Probability Quick Lesson: Class 12 Maths Chapter 13 (NCERT)

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The chapter in one minute

What the chapter is about, and what the exam asks.

Probability tells you how likely something is, as a number from 0 to 1. This chapter asks one new question: how does the chance change when you already know something?

If you know that $F$ has happened, only the outcomes inside $F$ still count. That is conditional probability, $P(E|F)$ ⁠, read "E given F".

The rest grows from it: the multiplication rule (one thing after another), independent events (knowing one tells you nothing about the other) and Bayes' theorem (from a result back to its cause). In recent CBSE sample papers, Bayes' theorem comes as a 4-mark case study or a 5-mark long answer.

Most exam questions on this chapter are one of eight kinds, and each kind has a fixed recipe. You will learn all eight below.

8question types

12formula cards

0 to 1where every probability lies: a quick check on each answer

Fractionskeep every answer exact

Formulas

The formulas you must know

Learn these cards. Every question in the chapter is built from them.

Equally likely outcomes

P(E) = \frac{n(E)}{n(S)}

Count the outcomes you want, and divide by all the outcomes.

Not E

P(E') = 1 - P(E)

Often the quick way in: find the chance that it does not happen.

Or: the addition rule

\begin{aligned} &P(A\cup B) \\ &= P(A) + P(B) \\ &\quad - P(A\cap B) \end{aligned}

$\cup$ means "or" and $\cap$ means "and". Take the overlap away once.

Conditional probability

P(E|F) = \frac{P(E\cap F)}{P(F)}

Read it "E given F". The event after the bar goes on the bottom, and $P(F)$ must not be 0.

Conditional, by counting

P(E|F) = \frac{n(E\cap F)}{n(F)}

For equally likely outcomes. $F$ becomes the new sample space.

Multiplication rule

\begin{aligned} &P(E\cap F) \\ &= P(E)\,P(F|E) \end{aligned}

One thing after another: the first, times the second given the first. $P(E)$ must not be 0.

Three in a row

\begin{aligned} &P(E\cap F\cap G) \\ &= P(E)\,P(F|E) \\ &\quad\times P(G|E\cap F) \end{aligned}

Keep multiplying, one condition at a time.

Independent events

P(E\cap F) = P(E)\,P(F)

This is the test for independence, and the rule to use once you know it.

Mutually exclusive events

A\cap B = \phi

They cannot happen together, so $P(A\cap B) = 0$ and $P(A\cup B) = P(A) + P(B)$ ⁠.

At least one (independent)

\begin{aligned} &P(A\cup B) \\ &= 1 - P(A')\,P(B') \end{aligned}

One minus the chance that neither happens.

Total probability

\begin{aligned} &P(A) = P(E_1)P(A|E_1) \\ &\quad + P(E_2)P(A|E_2) \end{aligned}

$E_1$ and $E_2$ are the causes, and they must form a partition: exactly one of them happens. Add one more term for each extra cause.

Bayes' theorem

\begin{aligned} &P(E_1|A) \\ &= \frac{P(E_1)P(A|E_1)}{P(A)} \end{aligned}

Top: one term of the total. Bottom: the whole total, from the card before this one.

Question types

The question types and how to crack each one

For each type: how to spot it, what to do, and one real NCERT question worked the exam way.

1. Given numbers: divide by the event after the bar

Spot it: the question gives $P(E)$ ⁠, $P(F)$ and $P(E\cap F)$ ⁠, and asks for $P(E|F)$ or $P(F|E)$ ⁠.

Write the formula first: $P(E|F) = \frac{P(E\cap F)}{P(F)}$ ⁠.
Put in the numbers and simplify. A fraction of decimals clears at once: $\frac{0.2}{0.3} = \frac{2}{3}$ ⁠.
For $P(F|E)$ ⁠, the bottom changes to $P(E)$ ⁠.

Exercise 13.1 Q1. Given $P(E) = 0.6$ ⁠, $P(F) = 0.3$ and $P(E\cap F) = 0.2$ ⁠, find $P(E|F)$ and $P(F|E)$ ⁠.

\begin{aligned} P(E|F) &= \frac{P(E\cap F)}{P(F)} \\ &= \frac{0.2}{0.3} = \frac{2}{3} \end{aligned}

\begin{aligned} P(F|E) &= \frac{P(E\cap F)}{P(E)} \\ &= \frac{0.2}{0.6} = \frac{1}{3} \end{aligned}

∴ $P(E|F) = \frac{2}{3}$ and $P(F|E) = \frac{1}{3}$

See a full exam answer · Book Example 1 · Go deeper: see it visually · Basics: dividing fractions

In past papers (7): click to see

Board 2025 Q7 · Board 2024 Q11 · Board 2023 Q12 · Board 2023 Q36(iii)(a) · Board 2022 T2 Q4 · Sample 2024-25 Q8 · Sample 2020-21 Q28(b)

2. Dice, coins and cards: shrink the sample space

Spot it: equally likely outcomes, and the words "given that" or "it is known that".

List $F$ ⁠, the outcomes you are told have happened. It is your new sample space.
Count the outcomes of $F$ that are also in $E$ ⁠.
Divide: $P(E|F) = \frac{n(E\cap F)}{n(F)}$ ⁠.

Exercise 13.1 Q14. Two dice are thrown, and the numbers are different. Find the probability that the sum is 4.

Let $F$ ⁠: the numbers are different, and $E$ ⁠: the sum is 4. Of the 36 outcomes, 6 are doubles, so

n(F) = 36 - 6 = 30

$E\cap F = \{(1, 3), (3, 1)\}$ ⁠. The double $(2, 2)$ is not in $F$ ⁠.

\begin{aligned} P(E|F) &= \frac{n(E\cap F)}{n(F)} \\ &= \frac{2}{30} \end{aligned}

∴ $P(E|F) = \frac{1}{15}$

See a full exam answer · Book Examples 2, 3 and 6 · Go deeper: see it visually · Basics: counting outcomes

In past papers (2): click to see

Board 2026 Q19 · Sample 2026-27 Q17

3. One after the other: the multiplication rule

Spot it: draws "without replacement", or a second event that depends on the first.

Name the events in order: $E$ first, $F$ second.
Find $P(E)$ ⁠, then $P(F|E)$ ⁠. After the first draw there is one item fewer.
Multiply: $P(E\cap F) = P(E)\,P(F|E)$ ⁠.

Exercise 13.2 Q2. Two cards are drawn at random, without replacement, from a pack of 52. Find the probability that both are black.

Let $E$ ⁠: the first card is black, and $F$ ⁠: the second card is black.

$P(E) = \frac{26}{52}$ ⁠. After a black card, 25 of the 51 cards left are black, so $P(F|E) = \frac{25}{51}$ ⁠.

\begin{aligned} P(E\cap F) &= P(E)\,P(F|E) \\ &= \frac{26}{52}\times\frac{25}{51} \end{aligned}

∴ $P(E\cap F) = \frac{25}{102}$

See a full exam answer · Book Examples 8 and 9 · Go deeper: see it visually · Basics: tree diagrams

In past papers (3): click to see

Sample 2023-24 Q36(i) · Sample 2021-22 T2 Q6 · Sample 2020-21 Q18(ii)

4. Are they independent? Test the product

Spot it: "Are $A$ and $B$ independent?" or "Check whether the events are independent".

Find $P(A)$ ⁠, $P(B)$ and $P(A\cap B)$ ⁠. List the outcomes if you need to.
Work out $P(A)\,P(B)$ ⁠.
Equal to $P(A\cap B)$ ⁠: independent. Not equal: not independent. Write the conclusion in words.

Exercise 13.2 Q5. A die has 1, 2, 3 in red and 4, 5, 6 in green. $A$ ⁠: the number is even. $B$ ⁠: the number is red. Are $A$ and $B$ independent?

$A = \{2, 4, 6\}$ ⁠, $B = \{1, 2, 3\}$ and $A\cap B = \{2\}$ ⁠.

\begin{aligned} P(A)\,P(B) &= \frac{3}{6}\times\frac{3}{6} \\ &= \frac{1}{4} \end{aligned}

P(A\cap B) = \frac{1}{6} \ne \frac{1}{4}

∴ $A$ and $B$ are not independent.

See a full exam answer · Book Examples 10 to 12 · Go deeper: see it visually · Basics: events as sets

5. Mutually exclusive or independent: two different rules

Spot it: the question uses one of the two words, or asks you to compare them.

Mutually exclusive means they cannot happen together: $P(A\cap B) = 0$ ⁠.
Independent means one does not change the chance of the other: $P(A\cap B) = P(A)\,P(B)$ ⁠.
Put the right one into the addition rule, and solve.

Exercise 13.2 Q7. $P(A) = \frac{1}{2}$ ⁠, $P(A\cup B) = \frac{3}{5}$ and $P(B) = p$ ⁠. Find $p$ if $A$ and $B$ are (i) mutually exclusive, (ii) independent.

(i) Here $P(A\cap B) = 0$ ⁠, so $P(A\cup B) = P(A) + P(B)$ ⁠.

\frac{3}{5} = \frac{1}{2} + p

p = \frac{3}{5} - \frac{1}{2} = \frac{1}{10}

(ii) Here $P(A\cap B) = P(A)\,P(B) = \frac{p}{2}$ ⁠.

\begin{aligned} \frac{3}{5} &= \frac{1}{2} + p - \frac{p}{2} \\ &= \frac{1}{2} + \frac{p}{2} \end{aligned}

p = 2\left(\frac{3}{5} - \frac{1}{2}\right) = \frac{1}{5}

∴ (i) $p = \frac{1}{10}$ ⁠, (ii) $p = \frac{1}{5}$

See a full exam answer · Go deeper: see it visually · Basics: the addition rule

In past papers (1): click to see

Board 2024 Q31

6. At least one, or exactly one: think "not" first

Spot it: independent events, with "at least one", "the problem is solved", "neither" or "exactly one".

Write the chance that each one fails: $P(A') = 1 - P(A)$ ⁠.
At least one: $1 - P(A')\,P(B')$ ⁠.
Exactly one: $P(A)\,P(B') + P(A')\,P(B)$ ⁠.

Exercise 13.2 Q14. A and B try a problem independently, and solve it with probabilities $\frac{1}{2}$ and $\frac{1}{3}$ ⁠. Find the probability that (i) the problem is solved, (ii) exactly one of them solves it.

Let $A$ ⁠: A solves it, and $B$ ⁠: B solves it. Then $P(A') = \frac{1}{2}$ and $P(B') = \frac{2}{3}$ ⁠.

(i) The problem is solved if at least one of them solves it.

\begin{aligned} P(A\cup B) &= 1 - P(A')\,P(B') \\ &= 1 - \frac{1}{2}\times\frac{2}{3} \end{aligned}

(ii) Exactly one: A but not B, or B but not A.

\begin{aligned} &P(A\cap B') + P(A'\cap B) \\ &= \frac{1}{2}\times\frac{2}{3} + \frac{1}{2}\times\frac{1}{3} \\ &= \frac{1}{3} + \frac{1}{6} \end{aligned}

∴ (i) $P(A\cup B) = \frac{2}{3}$ ⁠, (ii) $P(A\cap B') + P(A'\cap B) = \frac{1}{2}$

See a full exam answer · Book Example 14 · Go deeper: see it visually · Basics: Venn diagrams

In past papers (8): click to see

Board 2026 Q37(iii) · Board 2023 Q30(b) · Sample 2026-27 Q18 · Sample 2025-26 Q31 · Sample 2023-24 Q14 · Sample 2022-23 Q14 · Sample 2022-23 Q38(i) · Sample 2020-21 Q15

7. Two routes to one result: total probability

Spot it: the result can happen in two (or more) ways, such as a ball from bag I or from bag II, and you want its overall chance.

Name the causes $E_1$ ⁠, $E_2$ and the result $A$ ⁠. The causes must form a partition: exactly one of them happens.
Write $P(E_1)$ ⁠, $P(E_2)$ ⁠, $P(A|E_1)$ and $P(A|E_2)$ ⁠.
Multiply along each route and add, as on the total probability card.

Exercise 13.3 Q1. An urn has 5 red and 5 black balls. A ball is drawn, its colour is noted, and it is put back with 2 more balls of that colour. A second ball is then drawn. Find the probability that it is red.

Let $E_1$ ⁠: the first ball is red, $E_2$ ⁠: the first ball is black, and $A$ ⁠: the second ball is red. $E_1$ and $E_2$ form a partition: the first ball is one colour or the other.

$P(E_1) = \frac{5}{10} = \frac{1}{2}$ and $P(E_2) = \frac{1}{2}$ ⁠.

After a red ball the urn has 7 red balls out of 12, so $P(A|E_1) = \frac{7}{12}$ ⁠. After a black ball it has 5 red out of 12, so $P(A|E_2) = \frac{5}{12}$ ⁠.

\begin{aligned} P(A) &= P(E_1)P(A|E_1) \\ &\quad + P(E_2)P(A|E_2) \\ &= \frac{1}{2}\cdot\frac{7}{12} + \frac{1}{2}\cdot\frac{5}{12} \\ &= \frac{7}{24} + \frac{5}{24} \end{aligned}

∴ $P(A) = \frac{12}{24} = \frac{1}{2}$

See a full exam answer · Book Example 15 · Go deeper: see it visually · Basics: tree diagrams

In past papers (11): click to see

Board 2026 Q31(a) · Board 2025 Q30(b) · Board 2025 Q38(i) · Board 2024 Q37(i) · Sample 2026-27 Q35(i) · Sample 2025-26 Q38 I. · Sample 2024-25 Q31(a)(ii) · Sample 2024-25 Q38(i) · Sample 2023-24 Q36(ii) · Sample 2021-22 T2 Q14(i) · Sample 2020-21 Q18(iii)

8. Bayes' theorem: from the result back to the cause

Spot it: the result has already happened (a red ball, a faulty item, an A grade), and you are asked for the chance that it came from one cause. It often comes as a case study or a 5-mark question.

Define the events: the causes $E_1$ ⁠, $E_2$ (a partition: exactly one happens) and the result $A$ ⁠.
List $P(E_1)$ ⁠, $P(E_2)$ ⁠, $P(A|E_1)$ ⁠, $P(A|E_2)$ as fractions.
Find $P(A)$ with the total probability card.
Write Bayes' formula, substitute, and give the answer.

Exercise 13.3 Q3. In a college, 60% of students live in the hostel and 40% are day scholars. 30% of hostel students and 20% of day scholars get an A grade. A student chosen at random has an A grade. Find the probability that the student lives in the hostel.

Let $E_1$ ⁠: the student lives in the hostel, $E_2$ ⁠: the student is a day scholar, and $A$ ⁠: the student has an A grade. $E_1$ and $E_2$ form a partition: each student is one or the other.

$P(E_1) = \frac{3}{5}$ ⁠, $P(E_2) = \frac{2}{5}$ ⁠, $P(A|E_1) = \frac{3}{10}$ and $P(A|E_2) = \frac{1}{5}$ ⁠.

By total probability,

\begin{aligned} P(A) &= \frac{3}{5}\cdot\frac{3}{10} + \frac{2}{5}\cdot\frac{1}{5} \\ &= \frac{9}{50} + \frac{4}{50} = \frac{13}{50} \end{aligned}

By Bayes' theorem,

\begin{aligned} P(E_1|A) &= \frac{P(E_1)P(A|E_1)}{P(A)} \\ &= \frac{9}{50}\times\frac{50}{13} \end{aligned}

∴ $P(E_1|A) = \frac{9}{13}$

See a full exam answer · Book Examples 16 to 21 · Go deeper: see it visually · Basics: dividing fractions

In past papers (12): click to see

Board 2025 Q38(ii) · Board 2024 Q37(ii) · Board 2022 T2 Q12 · Sample 2026-27 Q35(ii) · Sample 2025-26 Q38 II. · Sample 2024-25 Q38(ii) · Sample 2023-24 Q36(iii)(a) · Sample 2023-24 Q36(iii)(b) · Sample 2022-23 Q38(ii) · Sample 2021-22 T2 Q14(ii) · Sample 2020-21 Q18(iv) · Sample 2020-21 Q18(v)

Shortcuts

Shortcuts and quick tips

Small habits that save time and marks.

The word "given" tells you the bottom: in $P(E|F)$ ⁠, you divide by $P(F)$ ⁠.
Every probability lies between 0 and 1. An answer above 1 usually means the wrong bottom.
Turn decimals and percentages into fractions straight away: 30% is $\frac{3}{10}$ ⁠. Fractions keep every answer exact.
"At least one" is 1 minus "none". With two dice, the chance of at least one six is $1 - \left(\frac{5}{6}\right)^2 = \frac{11}{36}$ ⁠.
Draw a tree for one-after-the-other and Bayes questions: multiply along a branch, then add the branches you want.
Without replacement, the bottom drops by one each draw: $\frac{26}{52}$ ⁠, then $\frac{25}{51}$ ⁠.
In Bayes, the answers for all the causes add up to 1: $P(E_1|A) + P(E_2|A) = 1$ ⁠. It is a quick check.
Quick MCQ facts, when $P(A) \ne 0$ ⁠: if $A\subset B$ ⁠, then $P(B|A) = 1$ ⁠. If $A\cap B = \phi$ ⁠, then $P(B|A) = 0$ ⁠.
Mutually exclusive events with chances above 0 are never independent: $P(A\cap B) = 0$ ⁠, but $P(A)\,P(B)$ is above 0.
When $P(B)$ is 0, $P(A|B)$ is not defined, because you cannot divide by 0.

Watch out

Mistakes that cost marks

Each one is common, and each has an easy fix.

Dividing by the wrong event → in $P(E|F)$ ⁠, the bottom is $P(F)$ ⁠, the event after the bar.
Taking $P(A|B)$ and $P(B|A)$ as the same → both have $P(A\cap B)$ on top, but the bottoms are $P(B)$ and $P(A)$ ⁠.
Thinking mutually exclusive means independent → mutually exclusive: $P(A\cap B) = 0$ ⁠. Independent: $P(A\cap B) = P(A)\,P(B)$ ⁠.
Multiplying $P(A)\,P(B)$ for events that are not independent → use $P(A)\,P(B|A)$ ⁠.
Keeping the same bottom without replacement → one item fewer each time: $\frac{26}{52}$ ⁠, then $\frac{25}{51}$ ⁠.
Stopping at $P(A)$ in a Bayes question → $P(A)$ is only the bottom; divide the term for your cause by it.
Forgetting the overlap in "or" → $P(A\cup B) = P(A) + P(B) - P(A\cap B)$ ⁠.
Giving only a rounded decimal → give the exact fraction, such as $\frac{9}{13}$ ⁠.

Marks

How the examiner gives marks

From CBSE's own marking schemes. Write these lines and the marks follow.

Start by defining the events ("Let $E_1$ be the event that ..."). The marking scheme expects these lines in a probability answer.
Write the rule you use (conditional probability, the multiplication rule, total probability or Bayes' theorem) before you put the numbers in.
Show the substitution with the numbers in place, then the value. Each line that changes the working can earn its own part of the marks.
Conclusions in words carry marks: write "so $A$ and $B$ are independent" or "not independent".
A case study is marked part by part: answer (i), (ii) and (iii) separately, each labelled.
Give the final answer on its own line, as a fraction in lowest terms. A slip is penalised only once, so keep going. An MCQ earns 1 mark for the right option only.

Practice

Practise these first

Easiest first. Together they cover every type above.

Exercise 13.1 Q3 conditional probability from numbers, and the multiplication rule
Exercise 13.1 Q12 two children: shrink the sample space
Exercise 13.1 Q10 two dice, given one of them
Exercise 13.2 Q6 test the product for independence
Exercise 13.2 Q18 an MCQ: independent is not mutually exclusive
Exercise 13.2 Q11 and, or, neither, for independent events
Exercise 13.2 Q12 at least once: 1 minus none
Exercise 13.2 Q3 three draws without replacement
Example 15 total probability with two causes
Exercise 13.3 Q8 Bayes' theorem with two machines
Exercise 13.3 Q4 Bayes' theorem: knows the answer or guesses
Miscellaneous Q10 a ball moved between bags: total probability, then Bayes

Eight kinds of question, one at a time. Work through the practice list, check that each answer lies between 0 and 1, and you are ready for the exam.

Ready to go deeper? Open the interactive lesson

Probability, the quick way