Probability: Class 12 Maths · Chapter 13

Updating chances when you learn something new

Mathematics Part 2 · Companion notes, simulations and practice · Official NCERT chapter PDF ↗

Published 12 Oct 2026 · Updated 12 Oct 2026

100 min
1 simulation
66 practice questions
21 MCQs

What you will learn in Probability

Find the conditional probability P(E|F) by shrinking the sample space to F, or from P(E ∩ F)/P(F), and use its properties
Use the multiplication rule to find the probability that two or three events all happen, including draws without replacement
Decide whether two or three events are independent, and keep independent and mutually exclusive events apart
Recognise a partition of a sample space and use the theorem of total probability
Use Bayes' theorem to update a prior probability into a posterior one, with trees and natural frequencies as checks
Describe a random variable as a real-valued function on a sample space and list the values it takes

Original companion text aligned to NCERT Mathematics Class 12, Chapter 13, pp. 406–438. Not affiliated with or endorsed by NCERT.

13.1 Introduction

Will it rain on the day of the inter-house sports meet? Will the 8:15 local reach Dadar on time? Since Class 9 you have put a number between 0 and 1 on questions like these. In Class 11 you computed it in two ways: by counting equally likely outcomes, and from Kolmogorov's axioms, which treat probability as a function defined on events. You also met the addition rule, which gives $P (A \cup B)$ as $P (A) + P (B)$ minus the overlap $P (A \cap B)$ .

Real judgements rarely start from zero information. A forecaster revises the chance of rain once clouds gather; a doctor weighs a test report before naming an illness. This chapter is about probability when we learn something: conditional probability, the multiplication rule, independence, and Bayes' theorem, which reasons backwards from an effect to its likely cause. It closes with a first look at random variables. Unless a problem says otherwise, the outcomes of each experiment are equally likely.

Warm up with two dice. Ordered pairs from $(1, 1)$ to $(6, 6)$ give 36 equally likely outcomes, and six of them have a total above 9. The classical probability of a total above 9 is therefore $\frac{6}{36} = \frac{1}{6}$ . The lab below rolls the dice for you, so you can watch the relative frequency approach that value.

SimulationTwo dice: totals above 9

A two-dice probability lab. Buttons roll the pair once, 10, 100 or 1000 times. A bar chart shows how often each total from 2 to 12 has come up, beside its theoretical probability, and a readout compares the relative frequency of a total greater than 9 with the theoretical value one sixth. Nothing moves until the learner presses a roll button.

Fig. 13.1Each bar counts one total from 2 to 12. The tracked event is a total of 10, 11 or 12, with theoretical probability

\frac{1}{6}

Try this

Roll 100 at a time until you have about 1000 rolls. Is the share of totals above 9 close to $\frac{1}{6} \approx 0.17$ ?
Which total comes up most often? Count how many of the 36 ordered pairs give it.
Before Section 13.2: among rolls with a total above 9, what fraction do you expect to contain a 6?

Quick check

1 of 2

Two fair dice are rolled. What is the probability that the total is exactly 9? (A fraction such as 1/5 is fine.)

Show answer

Answer 0.1111

Four ordered pairs give 9:

(3, 6), (4, 5), (5, 4), (6, 3)

. So the probability is

\frac{4}{36} = \frac{1}{9}

. Counting

(4, 5)

and

(5, 4)

as one outcome would give the wrong denominator.

13.2 Conditional Probability

In a board game you need a 6 on either of two dice. The dice are still under the cup when a friend peeks and says, the total is at least 10. Is that good news? Before the hint, the chance of at least one 6 is $\frac{11}{36}$ , a little under a third. After it, only six outcomes remain: $(4, 6)$ , $(5, 5)$ , $(5, 6)$ , $(6, 4)$ , $(6, 5)$ and $(6, 6)$ . Five of these contain a 6, so the chance has jumped to $\frac{5}{6}$ (Fig. 13.2).

2/3Event $F$ , a total of 10 or more, leaves only 6 possible outcomes.

Fig. 13.2Learning that the total is at least 10 shrinks the sample space from 36 outcomes to 6.

A grid of 36 cells, one for each ordered pair of throws, with the first die down the side and the second die along the top; each cell shows the total. In the first step the 11 cells that contain a 6, the last row and the last column, are ringed, and the panel reads E, a 6 shows, 11 of 36 cells. In the second step the six cells with a total of 10, 11 or 12, in the bottom right corner, are outlined and the other cells fade; the panel says only 6 cells are possible now. In the third step the other cells almost vanish, leaving a new sample space of 6 cells, and 5 of them are ringed, so the conditional probability is five sixths.

Let us name the events. $E$ is at least one die shows 6 and $F$ is the total is 10 or more. Once we know $F$ has happened, every outcome outside $F$ is ruled out, and $F$ itself becomes the sample space. The outcomes of $E$ that survive are exactly those of $E \cap F$ . We write the new probability as $P (E ∣ F)$ , read the probability of $E$ given $F$ .

\begin{aligned} P (E ∣ F) \\ = \frac{n (E \cap F)}{n (F)} = \frac{5}{6} \end{aligned}

Divide the top and the bottom by $n (S) = 36$ and the counts become probabilities: $\frac{5 / 36}{6 / 36}$ . This second form does not need equally likely outcomes, so it serves as the definition.

Definition

Conditional probability

Take two events $E$ and $F$ of one sample space, with $P (F) \neq 0$ . The conditional probability of $E$ given $F$ is the share of $F$ 's probability that also lies in $E$ : $P (E ∣ F) = \frac{P (E \cap F)}{P (F)}$ .

ExampleTwo dice: $P (a 6 ∣ total \geq 10) = \frac{5 / 36}{6 / 36} = \frac{5}{6}$ .

Conditional probability

\begin{aligned} P (E ∣ F) \\ = \frac{P (E \cap F)}{P (F)} \\ if P (F) \neq 0 \end{aligned}

(13.1)

where

$P (E \cap F)$: the probability that $E$ and $F$ both happen
$P (F)$: the probability of the condition $F$

Why insist on $P (F) \neq 0$ ? Conditioning on an impossible event gives $\frac{0}{0}$ , which has no value. Notice too that the condition sits on the right of the bar and changes the denominator. Here $P (F ∣ E) = \frac{5 / 36}{11 / 36} = \frac{5}{11}$ , a different number from $P (E ∣ F) = \frac{5}{6}$ .

13.2.1 Properties of conditional probability

Fix an event $F$ with $P (F) \neq 0$ . The rule $E \mapsto P (E ∣ F)$ then behaves like an ordinary probability on the smaller sample space $F$ . Three properties make this precise.

PropertyProperty 1: the condition is certain

$P (S ∣ F) = P (F ∣ F) = 1$ .

PropertyProperty 2: unions given F

For any events $A$ and $B$ , $P (A \cup B ∣ F)$ equals $P (A ∣ F) + P (B ∣ F)$ minus $P (A \cap B ∣ F)$ . When $A$ and $B$ are disjoint the last term is zero.

PropertyProperty 3: complements given F

$P (E^{'} ∣ F) = 1 - P (E ∣ F)$ .

One more consequence: $E \cap F$ is part of $F$ , so $P (E \cap F) \leq P (F)$ , and therefore $0 \leq P (E ∣ F) \leq 1$ , as every probability must be.

Example 1NCERT Example 1 · p. 409

Reading the formula both ways

Events $A$ and $B$ have $P (A) = 0.45$ , $P (B) = 0.6$ and $P (A \cap B) = 0.24$ . Find $P (A ∣ B)$ and $P (B ∣ A)$ .

Try it yourself first. Then reveal the solution one step at a time.

Example 2NCERT Example 2 · p. 409

Three children, at least two girls

A family has three children. Assume each child is equally likely to be a girl or a boy, independently of the others. Given that at least two of the children are girls, find the probability that all three are girls.

Try it yourself first. Then reveal the solution one step at a time.

Example 3NCERT Example 3 · p. 410

A token that is a perfect square

Tokens numbered 1 to 30 are mixed in a bag for a lucky draw, and one is drawn at random. It turns out to be a perfect square. What is the probability that it is odd?

Try it yourself first. Then reveal the solution one step at a time.

Example 4NCERT Example 4 · p. 410

Bus riders in Class 12

A school has 1200 students. Of them, 520 come by the school bus, and 156 of the bus riders are in Class 12. A student chosen at random turns out to be a bus rider. Find the probability that the student is in Class 12.

Try it yourself first. Then reveal the solution one step at a time.

Example 5NCERT Example 5 · p. 411

After three tails

A fair coin is tossed four times. The first three tosses all show tails. Find the probability that the fourth toss shows a head.

Try it yourself first. Then reveal the solution one step at a time.

Example 6NCERT Example 6 · p. 411

A product of 12

A die is thrown twice and the product of the two numbers is 12. What is the probability that one of the throws was a 6?

Try it yourself first. Then reveal the solution one step at a time.

So far every outcome was equally likely. The definition $P (E \cap F) / P (F)$ works just as well when they are not; we then add the probabilities of outcomes instead of counting them. Experiments that happen in stages are best drawn as a tree diagram: one branch for each result of a stage, labelled with its probability given everything before it. Multiplying along a path gives the probability of that path (Fig. 13.3).

3/3No head leaves $\frac{2}{3}$ of the probability; the two 6-paths carry $\frac{1}{6}$ of it.

Fig. 13.3A staged experiment as a tree: multiply along a path, then add the paths you need.NCERT Fig. 13.2 · p. 412

A tree that starts at a dot on the left. Three branches lead to the die results 6, 5 and 1 to 4, with probabilities one sixth, one sixth and two thirds. From 6 and from 5, branches lead to red with one third and green with two thirds; from 1 to 4, branches lead to head and tail with one half each. In the second step the probability of each path appears beside its leaf: one eighteenth, one ninth, one eighteenth, one ninth, one third and one third. In the third step the head leaf fades, the five leaves with no head are outlined, a brace marks the two leaves after a 6 as E and F together, and the result reads P of E given F equals one quarter.

Example 7NCERT Example 7 · p. 412

A die, then a ball or a coin

Meera throws a die. If it shows 5 or 6, she draws a ball from a bag with 1 red and 2 green balls; otherwise she tosses a coin. Find the probability that the die showed 6, given that no head appeared.

Try it yourself first. Then reveal the solution one step at a time.

Quick check

1 of 4

Two dice are rolled and the total is 7. What is the probability that one of the dice shows 2?

\frac{1}{6}

\frac{11}{36}

\frac{1}{3}

\frac{1}{18}

Show answer

Answer (C) $\frac{1}{3}$

A total of 7 has six outcomes, and two of them,

(2, 5)

and

(5, 2)

, contain a 2:

\frac{2}{6} = \frac{1}{3}

. The value

\frac{1}{18}

P (E \cap F)

before dividing by

P (F)

, and

\frac{11}{36}

ignores the condition.

Practice · 17 questions

Practice 13.1: questions with answers

Short answerEasyNCERT Exercise 13.1, Q1
Events $E$ and $F$ have $P (E) = 0.5$ , $P (F) = 0.4$ and $P (E \cap F) = 0.15$ . Find $P (E ∣ F)$ and $P (F ∣ E)$ .
Show solution
Answer $\frac{3}{8}$ and $0.3$
$P (E ∣ F) = \frac{0.15}{0.4} = 0.375 = \frac{3}{8}$ .
$P (F ∣ E) = \frac{0.15}{0.5} = 0.3$ .
Quick answerEasyNCERT Exercise 13.1, Q2
If $P (B) = 0.8$ and $P (A \cap B) = 0.36$ , find $P (A ∣ B)$ .
Show answer
Answer 0.45
$P (A ∣ B) = \frac{0.36}{0.8} = 0.45$ .
3
Short answerMediumNCERT Exercise 13.1, Q3
$P (A) = 0.6$ , $P (B) = 0.45$ and $P (B ∣ A) = 0.5$ .
1. Find $P (A \cap B)$ .
  Show answer
  Answer 0.3
  Multiply $P (A)$ by $P (B ∣ A)$ : $0.6 \times 0.5 = 0.3$ .
2. Find $P (A ∣ B)$ .
  Show answer
  Answer 0.6667
  $P (A ∣ B) = \frac{0.3}{0.45} = \frac{2}{3}$ .
3. Find $P (A \cup B)$ .
  Show answer
  Answer 0.75
  $P (A \cup B) = 0.6 + 0.45 - 0.3 = 0.75$ .
Short answerMediumNCERT Exercise 13.1, Q4
Find $P (A \cup B)$ if $3 P (A) = P (B) = \frac{3}{5}$ and $P (A ∣ B) = \frac{1}{4}$ .
Show solution
Answer $\frac{13}{20}$
$P (B) = \frac{3}{5}$ and $P (A) = \frac{1}{5}$ .
$P (A \cap B) = P (B) P (A ∣ B) = \frac{3}{5} \cdot \frac{1}{4} = \frac{3}{20}$ .
So $P (A \cup B)$ is $\frac{4}{20} + \frac{12}{20} - \frac{3}{20} = \frac{13}{20}$ .
Short answerMediumNCERT Exercise 13.1, Q5
$P (A) = \frac{7}{15}$ , $P (B) = \frac{2}{5}$ and $P (A \cup B) = \frac{3}{5}$ . Find (a) $P (A \cap B)$ , (b) $P (A ∣ B)$ , (c) $P (B ∣ A)$ .
Show solution
Answer $\frac{4}{15}$ , $\frac{2}{3}$ , $\frac{4}{7}$
(a) Read the addition rule backwards: the overlap is $\frac{7}{15} + \frac{6}{15} - \frac{9}{15} = \frac{4}{15}$ .
(b) $P (A ∣ B) = \frac{4 / 15}{6 / 15} = \frac{2}{3}$ .
(c) $P (B ∣ A) = \frac{4 / 15}{7 / 15} = \frac{4}{7}$ .
6
Short answerMediumNCERT Exercise 13.1, Q6
A coin is tossed four times. Find $P (E ∣ F)$ in each case.
1. $E$ : tail on the fourth toss; $F$ : tails on each of the first three tosses.
  Show answer
  Answer 0.5
  $F = {T T T H, T T T T}$ and $E \cap F = {T T T T}$ , so $P (E ∣ F) = \frac{1}{2}$ .
2. $E$ : exactly two heads; $F$ : at least one head.
  Show answer
  Answer 0.4
  $F$ has $16 - 1 = 15$ outcomes. Exactly two heads happens in $(\binom{4}{2}) = 6$ of them, so $P (E ∣ F) = \frac{6}{15} = \frac{2}{5}$ .
3. $E$ : at least three heads; $F$ : head on the first toss.
  Show answer
  Answer 0.5
  Given a first head, we need at least two heads in the last three tosses: $3 + 1 = 4$ of 8 outcomes, so $P (E ∣ F) = \frac{1}{2}$ .
7
Short answerEasyNCERT Exercise 13.1, Q7
A coin is tossed and a die is thrown. Find $P (E ∣ F)$ .
1. $E$ : the die shows 6; $F$ : the coin shows a head.
  Show answer
  Answer 0.16667
  $F$ has six outcomes $(H, 1), \dots, (H, 6)$ and one of them has a 6: $\frac{1}{6}$ .
2. $E$ : a head and a number above 4; $F$ : the coin shows a tail.
  Show answer
  Answer 0
  $E$ needs a head and $F$ says tail, so $E \cap F = \emptyset$ and $P (E ∣ F) = 0$ .
Quick answerEasyNCERT Exercise 13.1, Q8
A die is thrown four times. $E$ : the fourth throw shows more than 4. $F$ : each of the first three throws shows 1. Find $P (E ∣ F)$ .
Show answer
Answer 0.3333
$F$ fixes the first three throws, leaving 6 equally likely values for the fourth; two of them (5, 6) are more than 4. So $P (E ∣ F) = \frac{2}{6} = \frac{1}{3}$ .
Short answerMediumNCERT Exercise 13.1, Q9
Four friends, Asha, Bina, Charu and Dev, sit at random in a row of four seats. $E$ : Asha sits at one end. $F$ : Bina sits in the second seat. Find $P (E ∣ F)$ .
Show solution
Answer $\frac{2}{3}$
Given $F$ , the other three friends fill seats 1, 3 and 4 in $3! = 6$ equally likely ways.
Asha is at an end (seat 1 or seat 4) in $2 \times 2! = 4$ of them.
So $P (E ∣ F) = \frac{4}{6} = \frac{2}{3}$ .
10
Short answerMediumNCERT Exercise 13.1, Q10
A white die and a blue die are rolled together.
1. Find the probability that the total is at least 8, given that the white die shows 4.
  Show answer
  Answer 0.5
  The blue die must show 4, 5 or 6: $\frac{3}{6} = \frac{1}{2}$ .
2. Find the probability that the total is 5, given that the blue die shows an odd number.
  Show answer
  Answer 0.1111
  $F$ has $3 \times 6 = 18$ outcomes. A total of 5 needs white 4 with blue 1, or white 2 with blue 3 (blue 5 would need white 0). So $P = \frac{2}{18} = \frac{1}{9}$ .
Short answerMediumNCERT Exercise 13.1, Q11
A fair die is thrown. Let $A = {2, 4, 6}$ , $B = {1, 2, 3, 4}$ and $C = {3, 4, 5}$ . Find (a) $P (A ∣ B)$ and $P (B ∣ A)$ , (b) $P (A ∣ C)$ and $P (C ∣ A)$ , (c) $P (A \cup B ∣ C)$ and $P (A \cap B ∣ C)$ .
Show solution
Answer(a) $\frac{1}{2}$ , $\frac{2}{3}$ (b) $\frac{1}{3}$ , $\frac{1}{3}$ (c) $\frac{2}{3}$ , $\frac{1}{3}$
(a) $A \cap B = {2, 4}$ : $P (A ∣ B) = \frac{2}{4} = \frac{1}{2}$ and $P (B ∣ A) = \frac{2}{3}$ .
(b) $A \cap C = {4}$ : $P (A ∣ C) = \frac{1}{3}$ and $P (C ∣ A) = \frac{1}{3}$ .
(c) $A \cup B = {1, 2, 3, 4, 6}$ meets $C$ in ${3, 4}$ , so $P (A \cup B ∣ C) = \frac{2}{3}$ . $A \cap B = {2, 4}$ meets $C$ in ${4}$ , so $P (A \cap B ∣ C) = \frac{1}{3}$ .
12
Short answerEasyNCERT Exercise 13.1, Q12
A cricket team wins or loses each of its next two matches with equal chance, independently (no ties). Find the probability that it wins both, given that
1. it wins the first match;
  Show answer
  Answer 0.5
  Given ${W W, W L}$ , only $W W$ works: $\frac{1}{2}$ .
2. it wins at least one match.
  Show answer
  Answer 0.3333
  Given ${W W, W L, L W}$ , only $W W$ works: $\frac{1}{3}$ .
Quick answerEasyNCERT Exercise 13.1, Q13
A school library has 240 Hindi fiction books, 160 Hindi non-fiction books, 360 English fiction books and 440 English non-fiction books. A book is picked at random and it is in English. What is the probability that it is fiction?
Show answer
Answer 0.45
There are $360 + 440 = 800$ English books and 360 of them are fiction: $\frac{360}{800} = \frac{9}{20} = 0.45$ .
Quick answerMediumNCERT Exercise 13.1, Q14
Two dice are thrown and the two numbers are different. Find the probability that their product is 12.
Show answer
Answer 0.1333
Different numbers: $36 - 6 = 30$ outcomes. Product 12: $(2, 6), (6, 2), (3, 4), (4, 3)$ , all with different numbers. So $P = \frac{4}{30} = \frac{2}{15}$ .
Short answerHardNCERT Exercise 13.1, Q15
A die is thrown. If it shows an even number, it is thrown again; if it shows an odd number, a coin is tossed. Find the probability that a head appears, given that the first throw is less than 3.
Show solution
Answer $\frac{1}{4}$
Let $F$ be first throw less than 3, so $P (F) = \frac{2}{6} = \frac{1}{3}$ .
A head needs a first throw of 1 (odd, so a coin is tossed) and then a head: $P (E \cap F) = \frac{1}{6} \cdot \frac{1}{2} = \frac{1}{12}$ . A first throw of 2 leads to a die, never a coin.
$P (E ∣ F) = \frac{1 / 12}{1 / 3} = \frac{1}{4}$ .
Quick answerEasyNCERT Exercise 13.1, Q16
$P (A) = 0.7$ and $B$ is an impossible event. Then $P (A ∣ B)$ is
$0$ $0.7$ $1$ not defined
Show answer
Answer (D) not defined
$P (B) = 0$ , and the definition needs $P (B) \neq 0$ ; the fraction would be $\frac{0}{0}$ . So $P (A ∣ B)$ is not defined.
Quick answerMediumNCERT Exercise 13.1, Q17
$A$ and $B$ are events with $P (A ∣ B) = P (B ∣ A) \neq 0$ . Which of these must be true?
$A \subset B$ $A \cap B = \emptyset$ $P (A) = P (B)$ $A \cup B = S$
Show answer
Answer (C) $P (A) = P (B)$
Both fractions share the numerator $P (A \cap B)$ , which is not zero here. Equal fractions with the same non-zero numerator have equal denominators, so $P (A) = P (B)$ . The events need not be equal or nested, and disjoint events would make both sides 0.

13.3 Multiplication Theorem on Probability

Often we want the chance that two things both happen: the first two toffees taken from a packet are orange, or a train is late and it is raining. The event both $E$ and $F$ is $E \cap F$ , sometimes written $E F$ . Turning the definition of conditional probability around lets us find it one stage at a time.

From (13.1), $P (E \cap F) = P (F) P (E ∣ F)$ whenever $P (F) \neq 0$ . Exchange the roles of the two events and use $F \cap E = E \cap F$ : the same probability is also $P (E)$ times $P (F ∣ E)$ , provided $P (E) \neq 0$ .

Multiplication rule

\begin{aligned} P (E \cap F) \\ = P (E) P (F ∣ E) \\ = P (F) P (E ∣ F) \end{aligned}

(13.2)

where

$P (F ∣ E)$: the chance of $F$ once $E$ has happened

This is the multiplication rule of probability. On a tree diagram it is the instruction multiply along the path: the first factor is the chance of the first branch, and the second is the chance of the next branch given the first (Fig. 13.4).

3/3Both orange: multiply along that path, $\frac{4}{12} \cdot \frac{3}{11} = \frac{1}{11}$ in all.

Fig. 13.4Without replacement the second branch probabilities depend on the first draw; multiply along a path.NCERT Example 8 · p. 415

A tree for drawing two toffees from a packet of 8 mango and 4 orange toffees. In the first step only the first draw is shown: mango with 8 out of 12 and orange with 4 out of 12. In the second step the second draw appears: after a mango, mango has 7 out of 11 and orange 4 out of 11; after an orange, mango has 8 out of 11 and orange 3 out of 11. In the third step the path orange then orange is drawn thick, the other branches fade, and the four path products are shown: 14 over 33, 8 over 33, 8 over 33 and 4 over 12 times 3 over 11, which is 1 over 11.

Example 8NCERT Example 8 · p. 415

Two orange toffees

A packet holds 8 mango and 4 orange toffees. Two toffees are taken out one after the other, and the first is not put back. What is the probability that both are orange?

Try it yourself first. Then reveal the solution one step at a time.

For three events we apply the rule twice. Each new factor is conditioned on everything that has already happened, and the pattern continues for four or more events.

Three events

\begin{aligned} P (E \cap F \cap G) \\ = P (E) P (F ∣ E) \\ \times P (G ∣ E \cap F) \end{aligned}

(13.3)

Example 9NCERT Example 9 · p. 416

A raffle draw

Six of the 20 tickets in a school raffle box win a prize. Three tickets are drawn one after another without replacement. Find the probability that the first two tickets win and the third does not.

Try it yourself first. Then reveal the solution one step at a time.

Quick check

1 of 3

A box has 3 red and 4 blue pens. Two pens are taken out, one after the other, and the first is not returned. Find the chance that both are red.

Show answer

Answer 0.142857

\frac{3}{7} \cdot \frac{2}{6} = \frac{1}{7} \approx 0.143

. Using

\frac{3}{7} \cdot \frac{3}{7}

treats the first pen as if it had been put back.

13.4 Independent Events

Draw a ticket at random from a box numbered 1 to 20. Let $E$ be the number is even and $F$ be the number is a multiple of 5. Then $P (E) = \frac{1}{2}$ . If you are told that $F$ happened, the ticket is one of 5, 10, 15 and 20, and two of these are even, so $P (E ∣ F) = \frac{2}{4} = \frac{1}{2}$ . The news about $F$ did not change the chance of $E$ . It works the other way too: $P (F) = \frac{4}{20} = \frac{1}{5}$ and $P (F ∣ E) = \frac{2}{10} = \frac{1}{5}$ .

Now use a box numbered 1 to 15 with the same two descriptions. This time $P (E) = \frac{7}{15}$ , but among 5, 10 and 15 only 10 is even, so $P (E ∣ F) = \frac{1}{3}$ . The events sound identical in both boxes, yet only the first pair is unaffected by each other. Independence is a fact about probabilities, not about whether two events feel connected.

Definition

Independent events

Events $E$ and $F$ are independent if $P (F ∣ E) = P (F)$ (when $P (E) \neq 0$ ) and $P (E ∣ F) = P (E)$ (when $P (F) \neq 0$ ): learning that one has happened leaves the probability of the other unchanged.

ExampleTickets 1–20: even and multiple of 5.

By the multiplication rule, $P (E \cap F) = P (E) P (F ∣ E)$ , and for independent events the second factor is simply $P (F)$ . The resulting product form needs no condition on $P (E)$ or $P (F)$ , so it is the working definition.

Independent events

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

(13.4)

Events that fail this test are called dependent: $P (E \cap F) \neq P (E) P (F)$ . Two experiments are called independent when every event of the first is independent of every event of the second, as with a coin toss and a die throw, or two separate throws of one die.

An area picture helps (Fig. 13.5). Draw the sample space as a unit square in which area stands for probability, and let $E$ be its left half. $F$ is independent of $E$ exactly when $F$ covers the same fraction of $E$ as it does of the whole square.

2/4 $F$ crosses $E$ in proportion: the overlap, $\frac{1}{2} \cdot \frac{1}{5} = \frac{1}{10}$ , is the product.

Fig. 13.5Independent events cross each other in proportion; exclusive events do not meet at all.NCERT Fig. 13.3 · p. 420

A unit square stands for the sample space, with area equal to probability. In the first step the left half is shaded as event E, with probability one half. In the second step a band along the bottom, one fifth of the height, is event F; it crosses E, and the overlap is hatched, with area one tenth, equal to P of E times P of F, so the panel says independent. In the third step the complement of F, the top four fifths, is outlined, and its overlap with E has area two fifths, again the product, so E and F prime are independent too. In the fourth step F is redrawn as a box of the same area entirely inside the right half, so it does not meet E: the overlap is 0 while the product is one tenth, and the panel says exclusive, so dependent.

Compare

Independent or mutually exclusive?

Independent

Defined through probabilities: $P (E \cap F) = P (E) P (F)$ .
Can share outcomes; with positive probabilities they must.
Knowing one tells you nothing new about the other.

Mutually exclusive

Defined through the sets: $E \cap F = \emptyset$ .
Share no outcome at all.
Knowing one happened tells you the other did not.

So two events with non-zero probabilities can never be both. If $E \cap F = \emptyset$ then $P (E \cap F) = 0$ , while $P (E) P (F) > 0$ . Mutually exclusive events of this kind are always dependent.

Example 10NCERT Example 10 · p. 419

One die, two descriptions

A die is thrown. Let $A$ be the number is at most 2 and $B$ be the number is odd. Are $A$ and $B$ independent? What changes if $A$ is at most 3 instead?

Try it yourself first. Then reveal the solution one step at a time.

Example 11NCERT Example 11 · p. 419

A six first, and a total of 7

Two dice are thrown. Let $A$ be the first die shows 6 and $B$ be the total is 7. Check whether $A$ and $B$ are independent.

Try it yourself first. Then reveal the solution one step at a time.

Example 12NCERT Example 12 · p. 419

Which pairs are independent?

Two dice are thrown. $E$ : a double (both dice equal). $F$ : the total is 8. $G$ : the first die is even. Which of the pairs $(E, F)$ , $(E, G)$ , $(F, G)$ are independent?

Try it yourself first. Then reveal the solution one step at a time.

Example 13NCERT Example 13 · p. 420

Complements of independent events

Show that independence survives taking a complement: whenever $E$ and $F$ are independent, so are $E$ and $F^{'}$ .

Try it yourself first. Then reveal the solution one step at a time.

Example 14NCERT Example 14 · p. 421

At least one of two

Show that if $A$ and $B$ are independent, the probability that at least one of them happens is $1 - P (A^{'}) P (B^{'})$ . A jewellery shop has two alarms that work independently; in a break-in they stay silent with probabilities 0.1 and 0.2. Find the probability that at least one rings.

Try it yourself first. Then reveal the solution one step at a time.

13.5 Bayes' Theorem

Two boxes of pens sit on a teacher's desk. Box I holds 4 blue and 2 black pens; Box II holds 3 blue and 5 black. A student picks a box at random and pulls out a pen without looking. The forward question is easy: if the pen came from Box II, it is black with probability $\frac{5}{8}$ . Now turn it round. You see only that the pen is black. How likely is it that it came from Box II? We are reasoning backwards, from an observed effect to its possible cause.

The English minister and mathematician Thomas Bayes found how to answer such inverse questions. His essay was read to the Royal Society in 1763, two years after his death, by his friend Richard Price. Two ideas come first: a partition of the sample space, and the theorem of total probability.

13.5.1 Partition of a sample space

Definition

Partition of a sample space

We say that events $E_{1}, E_{2}, \dots, E_{n}$ partition $S$ when (a) no two of them overlap, $E_{i} \cap E_{j} = \emptyset$ for $i \neq j$ ; (b) their union is all of $S$ ; and (c) each one has positive probability, $P (E_{i}) > 0$ .

ExampleFor one die: ${1, 2}$ , ${3, 4, 5}$ , ${6}$ .

Whatever the outcome, exactly one part of a partition happens. In the pen problem, Box I is chosen and Box II is chosen are the two parts; the three suppliers of Example 19 give a partition with three parts. Any event $E$ with $0 < P (E) < 1$ , taken with $E^{'}$ , is a partition too, and two events $E$ and $F$ give a finer one: those of the four pieces $E \cap F$ , $E \cap F^{'}$ , $E^{'} \cap F$ , $E^{'} \cap F^{'}$ that have positive probability. Since one sample space can be cut up in many ways, we choose the partition whose parts are the possible causes.

Fig. 13.6 draws a partition into three events as vertical strips whose widths are $P (E_{1})$ , $P (E_{2})$ and $P (E_{3})$ . Another event $A$ then cuts a piece out of each strip, and the height of the piece in strip $E_{i}$ is $P (A ∣ E_{i})$ .

3/4Adding the pieces of $A$ , $0.15 + 0.18 + 0.04$ , gives $P (A) = 0.37$ in all.

Fig. 13.6A partition as strips: total probability adds the pieces of A; Bayes' theorem takes one piece's share.NCERT Fig. 13.4 · p. 424

A rectangle stands for the sample space, with area equal to probability. In the first step it is divided into three vertical strips E1, E2 and E3 whose widths are 0.5, 0.3 and 0.2, labelled under the rectangle. In the second step a hatched piece of event A appears at the bottom of each strip, with heights 0.3, 0.6 and 0.2, which are the conditional probabilities of A in each strip. In the third step the pieces show their areas 0.15, 0.18 and 0.04, and the label reads P of A equals 0.37. In the fourth step only the middle piece stays bright and the label reads P of E2 given A equals 0.18 over 0.37, about 0.49.

13.5.2 Theorem of total probability

TheoremTheorem of total probability

Suppose the events $E_{1}, E_{2}, \dots, E_{n}$ partition $S$ , and $A$ is any event. Then $P (A)$ equals the sum of the $n$ products $P (E_{j}) P (A ∣ E_{j})$ , one for each part.

Total probability

\begin{aligned} P (A) \\ = \sum_{j = 1}^{n} P (E_{j}) \\ \times P (A ∣ E_{j}) \end{aligned}

(13.5)

In the strip picture each piece of $A$ is a rectangle of width $P (E_{j})$ and height $P (A ∣ E_{j})$ , so the theorem adds up areas. On a tree diagram it adds the probabilities of all the paths that end in $A$ (Fig. 13.7).

3/3Either route can end on time, so add the two paths: $0.12 + 0.595 = 0.715$ in all.

Fig. 13.7Total probability on a tree: add every path that ends in the event.NCERT p. 425

A tree with two first branches: fog with 0.3 and clear with 0.7. After fog the train is on time with 0.4 and late with 0.6; after a clear morning it is on time with 0.85 and late with 0.15. In the second step the path probabilities appear beside the leaves: 0.12, 0.18, 0.595 and 0.105. In the third step the two on-time paths are highlighted, the late paths fade, and the label reads sum 0.715.

Example 15NCERT Example 15 · p. 425

Fog and the morning train

On a December morning in Delhi there is dense fog with probability 0.3. A certain train arrives on time with probability 0.4 on a foggy morning and 0.85 on a clear one. Find the probability that the train arrives on time.

Try it yourself first. Then reveal the solution one step at a time.

Bayes' theorem

TheoremBayes' theorem

Suppose the events $E_{1}, E_{2}, \dots, E_{n}$ partition $S$ and $A$ is an event with $P (A) \neq 0$ . Then for each $i$ , $P (E_{i} ∣ A)$ equals $P (E_{i}) P (A ∣ E_{i})$ divided by $\sum_{j = 1}^{n} P (E_{j}) P (A ∣ E_{j})$ .

Bayes' theorem

\begin{aligned} P (E_{i} ∣ A) \\ = \frac{P (E_{i}) P (A ∣ E_{i})}{P (A)} \end{aligned}

(13.6)

where

$P (A)$: the total probability of the evidence, $\sum_{j} P (E_{j}) P (A ∣ E_{j})$ , from (13.5)

Read Bayes' theorem as a share. The numerator is the probability of the one path through $E_{i}$ that ends in $A$ ; the denominator is the probability of all the paths that end in $A$ . In the last step of Fig. 13.6 it is the area of one piece of $A$ divided by the area of all of $A$ .

Example 16NCERT Example 16 · p. 426

Which box did the pen come from?

Box I holds 4 blue and 2 black pens; Box II holds 3 blue and 5 black. A box is chosen at random and a pen drawn from it is black. Find the probability that it came from Box II.

Try it yourself first. Then reveal the solution one step at a time.

Example 17NCERT Example 17 · p. 426

Three pouches of coins

Three identical pouches each hold three coins. Pouch I has 3 gold coins, Pouch II has 1 gold and 2 silver, and Pouch III has 3 silver. A pouch is chosen at random and one coin taken from it is gold. What is the probability that the two coins still in that pouch are both gold?

Try it yourself first. Then reveal the solution one step at a time.

Example 18NCERT Example 18 · p. 427

A positive screening test

A screening test is used for a condition that affects 1% of a population. If a person has the condition, the test is positive with probability 0.98; if not, it still gives a false positive with probability 0.03. A person chosen at random tests positive. What is the probability that the person has the condition?

Try it yourself first. Then reveal the solution one step at a time.

3/3Only 98 of the 395 positive tests belong to ill people: about 0.25.

Fig. 13.8Natural frequencies make Bayes' theorem visible: most positives here come from the large healthy group.NCERT Example 18 · p. 427

A tree of counts. At the top is a box for 10 000 people. It splits into 100 ill and 9 900 well. In the second step each splits by test result: the 100 ill give 98 positive and 2 negative tests; the 9 900 well give 297 positive and 9 603 negative tests. In the third step the negative boxes fade and a bar at the bottom shows all 395 positive tests, a short part for the 98 ill people and a long part for the 297 well people.

The surprise in Example 18 is common in medicine, airport security and spam filters. When a condition is rare, even a good test meets far more healthy people than ill ones, so its few false alarms can outnumber its true alarms. The lab below lets you move the three numbers and watch the posterior respond.

InteractiveBayes lab

An interactive Bayes calculator for a screening test. Three sliders set the prior probability of being ill (0.1% to 99.9% on a log scale), the chance of a positive test when ill, and the chance of a positive test when well. Preset buttons load Example 18, a very rare condition and a common one. A tree of counts for 10 000 people shows how many are ill and well and how many in each group test positive or negative, and a bar splits the positive tests into ill and well people. A status panel gives the total probability of a positive test, the posterior probability of being ill given a positive test, and the probability of being well given a negative test. A Test again button replaces the prior with the current posterior. Nothing moves until the learner acts.

Fig. 13.9Drag the three sliders and watch the posterior. Test again uses the answer as the new prior, assuming the second test is independent of the first, given the true state.

Example 19NCERT Example 19 · p. 428

Which supplier made the faulty battery?

A phone factory buys batteries from three suppliers: 45% from the first, 35% from the second and 20% from the third. Their batteries are faulty 2%, 3% and 4% of the time. A battery picked at random from the store is faulty. Find the probability that it came from the second supplier.

Try it yourself first. Then reveal the solution one step at a time.

Example 20NCERT Example 20 · p. 429

Late for school

Tanvi gets to school by metro on half of all days, by auto-rickshaw on 30% of days and by cycle on the rest. She is late with probability $\frac{1}{10}$ by metro, $\frac{1}{4}$ by auto and $\frac{1}{20}$ by cycle. One day she arrives late. What is the probability that she came by auto?

Try it yourself first. Then reveal the solution one step at a time.

Example 21NCERT Example 21 · p. 430

A report that may be false

Ravi tells the truth 4 times out of 5. He throws a die and tells you the number was greater than 4. Find the probability that it really was. (When Ravi lies, he reports the other kind of result.)

Try it yourself first. Then reveal the solution one step at a time.

Quick check

1 of 4

Box I has 1 red and 3 white balls; Box II has 3 red and 1 white. A box is picked at random and the ball drawn from it is red. Find the chance that it came from Box II.

Show answer

Answer 0.75

\frac{\frac{1}{2} \cdot \frac{3}{4}}{\frac{1}{2} \cdot \frac{1}{4} + \frac{1}{2} \cdot \frac{3}{4}} = \frac{3}{4}

. The forward probability

P (red ∣ Box II)

is also

\frac{3}{4}

here, but only because the boxes mirror each other.

Random variables

Many experiments end in a number we care about more than the raw outcome: the total on two dice, the number of faulty items in a carton, the money won at a fair stall. Toss a coin three times. The sample space has 8 outcomes, and to each one we can attach the number of heads it contains.

Definition

Random variable

A rule that attaches a real number to every outcome of a random experiment; in other words, a function from the sample space $S$ to the real numbers $𝐑$ .

ExampleThree tosses: $X$ = the number of heads, so $X (H H T) = 2$ .

For the three tosses, $X (T T T) = 0$ , $X (T H T) = 1$ , $X (H H T) = 2$ and $X (H H H) = 3$ (Fig. 13.10). One sample space can carry many random variables. Suppose a game pays ₹10 for each head and takes ₹5 for each tail. Your winnings are $Y = 10 X - 5 (3 - X)$ , that is $Y = 15 X - 15$ rupees. So $Y (T T T) = - 15$ , $Y (H T T) = 0$ , $Y (H H T) = 15$ and $Y (H H H) = 30$ . $X$ and $Y$ are different functions on the same $S$ .

2/3 $X$ sends each outcome to its number of heads.

Fig. 13.10A random variable sends each outcome to a number; another rule on the same outcomes gives another random variable.NCERT p. 431

Along the top, the eight outcomes of three tosses sit in a shaded box labelled S: TTT, HTT, THT, TTH, HHT, HTH, THH and HHH. In the second step an arrow runs from each outcome down to a number line with ticks 0, 1, 2 and 3: TTT to 0, the three outcomes with one head to 1, the three with two heads to 2, and HHH to 3, with the note count the heads. In the third step the same arrows stay but the tick labels change to minus 15, 0, 15 and 30, with the note 10 rupees per head, minus 5 rupees per tail.

Looking ahead

Looking ahead: a probability distribution

List the values of $X$ with their probabilities and you have the probability distribution of $X$ :

Table 13.1Number of heads in three tosses

$x$	$P (X = x)$
0	$\frac{1}{8}$
1	$\frac{3}{8}$
2	$\frac{3}{8}$
3	$\frac{1}{8}$

The probabilities add up to 1, because each outcome gives exactly one value. The current syllabus stops at the definition of a random variable; distributions, their mean and variance are studied in later courses.

Quick check

1 of 2

Two dice are thrown and

X

is the larger of the two numbers shown. What is

X ((2, 5))

2573

Show answer

Answer (B) 5

X

picks the larger number, 5. The total 7 and the difference 3 would be other random variables on the same sample space.

Practice · 14 questions

Practice 13.3: questions with answers

1
Short answerHardNCERT Exercise 13.3, Q1
A jar has 4 red and 6 green balls. Pick one ball at random, note its colour and return it to the jar along with 3 extra balls of that colour. Then pick a second ball.
1. Find the probability that the second ball is green.
  Show answer
  Answer 0.6
  $P (G_{2}) = \frac{6}{10} \cdot \frac{9}{13} + \frac{4}{10} \cdot \frac{6}{13} = \frac{54 + 24}{130} = \frac{3}{5}$ , the same as for the first ball.
2. Given that the second ball is green, find the probability that the first one was green.
  Show answer
  Answer 0.69231
  $P (G_{1} ∣ G_{2}) = \frac{(6 / 10) (9 / 13)}{3 / 5} = \frac{9}{13}$ .
Short answerEasyNCERT Exercise 13.3, Q2
Basket I holds 5 ripe and 3 raw mangoes; Basket II holds 2 ripe and 6 raw. A basket is chosen at random and a mango taken from it is ripe. Find the probability that it came from Basket I.
Show solution
Answer $\frac{5}{7}$
$P (ripe ∣ I) = \frac{5}{8}$ and $P (ripe ∣ I I) = \frac{2}{8}$ .
The equal priors cancel, so the posterior of Basket I is $\frac{5 / 8}{5 / 8 + 2 / 8} = \frac{5}{7}$ .
Short answerMediumNCERT Exercise 13.3, Q3
At a coaching centre 70% of the students study online and 30% attend in person. In a mock test 25% of the online students and 40% of the in-person students score above 90. A student who scored above 90 is picked at random. Find the probability that the student attends in person.
Show solution
Answer $\frac{24}{59} \approx 0.41$
Paths to above 90: online $0.7 \times 0.25 = 0.175$ ; in person $0.3 \times 0.4 = 0.12$ .
The two paths add to $0.295$ , so the posterior of in person is $\frac{0.12}{0.295} = \frac{24}{59}$ .
Short answerMediumNCERT Exercise 13.3, Q4
In a test, each multiple-choice question has 5 options. A student knows the answer with probability $\frac{2}{3}$ and otherwise guesses at random. Given that the student answered a question correctly, what is the probability that the student knew the answer?
Show solution
Answer $\frac{10}{11}$
$P (correct) = \frac{2}{3} \cdot 1 + \frac{1}{3} \cdot \frac{1}{5} = \frac{11}{15}$ .
The posterior of knew is $\frac{2 / 3}{11 / 15} = \frac{10}{11}$ .
Short answerHardNCERT Exercise 13.3, Q5
An airport scanner flags 95% of the bags that carry a banned item, and it also flags 2% of the bags that do not. One bag in 200 carries a banned item. A bag is flagged. What is the probability that it really carries a banned item?
Show solution
Answer $\frac{95}{493} \approx 0.19$
True alarms: $0.005 \times 0.95 = 0.00475$ . False alarms: $0.995 \times 0.02 = 0.0199$ .
All flagged bags: $0.00475 + 0.0199 = 0.02465$ . So the chance of a banned item is $\frac{0.00475}{0.02465} = \frac{95}{493}$ , about 0.19.
So about four flagged bags in five are false alarms, which is why a flagged bag is then searched by hand.
Short answerMediumNCERT Exercise 13.3, Q6
A bag holds three dice: an ordinary die, a die with 6 on three of its faces, and a die with 6 on every face. One die is chosen at random and thrown, and it shows 6. What is the probability that it is the die with 6 on every face?
Show solution
Answer $\frac{3}{5}$
With priors $\frac{1}{3}$ each, $P (6) = \frac{1}{3} (\frac{1}{6} + \frac{1}{2} + 1) = \frac{5}{9}$ .
The all-six die's posterior is $\frac{1 / 3}{5 / 9} = \frac{3}{5}$ .
Short answerMediumNCERT Exercise 13.3, Q7
A delivery company employs 300 cyclists, 500 motorbike riders and 200 van drivers. In a month the probability of a traffic fine is 0.02 for a cyclist, 0.05 for a rider and 0.1 for a van driver. One employee received a fine this month. Find the probability that the employee is a cyclist.
Show solution
Answer $\frac{2}{17}$
Expected fines: cyclists $300 \times 0.02 = 6$ , riders $500 \times 0.05 = 25$ , vans $200 \times 0.1 = 20$ ; 51 in all.
So the chance of a cyclist is $\frac{6}{51} = \frac{2}{17}$ .
Quick answerMediumNCERT Exercise 13.3, Q8
An office has two printers. Printer P prints 70% of the pages and printer Q the rest. A page from P is smudged with probability 0.03, and from Q with probability 0.05. A page picked from the out-tray is smudged. What is the probability that Q printed it?
Show answer
Answer 0.41667
Paths: $0.7 \times 0.03 = 0.021$ and $0.3 \times 0.05 = 0.015$ . So $P (Q ∣ smudged) = \frac{0.015}{0.036} = \frac{5}{12}$ .
Quick answerMediumNCERT Exercise 13.3, Q9
Two candidates, X and Y, stand for school captain; X wins with probability 0.55 and Y with 0.45. If X wins, a new sports period is introduced with probability 0.8; if Y wins, with probability 0.4. The sports period was introduced. What is the probability that Y won?
Show answer
Answer 0.29032
Paths: $0.55 \times 0.8 = 0.44$ and $0.45 \times 0.4 = 0.18$ . So $P (Y ∣ period) = \frac{0.18}{0.62} = \frac{9}{31}$ .
Short answerHardNCERT Exercise 13.3, Q10
Arjun throws a die. If it shows 1 or 2, he tosses a coin three times; otherwise he tosses it twice. He tells you he got exactly two heads. What is the probability that the die showed 1 or 2?
Show solution
Answer $\frac{3}{7}$
Paths: $\frac{1}{3} \cdot \frac{3}{8} = \frac{1}{8}$ through 1 or 2, and $\frac{2}{3} \cdot \frac{1}{4} = \frac{1}{6}$ through 3 to 6.
So the posterior of 1 or 2 is $\frac{1 / 8}{1 / 8 + 1 / 6} = \frac{3}{7}$ .
Short answerMediumNCERT Exercise 13.3, Q11
A bakery runs three shifts. The morning shift bakes 40% of the loaves, the afternoon shift 35% and the night shift 25%. The proportions of burnt loaves are 2%, 3% and 6%. A burnt loaf is found. What is the probability that the night shift baked it?
Show solution
Answer $\frac{30}{67} \approx 0.45$
Paths to burnt: $0.4 \times 0.02 = 0.008$ , $0.35 \times 0.03 = 0.0105$ , $0.25 \times 0.06 = 0.015$ ; total $0.0335$ .
So the night shift's posterior is $\frac{0.015}{0.0335} = \frac{30}{67}$ .
Short answerHardNCERT Exercise 13.3, Q12
One card is lost from a full pack of 52. Two cards are then drawn from the remaining 51 and both are red. Find the probability that the lost card was red.
Show solution
Answer $\frac{12}{25}$
Given a red card was lost: $P (two red) = \frac{25}{51} \cdot \frac{24}{50}$ . Given a black card was lost: $\frac{26}{51} \cdot \frac{25}{50}$ .
The priors $\frac{1}{2}$ and the common factor $\frac{1}{51 \cdot 50}$ cancel, leaving $\frac{25 \cdot 24}{25 \cdot 24 + 26 \cdot 25} = \frac{600}{1250}$ , which is $\frac{12}{25}$ .
Two red cards in the draw make it slightly less likely that a red card is missing.
Quick answerEasyNCERT Exercise 13.3, Q13
Meena tells the truth 2 times out of 3. A coin is tossed and she reports a tail. The probability that it really was a tail is
$\frac{2}{3}$ $\frac{1}{3}$ $\frac{1}{2}$ $\frac{5}{6}$
Show answer
Answer (A) $\frac{2}{3}$
$\frac{\frac{1}{2} \cdot \frac{2}{3}}{\frac{1}{2} \cdot \frac{2}{3} + \frac{1}{2} \cdot \frac{1}{3}} = \frac{2}{3}$ . With a fair coin the posterior equals her truth rate.
Quick answerMediumNCERT Exercise 13.3, Q14
$B \subset A$ and $P (A) \neq 0$ . Which statement is correct?
$P (B ∣ A) < P (B)$ $P (B ∣ A) = P (B)$ always $P (B ∣ A) \geq P (B)$ $P (A ∣ B) < 1$
Show answer
Answer (C) $P (B ∣ A) \geq P (B)$
$P (B ∣ A) = \frac{P (B)}{P (A)} \geq P (B)$ because $P (A) \leq 1$ . Also $P (A ∣ B) = 1$ whenever $P (B) \neq 0$ , so the last option is false.

Miscellaneous Examples

Example 22NCERT Example 22 · p. 433

Four jars of sweets

Four jars of sweets stand on a counter. Their contents are listed below. One jar is picked at random and one sweet is taken from it without looking. It is a laddoo. Find the probability that it came from Jar II. Which jar is least likely?

Table 13.2Sweets in each jar

	Laddoo	Barfi	Peda
Jar I	2	3	5
Jar II	4	1	1
Jar III	1	4	3
Jar IV	3	3	6

Try it yourself first. Then reveal the solution one step at a time.

Example 23NCERT Example 23 · p. 434

Ring toss at the mela

At a mela ring-toss stall, Aman and Bela take turns, Aman first, until one of them rings a bottle and wins. On each throw Aman succeeds with probability $\frac{1}{3}$ and Bela with probability $\frac{1}{4}$ , independently. Find each player's probability of winning.

Try it yourself first. Then reveal the solution one step at a time.

3/3Collecting the pieces, Aman wins with probability $\frac{2}{3}$ and Bela with $\frac{1}{3}$ in all.

Fig. 13.11Each round shares what is left in the same ratio, 2 to 1, so Aman's total is two thirds.NCERT Example 23 · p. 434

A long bar stands for total probability 1. In the first step its first third is shaded A, for Aman winning on his first throw, the next sixth is dashed B, for Bela winning on her first throw, and the remaining half is marked game goes on. In the second step the remaining half is shared the same way: A gets a sixth and B a twelfth, leaving a quarter. In the third step later rounds fill the bar with ever smaller A and B pieces, and a second bar below collects them: A wins two thirds and B wins one third.

Example 24NCERT Example 24 · p. 435

Two seeds from one packet

A nursery's seed packets come from a good batch with probability 0.75. In a good batch each seed sprouts with probability 0.9; in a poor batch, with probability 0.5. Seeds sprout independently of one another. You sow two seeds from one packet and both sprout. Find the probability that the packet is from a good batch.

Try it yourself first. Then reveal the solution one step at a time.

Quick check

1 of 2

In the ring-toss game of Example 23, what is the probability that Bela wins on her very first throw?

Show answer

Answer 0.166667

Aman must miss first (

\frac{2}{3}

) and then Bela must succeed (

\frac{1}{4}

\frac{2}{3} \cdot \frac{1}{4} = \frac{1}{6}

. The answer

\frac{1}{4}

forgets that Aman throws first.

Timeline

Historical note: from dice tables to axioms

c. 1564
Cardano counts dice outcomes
Girolamo Cardano's Liber de Ludo Aleae listed the favourable cases for throws of dice; it was printed only in 1663.
c. 1620
Galileo and three dice
Galileo explained why a total of 10 turns up more often than 9 with three dice: 27 of the 216 ordered outcomes against 25.
1654
Pascal and Fermat
Letters between Blaise Pascal and Pierre de Fermat solved how to divide the stakes of an unfinished game.
1657
The first printed textbook
Christiaan Huygens published De Ratiociniis in Ludo Aleae, a short book of problems on games of chance.
1713
Bernoulli's Ars Conjectandi
Jacob Bernoulli's book, published after his death, proved an early law of large numbers.
1763
Bayes and inverse probability
Thomas Bayes' essay on reasoning from effects back to causes was presented by Richard Price, two years after Bayes died.
1812
Laplace's treatise
Pierre-Simon Laplace set out the rule we now call Bayes' theorem in general form.
1933
Kolmogorov's axioms
Andrey Kolmogorov built probability on a few axioms about sets of outcomes, the approach you met in Class 11.

Recap

Probability: key points and summary

Knowing that $F$ happened makes $F$ the new sample space: $P (E ∣ F) = \frac{P (E \cap F)}{P (F)}$ , with $P (F) \neq 0$ (Section 13.2).
Given $F$ , probabilities obey the usual rules: $P (S ∣ F) = 1$ , $P (E^{'} ∣ F) = 1 - P (E ∣ F)$ , and the addition rule holds (Section 13.2.1).
$P (E ∣ F)$ and $P (F ∣ E)$ have the same numerator but different denominators; in general they differ.
Multiplication rule: $P (E \cap F) = P (E) P (F ∣ E)$ ; for three events, multiply in stages (Section 13.3).
A tree diagram multiplies along a path and adds across paths (Fig. 13.3).
$E$ and $F$ are independent when $P (E \cap F) = P (E) P (F)$ ; then complements are independent too (Section 13.4).
Mutually exclusive events with positive probabilities are dependent; three events need all four product checks.
A partition is a set of disjoint events with positive probabilities that together fill $S$ (Section 13.5.1).
Total probability: $P (A) = \sum_{j} P (E_{j}) P (A ∣ E_{j})$ (Section 13.5.2).
Bayes' theorem turns priors $P (E_{i})$ into posteriors $P (E_{i} ∣ A)$ by taking one path's share of $P (A)$ (Bayes' theorem).
With a rare condition, most positive tests can be false; count with natural frequencies to check (Example 18).
A random variable attaches a real number to each outcome; one sample space can carry many (Random variables).

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Miscellaneous Exercise on Chapter 13

Practice · 17 questions

Practice questions with answers (NCERT “Miscellaneous exercise”)

1
Short answerEasyNCERT Miscellaneous Exercise, Q1
$A$ and $B$ are events with $P (B) \neq 0$ . Find $P (A ∣ B)$ if
1. $B \subset A$ ;
  Show solution
  Answer1
  $A \cap B = B$ , so $P (A ∣ B) = \frac{P (B)}{P (B)} = 1$ .
2. $A \cap B = \emptyset$ ;
  Show solution
  Answer0
  $P (A \cap B) = 0$ , so $P (A ∣ B) = 0$ .
3. $A \subset B$ .
  Show solution
  Answer $\frac{P (A)}{P (B)}$
  $A \cap B = A$ , so $P (A ∣ B) = \frac{P (A)}{P (B)}$ , which is at least $P (A)$ .
2
Short answerMediumNCERT Miscellaneous Exercise, Q2
Two dice are thrown. Find the probability that both show 6, given that
1. at least one die shows 6;
  Show answer
  Answer 0.0909
  At least one 6 has 11 outcomes, and only $(6, 6)$ has two: $\frac{1}{11}$ .
2. the first die shows 6.
  Show answer
  Answer 0.16667
  Given a first 6, the second die must show 6: $\frac{1}{6}$ .
Short answerMediumNCERT Miscellaneous Exercise, Q3
About 8% of men and 0.5% of women are red–green colour-blind. A colour-blind person is chosen at random from a group with equal numbers of men and women. What is the probability that the person is a man?
Show solution
Answer $\frac{16}{17}$
The equal priors cancel. Men supply 0.08 and women 0.005 of the colour-blind paths, so the chance of a man is $\frac{0.08}{0.085} = \frac{16}{17}$ , about 0.94.
Short answerHardNCERT Miscellaneous Exercise, Q4
Each seed in a packet sprouts with probability 0.8, independently of the others. Six seeds are sown. Find the probability that at most 4 of them sprout.
Show solution
Answer $0.34464$
All 6 sprout: ${0.8}^{6} = 0.262144$ .
Exactly 5 sprout: choose the one that fails in 6 ways, each with probability ${0.8}^{5} \times 0.2$ , so $6 \times 0.065536 = 0.393216$ .
At most 4: $1 - 0.262144 - 0.393216 = 0.34464$ .
Quick answerMediumNCERT Miscellaneous Exercise, Q5
A leap year is chosen at random. What is the probability that it has 53 Fridays or 53 Saturdays?
Show answer
Answer 0.42857
The two extra days are one of 7 equally likely pairs (Sun–Mon, …, Sat–Sun). Thu–Fri, Fri–Sat and Sat–Sun contain a Friday or a Saturday: $\frac{3}{7}$ .
Short answerMediumNCERT Miscellaneous Exercise, Q6
Four boxes K, L, M and N each hold 10 marbles, coloured as in the table. A box is chosen at random and a marble drawn from it is green. Find the probability that it came from box K, from box L and from box M.
Table 13.3Marbles in each box
Green Red White
K 2 5 3
L 4 4 2
M 7 1 2
N 0 3 7
Show solution
Answer $\frac{2}{13}$ , $\frac{4}{13}$ , $\frac{7}{13}$
$P (green ∣ box)$ is 0.2, 0.4, 0.7 and 0 for K, L, M and N; the priors $\frac{1}{4}$ cancel.
Total $0.2 + 0.4 + 0.7 + 0 = 1.3$ , so the posteriors are $\frac{2}{13}$ , $\frac{4}{13}$ and $\frac{7}{13}$ (and 0 for N).
Short answerMediumNCERT Miscellaneous Exercise, Q7
A student's chance of catching a cold in winter is 40%. A daily walk multiplies this chance by 0.8; daily vitamin tablets multiply it by 0.6. Each student picks exactly one of the two, with equal probability. A student caught a cold. Find the probability that the student chose the walk.
Show solution
Answer $\frac{4}{7}$
Walk: $0.4 \times 0.8 = 0.32$ . Tablets: $0.4 \times 0.6 = 0.24$ .
The equal priors $\frac{1}{2}$ cancel, so the chance that the student chose the walk is $\frac{0.32}{0.32 + 0.24} = \frac{4}{7}$ .
Quick answerMediumNCERT Miscellaneous Exercise, Q8
Each entry of a $2 \times 2$ matrix is 0 or 1, chosen independently with probability $\frac{1}{2}$ each. What is the probability that the determinant is non-zero?
Show answer
Answer 0.375
There are 16 equally likely matrices. $a d - b c = 1$ needs $a = d = 1$ and $b c = 0$ : 3 matrices. $a d - b c = - 1$ likewise gives 3. So $\frac{6}{16} = \frac{3}{8}$ .
9
Short answerMediumNCERT Miscellaneous Exercise, Q9
A scooter fails to start if its battery (A) or its starter (B) fails. Records give: $P (A fails) = 0.25$ , $P (B fails but A does not) = 0.1$ , $P (both fail) = 0.05$ . Find
1. $P (A fails ∣ B has failed)$ ;
  Show answer
  Answer 0.33333
  $P (B fails) = 0.1 + 0.05 = 0.15$ , so $\frac{0.05}{0.15} = \frac{1}{3}$ .
2. $P (A fails but B does not)$ .
  Show answer
  Answer 0.2
  $0.25 - 0.05 = 0.2$ .
Short answerHardNCERT Miscellaneous Exercise, Q10
Bag I has 4 white and 2 black balls; Bag II has 3 white and 5 black. A ball picked at random from Bag I is transferred into Bag II. A ball is then taken from Bag II, and it is white. Find the probability that the transferred ball was white.
Show solution
Answer $\frac{8}{11}$
Moved white ( $\frac{4}{6}$ ): Bag II then has 4 white of 9. Moved black ( $\frac{2}{6}$ ): 3 white of 9.
Paths: $\frac{4}{6} \cdot \frac{4}{9} = \frac{16}{54}$ and $\frac{2}{6} \cdot \frac{3}{9} = \frac{6}{54}$ . So the chance that a white ball was moved is $\frac{16}{22} = \frac{8}{11}$ .
Quick answerEasyNCERT Miscellaneous Exercise, Q11
$P (A) \neq 0$ and $P (B ∣ A) = 0$ . Then
$A \subset B$ $B \subset A$ $P (A \cap B) = 0$ $P (B) = 0$
Show answer
Answer (C) $P (A \cap B) = 0$
$P (A \cap B) = P (A) \times 0 = 0$ . Nothing forces $P (B) = 0$ : $B$ can still happen when $A$ does not.
Quick answerMediumNCERT Miscellaneous Exercise, Q12
$P (A)$ and $P (B)$ are non-zero and $P (A ∣ B) < P (A)$ . Which is correct?
$P (B ∣ A) > P (B)$ $P (B ∣ A) = P (B)$ $A$ and $B$ are independent $P (B ∣ A) < P (B)$
Show answer
Answer (D) $P (B ∣ A) < P (B)$
Multiplying both sides by $P (B)$ shows that the overlap $P (A \cap B)$ is less than $P (A)$ times $P (B)$ . Dividing that by $P (A)$ gives $P (B ∣ A) < P (B)$ : bad news for one is bad news for the other.
Quick answerMediumNCERT Miscellaneous Exercise, Q13
$A$ and $B$ are events with $P (A) \neq 0$ and $P (A \cup B) = P (B)$ . Then
$P (A ∣ B) = 1$ $P (B ∣ A) = 1$ $P (A ∣ B) = 0$ $P (B ∣ A) = 0$
Show answer
Answer (B) $P (B ∣ A) = 1$
By the addition rule, the given equation says that $P (A)$ minus the overlap is zero. So $P (A \cap B) = P (A)$ , and dividing by $P (A)$ gives $P (B ∣ A) = 1$ .
14
Quick answerEasy
Put the steps of a Bayes' theorem calculation in order.
1. note priors, forward chances
2. divide one path by P(A)
3. add the paths to get P(A)
4. list the hypotheses
Show answer
Answer
list the hypotheses
note priors, forward chances
add the paths to get P(A)
divide one path by P(A)
The hypotheses must form a partition before their priors and forward probabilities mean anything; $P (A)$ is the denominator.
15
Quick answerEasy
Match each rule with its formula.
Tap an item on the left, then its match on the right.
Show answer
Answer
multiplication rule : $P (E) P (F ∣ E)$
independence : $P (E) P (F)$
complement given F : $1 - P (E ∣ F)$
total probability : $\sum_{j} P (E_{j}) P (A ∣ E_{j})$
The first two both equal $P (E \cap F)$ , the second only for independent events; the third is $P (E^{'} ∣ F)$ .
Quick answerEasy
Two mutually exclusive events, each with non-zero probability, are always .
Show answer
Answer dependent
Their overlap has probability 0 but the product of their probabilities is positive.
Quick answerMedium
If $P (A)$ and $P (B)$ are non-zero and $P (A ∣ B) = P (A)$ , then $P (B ∣ A) = P (B)$ .
TrueFalse
Show answer
Answer True
$P (A ∣ B) = P (A)$ means $P (A \cap B) = P (A) P (B)$ , which is symmetric in $A$ and $B$ , so $P (B ∣ A) = P (B)$ .

	Green	Red	White
K	2	5	3
L	4	4	2
M	7	1	2
N	0	3	7

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Our practice questions are written for ncert.ai and follow the NCERT exercises. Find the one that matches the question in your textbook; its answer is with it.

In your NCERT textbook	Practice question here
NCERT Exercise 13.1, Q1	Practice 13.1 · Q1
NCERT Exercise 13.1, Q2	Practice 13.1 · Q2
NCERT Exercise 13.1, Q3	Practice 13.1 · Q3
NCERT Exercise 13.1, Q4	Practice 13.1 · Q4
NCERT Exercise 13.1, Q5	Practice 13.1 · Q5
NCERT Exercise 13.1, Q6	Practice 13.1 · Q6
NCERT Exercise 13.1, Q7	Practice 13.1 · Q7
NCERT Exercise 13.1, Q8	Practice 13.1 · Q8
NCERT Exercise 13.1, Q9	Practice 13.1 · Q9
NCERT Exercise 13.1, Q10	Practice 13.1 · Q10
NCERT Exercise 13.1, Q11	Practice 13.1 · Q11
NCERT Exercise 13.1, Q12	Practice 13.1 · Q12
NCERT Exercise 13.1, Q13	Practice 13.1 · Q13
NCERT Exercise 13.1, Q14	Practice 13.1 · Q14
NCERT Exercise 13.1, Q15	Practice 13.1 · Q15
NCERT Exercise 13.1, Q16	Practice 13.1 · Q16
NCERT Exercise 13.1, Q17	Practice 13.1 · Q17
NCERT Exercise 13.2, Q1	Practice 13.2 · Q1
NCERT Exercise 13.2, Q2	Practice 13.2 · Q2
NCERT Exercise 13.2, Q3	Practice 13.2 · Q3
NCERT Exercise 13.2, Q4	Practice 13.2 · Q4
NCERT Exercise 13.2, Q5	Practice 13.2 · Q5
NCERT Exercise 13.2, Q6	Practice 13.2 · Q6
NCERT Exercise 13.2, Q7	Practice 13.2 · Q7
NCERT Exercise 13.2, Q8	Practice 13.2 · Q8
NCERT Exercise 13.2, Q9	Practice 13.2 · Q9
NCERT Exercise 13.2, Q10	Practice 13.2 · Q10
NCERT Exercise 13.2, Q11	Practice 13.2 · Q11
NCERT Exercise 13.2, Q12	Practice 13.2 · Q12
NCERT Exercise 13.2, Q13	Practice 13.2 · Q13
NCERT Exercise 13.2, Q14	Practice 13.2 · Q14
NCERT Exercise 13.2, Q15	Practice 13.2 · Q15
NCERT Exercise 13.2, Q16	Practice 13.2 · Q16
NCERT Exercise 13.2, Q17	Practice 13.2 · Q17
NCERT Exercise 13.2, Q18	Practice 13.2 · Q18
NCERT Exercise 13.3, Q1	Practice 13.3 · Q1
NCERT Exercise 13.3, Q2	Practice 13.3 · Q2
NCERT Exercise 13.3, Q3	Practice 13.3 · Q3
NCERT Exercise 13.3, Q4	Practice 13.3 · Q4
NCERT Exercise 13.3, Q5	Practice 13.3 · Q5
NCERT Exercise 13.3, Q6	Practice 13.3 · Q6
NCERT Exercise 13.3, Q7	Practice 13.3 · Q7
NCERT Exercise 13.3, Q8	Practice 13.3 · Q8
NCERT Exercise 13.3, Q9	Practice 13.3 · Q9
NCERT Exercise 13.3, Q10	Practice 13.3 · Q10
NCERT Exercise 13.3, Q11	Practice 13.3 · Q11
NCERT Exercise 13.3, Q12	Practice 13.3 · Q12
NCERT Exercise 13.3, Q13	Practice 13.3 · Q13
NCERT Exercise 13.3, Q14	Practice 13.3 · Q14
NCERT Miscellaneous Exercise, Q1	Miscellaneous exercise · Q1
NCERT Miscellaneous Exercise, Q2	Miscellaneous exercise · Q2
NCERT Miscellaneous Exercise, Q3	Miscellaneous exercise · Q3
NCERT Miscellaneous Exercise, Q4	Miscellaneous exercise · Q4
NCERT Miscellaneous Exercise, Q5	Miscellaneous exercise · Q5
NCERT Miscellaneous Exercise, Q6	Miscellaneous exercise · Q6
NCERT Miscellaneous Exercise, Q7	Miscellaneous exercise · Q7
NCERT Miscellaneous Exercise, Q8	Miscellaneous exercise · Q8
NCERT Miscellaneous Exercise, Q9	Miscellaneous exercise · Q9
NCERT Miscellaneous Exercise, Q10	Miscellaneous exercise · Q10
NCERT Miscellaneous Exercise, Q11	Miscellaneous exercise · Q11
NCERT Miscellaneous Exercise, Q12	Miscellaneous exercise · Q12
NCERT Miscellaneous Exercise, Q13	Miscellaneous exercise · Q13

Quick check: MCQs on Probability

13.1 Introduction2 questions
13.2.1 Properties of conditional probability4 questions
13.3 Multiplication Theorem on Probability3 questions
13.4 Independent Events4 questions
Bayes' theorem4 questions
Random variables2 questions
Miscellaneous Examples2 questions

Glossary · 12

Key terms (English–Hindi)

Bayes' theoremबेज़-प्रमेयbez prameya: The rule that the posterior $P (E_{i} ∣ A)$ equals $P (E_{i}) P (A ∣ E_{i})$ divided by $P (A)$ : the chance of a cause once its effect is seen.First seen in the chapter →

conditional probabilityसप्रतिबंध प्रायिकताsapratibandh praayikta: The probability of $E$ once we know $F$ has happened: $P (E ∣ F) = P (E \cap F) / P (F)$ , with $P (F) \neq 0$ .First seen in the chapter →

hypothesisपरिकल्पनाparikalpana: In Bayes' theorem, one of the events $E_{i}$ of the partition: a possible cause of what was observed.First seen in the chapter →

independent eventsस्वतंत्र घटनाएँsvatantra ghatnaayen: Events with $P (E \cap F) = P (E) P (F)$ , so knowing one does not change the chance of the other. Otherwise they are dependent.

multiplication ruleप्रायिकता का गुणन नियमpraayikta ka gunan niyam: $P (E \cap F) = P (E) P (F ∣ E)$ : the chance of both is the chance of the first times the chance of the second given the first.First seen in the chapter →
mutually exclusive eventsपरस्पर अपवर्जी घटनाएँparaspar apavarji ghatnaayen: Events with no outcome in common, $E \cap F = \emptyset$ ; they cannot happen together.First seen in the chapter →

partition of a sample spaceप्रतिदर्श समष्टि का विभाजनpratidarsh samashti ka vibhaajan: Events $E_{1}, \dots, E_{n}$ that are pairwise disjoint, together make up $S$ , and each have non-zero probability.
posterior probability: $P (E_{i} ∣ A)$ , the probability of a hypothesis after the evidence $A$ is taken into account.First seen in the chapter →
prior probability: $P (E_{i})$ , the probability of a hypothesis before the evidence is taken into account.First seen in the chapter →

random variableयादृच्छिक चरyaadrichchhik char: A rule that gives a real number for each outcome of a random experiment: a function from $S$ to the real numbers.First seen in the chapter →

theorem of total probabilityसंपूर्ण प्रायिकता प्रमेयsampoorn praayikta prameya: For a partition $E_{1}, \dots, E_{n}$ : $P (A) = \sum_{j} P (E_{j}) P (A ∣ E_{j})$ .
tree diagramवृक्ष आरेखvriksh aarekh: A branching picture of a staged experiment; multiplying along a path gives the probability of that path.First seen in the chapter →

Probability: Class 12 Maths · Chapter 13

What you will learn in Probability

13.1 Introduction

13.2 Conditional Probability

13.2.1 Properties of conditional probability

Practice 13.1: questions with answers

13.3 Multiplication Theorem on Probability

13.4 Independent Events

Practice 13.2: questions with answers

13.5 Bayes' Theorem

13.5.1 Partition of a sample space

13.5.2 Theorem of total probability

Bayes' theorem

Random variables

Practice 13.3: questions with answers

Miscellaneous Examples

Probability: key points and summary

Miscellaneous Exercise on Chapter 13

Practice questions with answers (NCERT “Miscellaneous exercise”)

Find your NCERT exercise question

Quick check: MCQs on Probability

Key terms (English–Hindi)

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