Aᶜ are disjoint and together make S, so $P (A) + P (A^{c}) = 1$ .
Impossible event. $P (\emptyset) = 0$ , and no probability exceeds 1.
Monotonicity. If A is inside B, then $P (A) \leq P (B)$ . In particular $P (A \cap B) \leq P (A)$ : adding a condition can never make something more likely.
Unequal outcomes. When outcomes are not equally likely, give each a weight, and fix the weights by making them add to 1.

Pause and predict

A die is rigged so that each odd number is twice as likely as each even number. What is the probability of rolling more than 3? (CUET 2024)

JNU 2015 · foundation

A, B and C are mutually exclusive and exhaustive, with P(B) =

\frac{3}{2}

P(A) and P(C) =

\frac{1}{2}

P(B). Find P(A).

Write everything in terms of P(A) = x.
P(B) = 1.5x and P(C) = 0.75x.
Exhaustive and mutually exclusive means they add to 1.
$x + 1.5 x + 0.75 x = 3.25 x = 1 \Rightarrow x = \frac{4}{13}$

Answer

4/13.

DSE uses the same weighting idea in words. Five fibres 1 to 5 inches long, each breaking first with probability proportional to its length: the weights are k/15, so a fibre of at most 3 inches breaks first with probability (1 + 2 + 3)/15 = 2/5.

Interactive

The addition rule

Axiom 3 adds probabilities only for events that cannot happen together. For events that can, adding P(A) and P(B) counts the overlap twice, so we subtract it once. Move the three sliders: every number below updates, and the Venn shades whichever set you pick.

Addition rule, two events

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

P(A) = 0.40P(B) = 0.70P(A ∩ B) = 0.20can only be between 0.10 and 0.40

P(A) + P(B) = 1.10 > 1, so A and B must overlap by at least 0.10. There is not enough room in the sample space for them to be mutually exclusive.

P(A ∪ B) = 0.90

The sample space, to scale

A sits at the left. B is slid as far right as the overlap allows. Push P(A) + P(B) past 1 and B runs out of room.

P(A ∪ B) = P(A) + P(B) − P(A ∩ B): 0.90
P(A ∩ Bᶜ) = P(A) − P(A ∩ B): 0.20
P(neither) = 1 − P(A ∪ B): 0.10
P(A) × P(B): 0.28

Not mutually exclusiveNot independent

Three exam questions, one rule each way round. CUET 2018: 60% want Business Economics, 70% Financial Economics, 45% both; at least one is 0.60 + 0.70 − 0.45 = 0.85. CUET 2019: P(A) = 0.25, P(B) = 0.5, P(A ∩ B) = 0.14; neither is 1 − (0.25 + 0.5 − 0.14) = 0.39. UGC NET 2024: P(Aᶜ) = 0.4 and P(A ∩ B) = 0.2 give P(A ∩ Bᶜ) = 0.6 − 0.2 = 0.4: the part of A outside B.

Definitions

Mutually exclusive and exhaustive

Events are mutually exclusive (disjoint) if no outcome belongs to more than one of them: they cannot happen together, so P(A ∩ B) = 0 and the addition rule loses its last term. They are exhaustive if together they cover the whole sample space: one of them must happen. Events that are both form a partition, and their probabilities add to exactly 1.

With two dice, “sum is 10” and “sum below 7” are mutually exclusive: no pair can do both. “A double” and “sum is 10” are not: (5,5) is in both. JNU asked exactly this in 2011. And a woman who buys a dress at shop 1, 2 or 3 with probabilities 0.22, 0.18 and 0.35 buys at one of them with probability 0.75, because she can’t buy the one dress in two shops (Hyderabad 2012).

Pause and predict

P(A) = 0.6 and P(B) = 0.5. Could A and B be mutually exclusive?

Mutually exclusive is not independent

Mutually exclusive means the events cannot happen together. Independent (Lesson 3) means one happening does not change the chance of the other. These are close to opposites: if A and B are mutually exclusive and both possible, learning that A happened tells you B certainly did not, so they are not independent. Papers test this confusion every year.

Inclusion–exclusion

Three events at once

For three events, adding the three probabilities counts each pairwise overlap twice, so subtract the pairs; but that removes the triple overlap once too often, so add it back. Include, exclude, include.

Addition rule, 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)

IIT JAM 2024 · foundation

Each of 32 students takes at least one of English, Hindi and Mathematics. 16 take English, 8 Hindi and 18 Mathematics; 4 take English and Hindi, 5 Hindi and Mathematics, and 5 English and Mathematics. How many take Mathematics only?

Find the triple overlap first.
Everyone takes at least one, so the union is 32: 32 = 16 + 8 + 18 − 4 − 5 − 5 + x, which gives x = 4.
Fill the Venn from the centre out.
Mathematics and Hindi only: 5 − 4 = 1. Mathematics and English only: 5 − 4 = 1.
Mathematics only is what is left.
18 − 4 − 1 − 1 = 12. Equivalently 18 − 5 − 5 + 4.

Answer

12.

DSE 2008 · advanced

Of poor rural households, 40% own a radio, 15% a television and 60% a bicycle; 5% own a radio and a TV, 26% a radio and a bicycle, 5% a TV and a bicycle, and 1% all three. A household that owns exactly one of the three is picked. What is the probability that it is a bicycle?

Radio only.
40 − 5 − 26 + 1 = 10%. (Subtract both pairs, add back the triple you removed twice.)
TV only and bicycle only.
15 − 5 − 5 + 1 = 6%; 60 − 26 − 5 + 1 = 30%.
Given exactly one, the bicycle's share.
$\frac{30}{10 + 6 + 30} = \frac{30}{46} = \frac{15}{23}$

Answer

15/23. The last step is a conditional probability, the subject of Lesson 3: restrict to the households that own exactly one.

Solving the opening question

How big can the overlap be?

The overlap cannot be bigger than either event: it sits inside both. And it cannot be too small, because the union cannot exceed 1. Since $P (A \cup B) = P (A) + P (B) - P (A \cap B) \leq 1$ , the overlap must be at least $P (A) + P (B) - 1$ .

Bounds on an intersection and a union

max {0, P (A) + P (B) - 1} \leq P (A \cap B) \leq min {P (A), P (B)}

max {P (A), P (B)} \leq P (A \cup B) \leq min {1, P (A) + P (B)}

The opening question · DSE 2009

P(A) = 0.4 and P(B) = 0.7. Find the largest and smallest possible P(A ∩ B).

Largest: A sits entirely inside B.
P(A ∩ B) = min(0.4, 0.7) = 0.4.
Smallest: push them apart as far as the space allows.
0.4 + 0.7 − 1 = 0.1. There is only 1 of room, so at least 0.1 must be shared.
See it.
Set the lab to 0.40 and 0.70 and use the two overlap buttons: the bar shows B running out of room.

Answer

Maximum 0.4, minimum 0.1.

Watch out for

The mistakes examiners are fishing for

Adding without subtracting the overlap. P(A ∪ B) = P(A) + P(B) only when A and B are mutually exclusive.
The conjunction fallacy. JNU 2012 described Linda, a philosophy graduate and activist, and asked which is more likely: “bank teller” or “bank teller and feminist”. Whatever the story, P(A ∩ B) ≤ P(A), so the first is at least as likely.
Numbers that cannot exist. UGC NET December 2025 said 3% of men have cancer, 15% smoke, and 10% smoke or have cancer, then asked for P(both). The key applies the addition rule: 0.03 + 0.15 − 0.10 = 0.08. But “smoker or cancer” includes every smoker, so it cannot be below 15%: the data are impossible. If this happens in your paper, choose the answer the formula gives, and don’t lose time doubting yourself.
Treating unequal outcomes as equal. A rigged die or unequal class sizes need weights that add to 1.

Revise

The lesson on one screen

Translate: or = ∪, and = ∩, not = complement, neither = (A ∪ B)ᶜ = Aᶜ ∩ Bᶜ, not both = Aᶜ ∪ Bᶜ.
Axioms: P(A) ≥ 0; P(S) = 1; disjoint events add.
Consequences: P(Aᶜ) = 1 − P(A); P(∅) = 0; if A ⊆ B then P(A) ≤ P(B); so P(A ∩ B) ≤ P(A).
Addition rule: P(A ∪ B) = P(A) + P(B) − P(A ∩ B). Three events: add singles, subtract pairs, add the triple.
A but not B: P(A ∩ Bᶜ) = P(A) − P(A ∩ B).
Mutually exclusive: Can't happen together: P(A ∩ B) = 0. Impossible if P(A) + P(B) > 1. Not the same as independent.
Partition: Mutually exclusive and exhaustive: probabilities add to 1.
Bounds: max(0, a + b − 1) ≤ P(A ∩ B) ≤ min(a, b).

Practice

Check yourself

0 of 8 correct so far

Use the rules

Overlap

P(A) = 0.3, P(B) = 0.5 and P(A ∪ B) = 0.6. Find P(A ∩ B).

P(A ∩ B)

A but not B

With the same numbers, find P(A ∩ Bᶜ), the chance that A happens but B does not.

P(A ∩ Bᶜ)

Neither

In a town, 60% read newspaper X, 50% read Y and 30% read both. What share reads neither?

Three subjects

Of 100 students, 50 take Economics, 40 Maths and 30 Statistics; 20 take Economics and Maths, 10 Maths and Statistics, 15 Economics and Statistics, and 5 all three. How many take at least one?

Students

Quick questions

1A and B are mutually exclusive, with P(A) = 0.3 and P(B) = 0.4. P(A ∪ B) is

2(A ∪ B)ᶜ equals

3P(A) = 0.8 and P(B) = 0.7. The smallest possible value of P(A ∩ B) is

4Which of these holds for any two events?

Real questions

Past papers, by level

Foundation

Definitions and one-step problems. Most CUET and UGC NET questions sit here.

12 questionsPractise →

Standard

Two or three steps, up to the hardest JAM, JNU and Hyderabad questions.

10 questionsPractise →

Delhi School level

This topic’s hardest questions, from the DSE entrance, are worked through in Lesson 9: Counting and the rules.

18 questionsLesson 9 →

All 22 past-paper questions for this lesson, easiest first: practise the full set →

These are questions from the TestStop bank. A free account answers 10 a day, each with its worked solution; All-Access opens every set with no daily limit.

← Lesson 1Sample spaces and counting Lesson 3 →Conditional probability and independence

The rules of probability

Events are sets

Three axioms, and everything that follows

The addition rule

Mutually exclusive and exhaustive

Three events at once

How big can the overlap be?

The mistakes examiners are fishing for

The lesson on one screen

Check yourself

Use the rules

Quick questions

Past papers, by level

Foundation

Standard

Delhi School level

Read next

The rules of probability

Events are sets

Three axioms, and everything that follows

The addition rule

Mutually exclusive and exhaustive

Three events at once

How big can the overlap be?

The mistakes examiners are fishing for

The lesson on one screen

Check yourself

Use the rules

Quick questions

Past papers, by level

Foundation

Standard

Delhi School level

Read next