Conditional events

This chapter delves into the core concepts of conditional probability, the multiplication rule, and the crucial distinction between independent and dependent events. A comprehensive understanding of these principles is essential for accurately modeling real-world probabilistic scenarios and is consistently a key area evaluated in CMI examinations.

---

Chapter Contents

| Topic |

|---|-------| | 1 | Conditional probability definition | | 2 | Multiplication rule | | 3 | Independence | | 4 | Dependent event reasoning |

---

We begin with Conditional probability definition.

Part 1: Conditional probability definition

Conditional Probability Definition

Overview

Conditional probability measures the probability of an event after new information is known. This is one of the most important ideas in probability because many real problems do not ask for the chance of an event in isolation; they ask for the chance of an event given that something else has already happened. In exam problems, the main difficulty is choosing the correct reduced sample space. ---

Learning Objectives

❗ By the End of This Topic

After studying this topic, you will be able to:

State the definition of conditional probability.

Interpret the condition as a restriction of the sample space.

Compute $P(A \mid B)$ from given probabilities.

Use the multiplication relation

\qquad P(A \cap B)=P(B)\,P(A \mid B)

Distinguish conditional probability from ordinary probability.

---

Definition

📖 Conditional Probability

If $P(B)>0$ , then the conditional probability of $A$ given $B$ is

$\qquad P(A \mid B)=\dfrac{P(A \cap B)}{P(B)}$

This means:

we now look only inside the event $B$

among those outcomes, we ask what fraction also lie in $A$

::: ---

Interpretation

❗ Reduced Sample Space

The event $B$ becomes the new universe of possible outcomes.

So conditional probability is not about changing the event $A$ ; it is about changing the sample space to $B$ .

This is the central idea students often miss. ---

Immediate Consequences

📐 Multiplication Rule

From the definition,

$\qquad P(A \cap B)=P(B)\,P(A \mid B)$

Also,

$\qquad P(A \cap B)=P(A)\,P(B \mid A)$

These are among the most useful identities in probability. ---

Conditional Probability and Independence

📖 Independence

Events $A$ and $B$ are independent if

$\qquad P(A \cap B)=P(A)P(B)$

When

P(B)>0

, this is equivalent to

\qquad P(A \mid B)=P(A)

::: So for independent events, knowing

B

does not change the probability of

A

. ---

Minimal Worked Examples

Example 1 A card is chosen from a standard deck. Let:

$A$ = the card is a king

$B$ = the card is a face card

Then

\qquad P(A \cap B)=P(A)=\dfrac{4}{52}

Also

\qquad P(B)=\dfrac{12}{52}

\qquad P(A \mid B)=\dfrac{4/52}{12/52}=\dfrac{4}{12}=\dfrac{1}{3}

--- Example 2 A fair die is rolled. Let:

$A$ = outcome is even

$B$ = outcome is greater than $3$

Then

\qquad B=\{4,5,6\}

Inside this reduced sample space, the even outcomes are

\qquad \{4,6\}

\qquad P(A \mid B)=\dfrac{2}{3}

This matches the formula:

\qquad P(A \cap B)=\dfrac{2}{6},\quad P(B)=\dfrac{3}{6}

hence

\qquad P(A \mid B)=\dfrac{2/6}{3/6}=\dfrac{2}{3}

---

Common Patterns

💡 Typical Exam Patterns

condition given explicitly:

\qquad P(A \mid B)

reverse computation from

\qquad P(A \cap B)

and

P(B)

card, dice, and subset problems with reduced sample spaces

checking independence by testing whether

\qquad P(A \mid B)=P(A)

---

Common Mistakes

⚠️ Avoid These Errors

❌ using $\dfrac{P(A)}{P(B)}$ instead of $\dfrac{P(A \cap B)}{P(B)}$

✅ numerator must be the overlap

❌ forgetting the condition changes the sample space

✅ work inside

B

❌ applying the formula when $P(B)=0$

✅ conditional probability needs

P(B)>0

❌ mixing $P(A \mid B)$ with $P(B \mid A)$

✅ these are usually different

---

CMI Strategy

💡 How to Attack Conditional Probability Questions

Identify the conditioning event first.

Restrict attention to that event.

Count or compute the overlap with the target event.

\qquad P(A \mid B)=\dfrac{P(A \cap B)}{P(B)}

If the sample space is small, list outcomes directly.

If given probabilities are numerical, use the formula directly.

---

Practice Questions

:::question type="MCQ" question="If

P(B)>0

, then

P(A \mid B)

is equal to" options=["

\dfrac{P(A)}{P(B)}

","

\dfrac{P(A \cap B)}{P(B)}

","

\dfrac{P(B)}{P(A \cap B)}

","

P(A \cap B)

"] answer="B" hint="Recall the definition exactly." solution="By definition,

\qquad P(A \mid B)=\dfrac{P(A \cap B)}{P(B)}

provided

P(B)>0

. Hence the correct option is

\boxed{B}

." ::: :::question type="NAT" question="A fair die is rolled. Let

A

be the event 'even' and

B

be the event 'greater than

2

'. Find

P(A \mid B)

." answer="1/2" hint="Restrict the sample space to

B

." solution="The event

\qquad B=\{3,4,5,6\}

Within this reduced sample space, the even outcomes are

\qquad \{4,6\}

\qquad P(A \mid B)=\dfrac{2}{4}=\dfrac{1}{2}

Hence the answer is

\boxed{\dfrac{1}{2}}

." ::: :::question type="MSQ" question="Which of the following statements are true?" options=["

P(A \mid B)=\dfrac{P(A \cap B)}{P(B)}

when

P(B)>0

","Conditional probability uses a reduced sample space","If

A

and

B

are independent, then

P(A \mid B)=P(A)

","In general

P(A \mid B)=P(B \mid A)

"] answer="A,B,C" hint="Recall definition and independence." solution="1. True.

True.

False in general.

Hence the correct answer is

\boxed{A,B,C}

." ::: :::question type="SUB" question="A card is chosen from a standard deck. Given that the card is a face card, find the probability that it is a queen." answer="

\dfrac{1}{3}

" hint="There are

12

face cards and

4

queens among them." solution="Let

\qquad A=\{\text{queen}\},\quad B=\{\text{face card}\}

There are

12

face cards in a standard deck and

4

queens, all of which are face cards. So

\qquad P(A \mid B)=\dfrac{4}{12}=\dfrac{1}{3}

Therefore the required probability is

\boxed{\dfrac{1}{3}}

." ::: ---

Summary

❗ Key Takeaways for CMI

Conditional probability means probability inside a reduced sample space.

The definition is

\qquad P(A \mid B)=\dfrac{P(A \cap B)}{P(B)}

The multiplication rule comes directly from the definition.

Independence means the condition does not change the probability.

Clear identification of the conditioning event is the whole game.

---

💡 Next Up

Proceeding to Multiplication rule.

---

Part 2: Multiplication rule

Multiplication Rule

Overview

The multiplication rule is the main tool for finding the probability that several events happen together. It becomes especially important in sequential experiments, where the probability of a full path is built step by step. In exam problems, the rule often appears in card drawing, ball drawing, repeated trials, and multi-stage decisions. ---

Learning Objectives

❗ By the End of This Topic

After studying this topic, you will be able to:

Use the multiplication rule for two events.

Extend it to three or more events.

Handle both independent and dependent sequential processes.

Compute probabilities of ordered event sequences.

Recognize when multiplication must be combined with case-splitting or addition.

---

Core Rule

📐 Multiplication Rule for Two Events

For events $A$ and $B$ with $P(A)>0$ ,

$\qquad P(A\cap B)=P(A)\,P(B\mid A)$

This means:

first event happens,

then second event happens given the first.

:::

📐 Symmetric Form

Also,

$\qquad P(A\cap B)=P(B)\,P(A\mid B)$

Both forms are valid. ---

Chain Rule

📐 For Several Events

For events $A_1,A_2,\dots,A_n$ ,

$\qquad P(A_1\cap A_2\cap \cdots \cap A_n) <br>= P(A_1)\,P(A_2\mid A_1)\,P(A_3\mid A_1\cap A_2)\cdots$

This is the standard way to compute the probability of a full sequence. ::: ---

Independent Case

❗ Special Simplification

If $A$ and $B$ are independent, then

$\qquad P(B\mid A)=P(B)$

so the multiplication rule becomes

$\qquad P(A\cap B)=P(A)P(B)$

But in dependent settings, you must use the conditional form. ::: ---

Ordered vs Unordered

⚠️ Common Exam Distinction

The multiplication rule naturally computes ordered sequences.

For example:

first red then blue,

head then tail,

first chosen is a boy and second chosen is a girl.

If the event can happen in several orders, then you often need to add the probabilities of those paths.

---

Minimal Worked Examples

Example 1 A bag contains

3

red,

2

green, and

1

blue ball. Two balls are drawn without replacement. The probability that the first is red and the second is green is

\qquad \dfrac36 \cdot \dfrac25 = \dfrac15

--- Example 2 A fair coin is tossed three times. The probability of getting HHT is

\qquad \dfrac12 \cdot \dfrac12 \cdot \dfrac12 = \dfrac18

because the tosses are independent. ---

Multiplication + Addition Together

💡 Very Common Pattern

If an event can happen in several disjoint ways, then:

use multiplication rule for each ordered path,

add the path probabilities.

Example:
To get exactly one black ball in two draws, you may have:

black then non-black

non-black then black

So the total probability is the sum of the two path probabilities.

---

Common Mistakes

⚠️ Avoid These Errors

❌ Multiplying unconditional probabilities in a dependent setting.

❌ Forgetting to condition later steps on earlier steps.

❌ Using multiplication when the paths should first be split into cases.

❌ Treating unordered events as a single ordered path.

---

CMI Strategy

💡 How to Solve These Fast

Write the event in a clear time order.

Compute each step using the correct current probability.

Multiply along the path.

If several valid paths exist, add them at the end.

Check whether the process is with replacement or without replacement.

---

Practice Questions

:::question type="MCQ" question="A bag contains

3

red,

2

green, and

1

blue ball. Two balls are drawn without replacement. The probability that the first is red and the second is green is" options=["

\dfrac16

","

\dfrac15

","

\dfrac25

","

\dfrac13

"] answer="B" hint="Multiply the probability of the first step by the conditional probability of the second." solution="The probability that the first ball is red is

\qquad \dfrac36

After removing a red ball,

5

balls remain, of which

2

are green. So

\qquad P(\text{second green}\mid \text{first red})=\dfrac25

Hence the required probability is

\qquad \dfrac36\cdot \dfrac25=\dfrac15

So the correct option is

\boxed{B}

." ::: :::question type="NAT" question="A fair coin is tossed four times. Find the probability of the ordered outcome HTTH." answer="1/16" hint="Use independence and multiply four factors of

1/2

." solution="Each toss has probability

\dfrac12

, and the four tosses are independent. So

\qquad P(\text{HTTH})=\left(\dfrac12\right)^4=\dfrac{1}{16}

Hence the answer is

\boxed{\dfrac1{16}}

." ::: :::question type="MSQ" question="Which of the following statements are true?" options=["For events

A,B

with

P(A)>0

P(A\cap B)=P(A)P(B\mid A)

","For independent events

A,B

P(A\cap B)=P(A)P(B)

","The multiplication rule is naturally suited to sequential experiments","For dependent events, one may always replace

P(B\mid A)

P(B)

"] answer="A,B,C" hint="One statement ignores dependence." solution="1. True.

True.

False. That replacement is only valid for independent events.

Hence the correct answer is

\boxed{A,B,C}

." ::: :::question type="SUB" question="A box contains

4

red,

3

blue, and

2

green balls. Three balls are drawn without replacement. Find the probability that the three drawn balls are of three different colours." answer="2/7" hint="Count all colour orders and use the multiplication rule." solution="To get three different colours, we must draw one red, one blue, and one green in some order. There are

3!=6

possible colour orders. Let us compute one order, say red-blue-green:

\qquad \dfrac49 \cdot \dfrac38 \cdot \dfrac27 = \dfrac{24}{504}

Each of the

6

colour orders gives the same value, so the total probability is

\qquad 6 \cdot \dfrac{24}{504} = \dfrac{144}{504} = \dfrac27

Hence the required probability is

\boxed{\dfrac27}

." ::: ---

Summary

❗ Key Takeaways for CMI

The multiplication rule computes the probability of several events happening together.

In dependent settings, later probabilities must be conditional.

For independent events, the rule simplifies to plain multiplication.

Ordered paths are multiplied; multiple disjoint paths are then added.

The multiplication rule is one of the most important tools in sequential probability.

---

💡 Next Up

Proceeding to Independence.

---

Part 3: Independence

Independence

Overview

Independence is one of the most important ideas in probability, but also one of the most misunderstood. Two events are independent if knowing that one happened does not change the probability of the other. In CMI-style questions, the difficulty often comes from hidden conditioning, multi-stage experiments, and situations where events look symmetric but are actually dependent because of a common first step. This topic is especially important for conditional probability, because many problems ask whether events are independent unconditionally, conditionally, or neither. ---

Learning Objectives

❗ By the End of This Topic

After studying this topic, you will be able to:

test whether two events are independent,

compute conditional probabilities involving independent events,

distinguish unconditional independence from conditional independence,

handle multi-stage experiments with a hidden first random choice,

avoid common mistakes such as confusing disjointness with independence.

---

Core Definition

📖 Independence of Two Events

Two events $A$ and $B$ are independent if

$\qquad \mathbb{P}(A \cap B) = \mathbb{P}(A)\mathbb{P}(B)$

provided all probabilities are taken in the same experiment.

📐 Equivalent Conditional Form

If $\mathbb{P}(B) > 0$ , then independence of $A$ and $B$ is equivalent to

$\qquad \mathbb{P}(A \mid B) = \mathbb{P}(A)$

Similarly, if $\mathbb{P}(A) > 0$ , it is also equivalent to

$\qquad \mathbb{P}(B \mid A) = \mathbb{P}(B)$

So independence means that learning one event does not change the probability of the other. ---

How to Test Independence

💡 Fast Test

To check whether $A$ and $B$ are independent:

compute $\mathbb{P}(A)$ ,

compute $\mathbb{P}(B)$ ,

compute $\mathbb{P}(A \cap B)$ ,

compare with $\mathbb{P}(A)\mathbb{P}(B)$ .

If they are equal, the events are independent. If not, they are dependent. ---

Independence Is Not Disjointness

⚠️ Very Important Distinction

Disjoint events satisfy

$\qquad A \cap B = \varnothing$

so

$\qquad \mathbb{P}(A \cap B)=0$

If $A$ and $B$ are disjoint and both have positive probability, then they cannot be independent, because

$\qquad 0 \ne \mathbb{P}(A)\mathbb{P}(B)$

So:

disjointness means the events cannot occur together,

independence means occurrence of one gives no information about the other.

These are completely different ideas. ---

Independence and Complements

📐 Complement Facts

If $A$ and $B$ are independent, then the following pairs are also independent:

$A$ and $B^c$

$A^c$ and $B$

$A^c$ and $B^c$

This is very useful in computations. For example,

\qquad \mathbb{P}(A^c \cap B^c) = (1-\mathbb{P}(A))(1-\mathbb{P}(B))

when

A

and

B

are independent. ---

Multi-Stage Experiments

❗ Where Students Make Mistakes

In many problems, two visible outcomes are generated after a hidden first random choice.

In such cases:

the visible outcomes may be independent given the first stage,

but not independent overall.

This is one of the most common exam traps.

A common cause can create dependence. ---

Conditional Independence

📖 Conditional Independence

Events $A$ and $B$ are conditionally independent given an event or variable $C$ if

$\qquad \mathbb{P}(A \cap B \mid C) = \mathbb{P}(A \mid C)\mathbb{P}(B \mid C)$

This does not automatically imply that $A$ and $B$ are independent without conditioning.

---

Minimal Worked Example: Hidden Common Cause

Example 1 A fair coin is tossed.

If heads occurs, both daughters win.

If tails occurs, each daughter independently wins with probability $\dfrac{1}{3}$ .

Let

A

be the event that daughter

1

wins and

B

the event that daughter

2

wins. Then

\qquad \mathbb{P}(A) = \dfrac{1}{2}\cdot 1 + \dfrac{1}{2}\cdot \dfrac{1}{3} = \dfrac{2}{3}

Similarly,

\qquad \mathbb{P}(B) = \dfrac{2}{3}

Now

\qquad \mathbb{P}(A \cap B) = \dfrac{1}{2}\cdot 1 + \dfrac{1}{2}\cdot \dfrac{1}{3}\cdot \dfrac{1}{3}

\qquad = \dfrac{1}{2} + \dfrac{1}{18} = \dfrac{5}{9}

But

\qquad \mathbb{P}(A)\mathbb{P}(B) = \dfrac{2}{3}\cdot \dfrac{2}{3} = \dfrac{4}{9}

Since

\qquad \dfrac{5}{9} \ne \dfrac{4}{9}

the events

A

and

B

are not independent. However, given tails, the daughters' die throws are independent. This is a classic example of conditional independence but not unconditional independence. ---

Standard Probability Identities

📐 Useful Formulas

For any events $A,B$ with $\mathbb{P}(B)>0$ :

$\qquad \mathbb{P}(A \mid B) = \dfrac{\mathbb{P}(A \cap B)}{\mathbb{P}(B)}$

Also,

$\qquad \mathbb{P}(A \cap B) = \mathbb{P}(A \mid B)\mathbb{P}(B)$

If $A$ and $B$ are independent, then

$\qquad \mathbb{P}(A \cap B) = \mathbb{P}(A)\mathbb{P}(B)$

---

Pairwise vs Mutual Independence

📐 Three Events

For three events $A,B,C$ :

pairwise independence means each pair is independent,

mutual independence means

- each pair is independent, and
-

\mathbb{P}(A \cap B \cap C)=\mathbb{P}(A)\mathbb{P}(B)\mathbb{P}(C)

Mutual independence is stronger than pairwise independence. ---

Common Patterns

📐 Patterns to Recognize

two independent tosses or throws,

repeated Bernoulli trials,

hidden first-stage random choice causing dependence,

checking independence by direct calculation,

conditional-probability questions asking whether one event changes another.

---

Common Mistakes

⚠️ Avoid These Errors

❌ assuming symmetry implies independence,

✅ always check the product rule

❌ confusing disjoint events with independent events,

✅ disjoint positive-probability events are never independent

❌ treating conditional independence as ordinary independence,

✅ hidden common causes can destroy unconditional independence

❌ checking only $\mathbb{P}(A \mid B)=\mathbb{P}(A)$ without ensuring $\mathbb{P}(B)>0$ ,

✅ use the condition properly

---

CMI Strategy

💡 How to Solve These Problems

Define the events clearly.

If the experiment has stages, split according to the first stage.

Compute marginals and intersections separately.

Check independence only after exact calculation.

In conditional questions, write everything in terms of intersection over conditioning event.

---

Practice Questions

:::question type="MCQ" question="If

A

and

B

are independent with

\mathbb{P}(A)=\dfrac{1}{2}

and

\mathbb{P}(B)=\dfrac{1}{3}

, then

\mathbb{P}(A\cap B)

equals" options=["

\dfrac{1}{6}

","

\dfrac{2}{5}

","

\dfrac{5}{6}

","

\dfrac{1}{2}

"] answer="A" hint="Use the product rule for independent events." solution="Since

A

and

B

are independent,

\qquad \mathbb{P}(A\cap B)=\mathbb{P}(A)\mathbb{P}(B)=\dfrac{1}{2}\cdot\dfrac{1}{3}=\dfrac{1}{6}

. Hence the correct option is

\boxed{A}

." ::: :::question type="NAT" question="Suppose

A

and

B

are independent,

\mathbb{P}(A)=0.4

, and

\mathbb{P}(B)=0.5

. Find

\mathbb{P}(A^c\cap B)

." answer="0.3" hint="Use complement independence or subtract

\mathbb{P}(A\cap B)

from

\mathbb{P}(B)

." solution="Because

A

and

B

are independent,

\qquad \mathbb{P}(A\cap B)=0.4\cdot 0.5=0.2

\qquad \mathbb{P}(A^c\cap B)=\mathbb{P}(B)-\mathbb{P}(A\cap B)=0.5-0.2=0.3

Hence the answer is

\boxed{0.3}

." ::: :::question type="MSQ" question="Which of the following statements are true?" options=["If

A

and

B

are independent, then

\mathbb{P}(A\cap B)=\mathbb{P}(A)\mathbb{P}(B)

","If

A

and

B

are disjoint and both have positive probability, then they are independent","If

A

and

B

are independent, then

A

and

B^c

are independent","Conditional independence always implies unconditional independence"] answer="A,C" hint="Independence, disjointness, and conditional independence are different concepts." solution="1. True. This is the definition of independence.

False. If

A

and

B

are disjoint and both have positive probability, then

\mathbb{P}(A\cap B)=0

but

\mathbb{P}(A)\mathbb{P}(B)>0

True. Complements preserve independence.

False. Events can be conditionally independent but dependent overall.

Hence the correct answer is

\boxed{A,C}

." ::: :::question type="SUB" question="Suppose

A

and

B

are independent events with

\mathbb{P}(A)=p

and

\mathbb{P}(B)=q

. Prove that

A^c

and

B

are independent." answer="Since

\mathbb{P}(A^c\cap B)=\mathbb{P}(B)-\mathbb{P}(A\cap B)=q-pq=(1-p)q=\mathbb{P}(A^c)\mathbb{P}(B)

, the events are independent" hint="Write

A^c\cap B

B\setminus(A\cap B)

." solution="Since

A

and

B

are independent,

\qquad \mathbb{P}(A\cap B)=pq

Now

\qquad \mathbb{P}(A^c\cap B)=\mathbb{P}(B)-\mathbb{P}(A\cap B)

\qquad \mathbb{P}(A^c\cap B)=q-pq=q(1-p)

But

\qquad \mathbb{P}(A^c)=1-p

Hence

\qquad \mathbb{P}(A^c\cap B)=\mathbb{P}(A^c)\mathbb{P}(B)

Therefore

A^c

and

B

are independent." ::: ---

Summary

❗ Key Takeaways for CMI

Independence means $\mathbb{P}(A\cap B)=\mathbb{P}(A)\mathbb{P}(B)$ .

Equivalent conditional forms are valid when conditioning probability is positive.

Disjointness and independence are very different.

Hidden first-stage randomness can create dependence.

Conditional independence does not imply unconditional independence.

Always compute, do not guess.

---

💡 Next Up

Proceeding to Dependent event reasoning.

---

Part 4: Dependent event reasoning

Dependent Event Reasoning

Overview

Events are called dependent when the occurrence of one event changes the probability of another. This is one of the most important probability ideas in exam problems involving cards, balls, arrangements, and sequential choices without replacement. In CMI-style questions, the real test is whether you correctly update the sample space after earlier information is known. ---

Learning Objectives

❗ By the End of This Topic

After studying this topic, you will be able to:

Recognize when two events are dependent.

Update probabilities after new information is revealed.

Work correctly with “without replacement” situations.

Use conditional probability language in dependent settings.

Avoid treating dependent events as independent.

---

Core Idea

📖 Dependent Events

Two events $A$ and $B$ are dependent if knowing that $A$ has happened changes the probability of $B$ .

That is, dependence appears when

$\qquad P(B\mid A)\ne P(B)$

In practical problems, dependence usually arises because:

an object has been removed,

information has been revealed,

one choice affects the remaining possibilities.

::: ---

Standard Sequential View

📐 Without Replacement

When objects are chosen without replacement, later probabilities must be computed from the reduced set.

Example:
If a bag has $4$ red and $3$ blue balls, then after drawing one red ball, the composition becomes

$\qquad 3\text{ red and }3\text{ blue}$

So the probability of a blue ball on the next draw is not the original one. ---

Conditional Probability Language

📐 Interpretation

The expression

$\qquad P(B\mid A)$

means:
the probability of event $B$ given that event $A$ has already occurred.

In dependent-event reasoning, this is usually the correct language to use. ---

Typical Dependent Situations

💡 Common Exam Contexts

Dependent reasoning appears in:

drawing cards without replacement

drawing balls from an urn without replacement

arranging objects step by step

revealing partial information

“given that at least one” type conditions

---

Updating the Sample Space

❗ Most Important Habit

After an event occurs, do not continue using the old denominator automatically.

Always ask:

what outcomes are still possible now?

how many favourable outcomes remain now?

This is the main difference between dependent and independent reasoning. ---

Minimal Worked Examples

Example 1 A bag has

5

white and

4

black balls. One ball is drawn and not replaced. Given that the first ball was black, the probability that the second is black is

\qquad \dfrac{3}{8}

because now only

3

black remain out of

8

total balls. --- Example 2 A bag has

4

red and

3

blue balls. Given that the first ball drawn is red, the probability that the second ball is blue is

\qquad \dfrac{3}{6}=\dfrac12

This is not the original blue probability

\dfrac37

. ---

“At Least One” Conditioning

⚠️ Very Common Trap

Statements like

“given that at least one selected ball is black”

“given that at least one toss is a head”

change the sample space.

You must condition on the reduced event, not on the original total sample space.

This is one of the most common sources of error in dependent-event problems. ---

Common Mistakes

⚠️ Avoid These Errors

❌ Treating “without replacement” as if probabilities stay unchanged.

❌ Multiplying unconditional probabilities in a dependent setting.

❌ Ignoring the information in “given that” statements.

❌ Forgetting to update the denominator after one event occurs.

❌ Using symmetry where the events are not symmetric after conditioning.

---

CMI Strategy

💡 How to Solve These Fast

Identify what information is already known.

Rewrite the state of the system after that information.

Count the remaining favourable outcomes.

Count the remaining total outcomes.

Only then compute the probability.

---

Practice Questions

:::question type="MCQ" question="A bag contains

4

red and

3

blue balls. Two balls are drawn without replacement. Given that the first ball is red, the probability that the second is blue is" options=["

\dfrac37

","

\dfrac12

","

\dfrac47

","

\dfrac38

"] answer="B" hint="Update the composition after the first red ball is removed." solution="After one red ball is removed, the bag contains

3

red and

3

blue balls, so the probability that the second ball is blue is

\qquad \dfrac{3}{6}=\dfrac12

Hence the correct option is

\boxed{B}

." ::: :::question type="NAT" question="A bag contains

5

white and

4

black balls. Two balls are drawn without replacement. Given that the first ball drawn is black, find the probability that the second ball is also black." answer="3/8" hint="Condition on the updated bag after one black ball is removed." solution="After one black ball is drawn, the bag contains

5

white and

3

black balls, so there are

8

balls left in total. Hence the probability that the second ball is black is

\qquad \dfrac{3}{8}

So the answer is

\boxed{\dfrac38}

." ::: :::question type="MSQ" question="Which of the following situations involve dependent events?" options=["Drawing two cards from a deck without replacement","Drawing one ball, replacing it, and then drawing another","Choosing two students one after another without replacement","Tossing a fair coin twice"] answer="A,C" hint="Check whether the first outcome changes the second probability." solution="1. Dependent, because the first card changes the deck.

Independent, because replacement restores the original state.

Dependent, because the first selection changes the pool.

Independent, because one toss does not affect the other.

Hence the correct answer is

\boxed{A,C}

." ::: :::question type="SUB" question="A bag contains

5

white balls and

4

black balls. Two balls are drawn without replacement. Given that at least one of the two balls is black, find the probability that both balls are black." answer="3/13" hint="Use conditional probability: both black divided by at least one black." solution="Let

\qquad A=

event that both drawn balls are black and

\qquad B=

event that at least one drawn ball is black. Then

\qquad P(A\mid B)=\dfrac{P(A)}{P(B)}

Total number of ways to choose

2

balls from

9

\qquad \binom92 = 36

Now

\qquad P(A)=\dfrac{\binom42}{36}=\dfrac{6}{36}

To find

P(B)

, it is easier to use the complement: the only way

B

fails is if both balls are white. So

\qquad P(B)=1-\dfrac{\binom52}{36}=1-\dfrac{10}{36}=\dfrac{26}{36}

Hence

\qquad P(A\mid B)=\dfrac{6/36}{26/36}=\dfrac{6}{26}=\dfrac{3}{13}

Therefore the required probability is

\boxed{\dfrac{3}{13}}

." ::: ---

Summary

❗ Key Takeaways for CMI

Dependent events require updating probabilities after information is known.

Without replacement is the standard source of dependence.

The phrase “given that” must change the sample space.

Conditional reasoning is usually the cleanest language for dependent events.

The hardest part is not the arithmetic — it is identifying the correct updated state.

Chapter Summary

❗ Conditional events — Key Points

Conditional probability, $P(A|B)$ , quantifies the likelihood of event A occurring given that event B has already occurred, formally defined as $P(A \cap B) / P(B)$ for $P(B) > 0$ .

The Multiplication Rule, $P(A \cap B) = P(A|B)P(B)$ , is fundamental for calculating the probability of the intersection of events, particularly useful in sequential processes or when events are dependent.

Two events A and B are independent if the occurrence of one does not affect the probability of the other. This is mathematically expressed as $P(A|B) = P(A)$ , or equivalently, $P(A \cap B) = P(A)P(B)$ .

Events are dependent if the probability of one event is influenced by the occurrence or non-occurrence of another, meaning $P(A|B) \neq P(A)$ .

Careful interpretation of problem statements is crucial to distinguish between joint probabilities ( $P(A \cap B)$ ), conditional probabilities ( $P(A|B)$ or $P(B|A)$ ), and marginal probabilities ( $P(A)$ or $P(B)$ ).

Tree diagrams are a powerful visual tool for breaking down complex scenarios involving sequences of dependent events, aiding in the application of the multiplication rule and calculation of conditional probabilities.

Chapter Review Questions

:::question type="MCQ" question="An urn contains 5 red balls and 3 blue balls. Two balls are drawn without replacement. What is the probability that the second ball drawn is red, given that the first ball drawn was blue?" options=[" $\frac{5}{7}$ "," $\frac{5}{8}$ "," $\frac{3}{8}$ "," $\frac{3}{7}$ "] answer=" $\frac{5}{7}$ " hint="Consider the composition of the urn after the first ball is drawn." solution="Let R1 be the event that the first ball is red, and B1 be the event that the first ball is blue. Let R2 be the event that the second ball is red.
We are looking for $P(R2 | B1)$ .
If the first ball drawn was blue, then there are 7 balls remaining in the urn: 5 red and 2 blue.
The probability of drawing a red ball next is $\frac{\text{Number of red balls remaining}}{\text{Total number of balls remaining}} = \frac{5}{7}$ .
So, $P(R2 | B1) = \frac{5}{7}$ ."
:::

:::question type="NAT" question="For two events A and B, $P(A) = 0.6$ , $P(B) = 0.5$ , and $P(A \cup B) = 0.8$ . Calculate $P(B|A)$ ." answer="0.5" hint="First, find $P(A \cap B)$ using the formula for the probability of the union of two events." solution="We know the formula for the probability of the union of two events:
$P(A \cup B) = P(A) + P(B) - P(A \cap B)$
Substituting the given values:
$0.8 = 0.6 + 0.5 - P(A \cap B)$
$0.8 = 1.1 - P(A \cap B)$
$P(A \cap B) = 1.1 - 0.8 = 0.3$

Now, we can calculate the conditional probability $P(B|A)$ using its definition:
$P(B|A) = \frac{P(A \cap B)}{P(A)}$
$P(B|A) = \frac{0.3}{0.6} = 0.5$ ."
:::

:::question type="MCQ" question="A fair coin is tossed twice. Let event E be 'the first toss is a head' and event F be 'the two tosses are the same (both heads or both tails)'. Are E and F independent?" options=["Yes, they are independent." ,"No, they are dependent." ,"Only if the coin is biased." ,"Cannot be determined without more information."] answer="Yes, they are independent." hint="List the sample space and the outcomes for E, F, and $E \cap F$ . Then check if $P(E \cap F) = P(E)P(F)$ ." solution="The sample space for two coin tosses is $S = \{HH, HT, TH, TT\}$ . Each outcome has a probability of $\frac{1}{4}$ .

Event E: 'the first toss is a head' $\implies E = \{HH, HT\}$ .
$P(E) = \frac{2}{4} = \frac{1}{2}$ .

Event F: 'the two tosses are the same' $\implies F = \{HH, TT\}$ .
$P(F) = \frac{2}{4} = \frac{1}{2}$ .

Event $E \cap F$ : 'the first toss is a head AND the two tosses are the same' $\implies E \cap F = \{HH\}$ .
$P(E \cap F) = \frac{1}{4}$ .

To check for independence, we compare $P(E \cap F)$ with $P(E)P(F)$ :
$P(E)P(F) = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4}$ .

Since $P(E \cap F) = P(E)P(F)$ ( $\frac{1}{4} = \frac{1}{4}$ ), events E and F are independent.
Therefore, the correct answer is 'Yes, they are independent.'."
:::

What's Next?

💡 Continue Your CMI Journey

This chapter on conditional events forms a crucial foundation for more advanced topics in probability. The principles learned here are directly applied in understanding Bayes' Theorem, which deals with updating probabilities based on new evidence. Furthermore, a solid grasp of conditional probability is essential before delving into the study of random variables, their probability distributions (e.g., binomial, Poisson, normal), and the concepts of expectation and variance, which are central to statistical inference and modeling.

Conditional events

Conditional events

Chapter Contents

| Topic |

Part 1: Conditional probability definition

Conditional Probability Definition

Overview

Learning Objectives

Definition

Interpretation

Immediate Consequences

Conditional Probability and Independence

Minimal Worked Examples

Common Patterns

Common Mistakes

CMI Strategy

Practice Questions

Summary

Part 2: Multiplication rule

Multiplication Rule

Overview

Learning Objectives

Core Rule

Chain Rule

Independent Case

Ordered vs Unordered

Minimal Worked Examples

Multiplication + Addition Together

Common Mistakes

CMI Strategy

Practice Questions

Summary

Part 3: Independence

Independence

Overview

Learning Objectives

Core Definition

How to Test Independence

Independence Is Not Disjointness

Independence and Complements

Multi-Stage Experiments

Conditional Independence

Minimal Worked Example: Hidden Common Cause

Standard Probability Identities

Pairwise vs Mutual Independence

Common Patterns

Common Mistakes

CMI Strategy

Practice Questions

Summary

Part 4: Dependent event reasoning

Dependent Event Reasoning

Overview

Learning Objectives

Core Idea

Standard Sequential View

Conditional Probability Language

Typical Dependent Situations

Updating the Sample Space

Minimal Worked Examples

“At Least One” Conditioning

Common Mistakes

CMI Strategy

Practice Questions

Summary

Chapter Summary

Chapter Review Questions

What's Next?

🎯 Key Points to Remember

Related Topics in Probability

Counting-based probability

Bayes-type reasoning

Events and sample space

Discrete random variables

More Resources

Study Notes

Short Notes

Test Series

Mock Tests

Previous Year Papers