Discrete random variables

This chapter establishes the foundational concepts of discrete random variables, encompassing their probability distributions, expectation, and variance. A thorough understanding of these principles is critical for subsequent advanced topics in probability and statistics, and is consistently evaluated in CMI examinations through both theoretical and applied problems.

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Chapter Contents

| Topic |

|---|-------| | 1 | Probability distribution | | 2 | Expectation | | 3 | Variance at school level | | 4 | Simple modelling problems |

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We begin with Probability distribution.

Part 1: Probability distribution

Probability Distribution

Overview

A probability distribution describes how probability is assigned to the possible values of a random variable. In CMI-style questions, this topic is not just about definitions. It includes checking whether a formula really defines a distribution, building distributions from experiments like coin tosses and die throws, and computing exact probabilities from repeated independent trials. ---

Learning Objectives

❗ By the End of This Topic

After studying this topic, you will be able to:

Define and use the probability mass function of a discrete random variable.

Check whether a given table or formula is a valid probability distribution.

Construct distributions from simple experiments.

Use Bernoulli and binomial ideas in repeated independent trials.

Compute probabilities such as “exactly $k$ ”, “at least one”, and “at most $k$ ”.

---

Core Idea

📖 Discrete Random Variable

A discrete random variable $X$ is a variable that takes only finitely many or countably many values.

Its probability distribution is given by the function

$\qquad p(x) = P(X=x)$

called the probability mass function or pmf.

---

Valid Probability Distribution

📐 PMF Conditions

For a discrete random variable $X$ with pmf $p(x)$ , the following must hold:

$\qquad p(x) \ge 0$ for every possible value of $x$

$\qquad \sum_x p(x) = 1$

❗ Interpretation

The first condition says probabilities cannot be negative.

The second condition says the total probability over all possible values must be exactly $1$ .

---

Distribution Table

A discrete distribution is often written in tabular form. |

x

x_1

x_2

x_3

\cdots

| |---|---|---|---|---| |

P(X=x)

p_1

p_2

p_3

\cdots

| where

\qquad p_1+p_2+p_3+\cdots = 1

---

Cumulative Distribution Function

📐 CDF

The cumulative distribution function of $X$ is

$\qquad F(x)=P(X\le x)$

For a discrete random variable, the cdf is a step function. Useful facts:

$F(x)$ is nondecreasing

$0 \le F(x) \le 1$

as $x\to -\infty$ , $F(x)\to 0$

as $x\to \infty$ , $F(x)\to 1$

---

Expectation and Variance

📐 Expectation

If $X$ has pmf $p(x)$ , then

$\qquad E[X] = \sum_x x\,p(x)$

📐 Variance

The variance of $X$ is

$\qquad \operatorname{Var}(X)=E[X^2]-(E[X])^2$

where

$\qquad E[X^2]=\sum_x x^2 p(x)$

Expectation gives the average value in the long run; variance measures spread. ---

Bernoulli Distribution

📖 Bernoulli Random Variable

A Bernoulli random variable takes only the values $0$ and $1$ .

If

$\qquad P(X=1)=p,\qquad P(X=0)=1-p$

then $X$ has Bernoulli distribution with parameter $p$ .

This models a single success-failure experiment. Examples:

success = “head occurs”

success = “die shows an even number”

success = “outcome divisible by $3$ ”

---

Binomial Distribution

📖 Binomial Distribution

If an experiment is repeated $n$ times independently, and each trial has success probability $p$ , then the number of successes $X$ has binomial distribution:

$\qquad P(X=k)=\binom{n}{k}p^k(1-p)^{n-k},\qquad k=0,1,2,\dots,n$

This is one of the most important discrete distributions for exam problems. ---

Standard Probability Forms

📐 Exactly, At Least One, At Most

If $X\sim \operatorname{Bin}(n,p)$ , then:

Exactly $k$ successes:

\qquad P(X=k)=\binom{n}{k}p^k(1-p)^{n-k}

At least one success:

\qquad P(X\ge 1)=1-P(X=0)=1-(1-p)^n

At most one success:

\qquad P(X\le 1)=P(X=0)+P(X=1)

The complement method is especially important for “one or more” questions. ---

PYQ-Style Example 1

One or more of the first three throws is $4$ Each throw is independent, and

\qquad P(\text{not }4)=\dfrac{5}{6}

So $\qquad P(\text{at least one }4\text{ in first three throws}) =1-\left(\dfrac{5}{6}\right)^3$ $\qquad =1-\dfrac{125}{216} =\dfrac{91}{216}$ ---

PYQ-Style Example 2

Exactly two of the last four throws are divisible by $3$ A die outcome is divisible by

3

if it is

3

6

, so

\qquad p = \dfrac{2}{6}=\dfrac{1}{3}

Let

X

be the number of such throws among the last four. Then

\qquad X\sim \operatorname{Bin}\left(4,\dfrac{1}{3}\right)

Hence

\qquad P(X=2)=\binom{4}{2}\left(\dfrac{1}{3}\right)^2\left(\dfrac{2}{3}\right)^2

$\qquad =6\cdot \dfrac{1}{9}\cdot \dfrac{4}{9} =\dfrac{24}{81} =\dfrac{8}{27}$ ---

Constructing a Distribution from an Experiment

💡 Standard Method

To build the distribution of a random variable:

identify all possible values of the variable

compute the probability of each value

check that all probabilities are nonnegative

check that their sum is $1$

Example A fair coin is tossed twice. Let

X

be the number of heads. Possible values:

\qquad 0,1,2

Probabilities:

$\qquad P(X=0)=\dfrac{1}{4}$

$\qquad P(X=1)=\dfrac{2}{4}=\dfrac{1}{2}$

$\qquad P(X=2)=\dfrac{1}{4}$

So the distribution is |

x

0

1

2

| |---|---|---|---| |

P(X=x)

\dfrac{1}{4}

\dfrac{1}{2}

\dfrac{1}{4}

| ---

Common Mistakes

⚠️ Avoid These Errors

❌ Forgetting that total probability must add to $1$

✅ Always check the sum

❌ Using binomial formula without independence

✅ Binomial model needs independent identical trials

❌ Mixing up “at least one” with “exactly one”

✅ Use complement for “at least one”

❌ Ignoring the support of the random variable

✅ First list all possible values clearly

❌ Treating expectation as always one of the possible values

✅ Expectation is an average, not necessarily an attained value

---

CMI Strategy

💡 How to Attack Probability Distribution Questions

First decide whether the question is about a pmf, cdf, or event probability.

If a formula is given, check nonnegativity and total sum $1$ .

For repeated trials, test whether the setup is Bernoulli/binomial.

For “one or more”, use complement before expanding anything.

For “exactly $k$ ”, use the binomial coefficient carefully.

Write the final answer in exact form whenever possible.

---

Practice Questions

:::question type="MCQ" question="Which of the following can be a valid probability mass function on the set

\{0,1,2\}

?" options=["

P(X=0)=0.2,\ P(X=1)=0.3,\ P(X=2)=0.4

","

P(X=0)=0.1,\ P(X=1)=0.4,\ P(X=2)=0.5

","

P(X=0)=-0.1,\ P(X=1)=0.6,\ P(X=2)=0.5

","

P(X=0)=\dfrac{1}{2},\ P(X=1)=\dfrac{1}{2},\ P(X=2)=\dfrac{1}{2}

"] answer="B" hint="Check nonnegativity and whether the probabilities add to

1

." solution="A valid pmf must have all probabilities nonnegative and total sum

1

. Option A sums to

0.9

, so it is invalid. Option B has nonnegative probabilities and

\qquad 0.1+0.4+0.5=1

so it is valid. Option C has a negative probability. Option D sums to

\dfrac{3}{2}

, so it is invalid. Hence the correct option is

\boxed{B}

." ::: :::question type="NAT" question="A fair die is thrown

3

times. Find the probability that

4

appears in one or more of these throws." answer="\\dfrac{91}{216}" hint="Use the complement event 'no throw shows

4

'." solution="The probability that a single throw does not show

4

\qquad \dfrac{5}{6}

So the probability that none of the three throws shows

4

\qquad \left(\dfrac{5}{6}\right)^3=\dfrac{125}{216}

Therefore the probability that

4

appears in one or more throws is

\qquad 1-\dfrac{125}{216}=\dfrac{91}{216}

Hence the answer is

\boxed{\dfrac{91}{216}}

." ::: :::question type="MSQ" question="Which of the following statements are true for a discrete random variable

X

with pmf

p(x)

?" options=["

p(x)\ge 0

for every possible value of

x

","

\sum_x p(x)=1

","

F(x)=P(X\le x)

is nondecreasing","

E[X]

must always be an integer"] answer="A,B,C" hint="Recall the basic definitions of pmf, cdf, and expectation." solution="1. True.

True.

False. The expectation need not be an integer.

Hence the correct answer is

\boxed{A,B,C}

." ::: :::question type="SUB" question="A fair die is thrown twice. Let

X

be the number of throws whose outcome is divisible by

3

. Find the probability distribution of

X

." answer="

P(X=0)=\dfrac{4}{9},\ P(X=1)=\dfrac{4}{9},\ P(X=2)=\dfrac{1}{9}

" hint="An outcome is divisible by

3

if it is

3

6

." solution="A die outcome is divisible by

3

if it is

3

6

, so the success probability on one throw is

\qquad p=\dfrac{2}{6}=\dfrac{1}{3}

Since there are two independent throws, the random variable

X

, the number of divisible-by-

3

outcomes, has binomial distribution:

\qquad X\sim \operatorname{Bin}\left(2,\dfrac{1}{3}\right)

Therefore:

\qquad P(X=0)=\left(\dfrac{2}{3}\right)^2=\dfrac{4}{9}

\qquad P(X=1)=\binom{2}{1}\left(\dfrac{1}{3}\right)\left(\dfrac{2}{3}\right)=\dfrac{4}{9}

\qquad P(X=2)=\left(\dfrac{1}{3}\right)^2=\dfrac{1}{9}

So the required distribution is

\qquad P(X=0)=\dfrac{4}{9},\quad P(X=1)=\dfrac{4}{9},\quad P(X=2)=\dfrac{1}{9}

." ::: ---

Summary

❗ Key Takeaways for CMI

A valid discrete distribution must have nonnegative probabilities summing to $1$ .

The pmf is $p(x)=P(X=x)$ , while the cdf is $F(x)=P(X\le x)$ .

Bernoulli distribution models one success-failure trial.

Binomial distribution models the number of successes in repeated independent identical trials.

“At least one” is often easiest by complement.

Exact algebraic probability expressions are preferred over decimal approximations.

---

💡 Next Up

Proceeding to Expectation.

---

Part 2: Expectation

Expectation

Overview

Expectation is the average or mean value of a random variable in the long run. In school-level and olympiad-style probability, expectation is one of the most powerful summary quantities because it often turns a complicated random process into a simple weighted sum. In exam problems, the main difficulty is modeling the random variable correctly before applying the formula. ---

Learning Objectives

❗ By the End of This Topic

After studying this topic, you will be able to:

Compute expectation for a discrete random variable.

Build a random variable from a probability problem.

Use linearity of expectation.

Interpret expectation even when the value itself need not be achievable.

Solve basic modelling questions involving expected counts and gains.

---

Core Idea

📖 Expectation of a Discrete Random Variable

If a discrete random variable $X$ takes values

$\qquad x_1,x_2,\dots,x_n$

with probabilities

$\qquad p_1,p_2,\dots,p_n$ ,

then the expectation of $X$ is

$\qquad E(X)=x_1p_1+x_2p_2+\cdots+x_np_n$

❗ Interpretation

Expectation is the weighted average of possible values of the random variable.

---

Probability Distribution Conditions

📐 What a Valid Distribution Must Satisfy

If $X$ takes values $x_i$ with probabilities $p_i$ , then:

$\qquad p_i \ge 0$

$\qquad \sum p_i = 1$

Only after checking this should expectation be computed. ::: ---

Linearity of Expectation

📐 Linearity

For random variables $X$ and $Y$ ,

$\qquad E(X+Y)=E(X)+E(Y)$

and for a constant $c$ ,

$\qquad E(cX)=cE(X)$

This remains true whether or not

X

and

Y

are independent. :::

❗ Why This Matters

Linearity allows us to compute expectations of complicated quantities by breaking them into simpler parts.

---

Expectation of Common Variables

📐 Useful Standard Values

If $X$ is Bernoulli with success probability $p$ , then

\qquad E(X)=p

If a fair die is rolled and $X$ is the face value, then

\qquad E(X)=\frac{1+2+3+4+5+6}{6}=\frac{7}{2}

---

Minimal Worked Examples

Example 1 A fair die is rolled. Let

X

be the outcome. Then

\qquad E(X)=\frac{1+2+3+4+5+6}{6}=\frac{21}{6}=\frac{7}{2}

--- Example 2 A random variable

X

takes values

0,1,2

with probabilities

\qquad \frac{1}{4},\ \frac{1}{2},\ \frac{1}{4}

Then

\qquad E(X)=0\cdot \frac{1}{4}+1\cdot \frac{1}{2}+2\cdot \frac{1}{4}=0+\frac{1}{2}+\frac{1}{2}=1

\qquad E(X)=1

::: ---

Modelling Expectation

💡 How to Build the Random Variable

In many questions, the main step is deciding what $X$ represents.

Examples:

number of heads in repeated tosses

score obtained in a game

number of defective items chosen

profit or loss in a scheme

Once

X

is defined clearly, expectation usually becomes routine. ---

Expected Value Need Not Be a Possible Outcome

⚠️ Important Concept

Expectation need not be one of the actual values taken by the random variable.

For example, in one roll of a fair die:

$\qquad E(X)=\frac{7}{2}$

but the die never actually shows $\frac{7}{2}$ .

---

Indicator Variable Idea

📐 Expected Count

If $I$ is an indicator variable that is $1$ when an event occurs and $0$ otherwise, then

$\qquad E(I)=P(I=1)$

This is often used to count expected numbers of successes.

---

CMI Strategy

💡 How to Attack Expectation Questions

Define the random variable clearly.

List its possible values and probabilities.

Check that probabilities sum to $1$ .

Compute the weighted average.

Use linearity when several pieces are involved.

---

Common Mistakes

⚠️ Avoid These Errors

❌ Averaging values without using probabilities

❌ Forgetting that probabilities must sum to $1$

❌ Confusing expectation with most likely value

❌ Assuming expectation must be one of the actual outcomes

---

Practice Questions

:::question type="MCQ" question="If a fair die is rolled once and

X

is the outcome, then

E(X)

is" options=["

3

","

\dfrac{7}{2}

","

4

","

\dfrac{21}{2}

"] answer="B" hint="Use the average of all six equally likely values." solution="We have

\qquad E(X)=\frac{1+2+3+4+5+6}{6}=\frac{21}{6}=\frac{7}{2}

Hence the correct option is

\boxed{B}

." ::: :::question type="NAT" question="A random variable

X

takes values

1,2,3

with probabilities

\dfrac{1}{2},\dfrac{1}{3},\dfrac{1}{6}

respectively. Find

E(X)

." answer="5/3" hint="Compute the weighted average." solution="We compute

\qquad E(X)=1\cdot \frac{1}{2}+2\cdot \frac{1}{3}+3\cdot \frac{1}{6}

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

So the answer is

\boxed{\frac{5}{3}}

." ::: :::question type="MSQ" question="Which of the following are true?" options=["Expectation is a weighted average","Expectation of a die roll need not be an integer","Probabilities in a distribution must sum to

1

","Expectation is always the most likely outcome"] answer="A,B,C" hint="Check definition and interpretation carefully." solution="1. True. 2. True, for example a fair die has expectation

\frac{7}{2}

. 3. True. 4. False. Expectation need not be the most likely value. Hence the correct answer is

\boxed{A,B,C}

." ::: :::question type="SUB" question="A fair coin is tossed three times. Let

X

be the number of heads obtained. Find

E(X)

." answer="

\dfrac{3}{2}

" hint="List the binomial probabilities or use linearity of expectation." solution="Let

X

be the number of heads in three tosses. Using linearity of expectation, write

\qquad X=I_1+I_2+I_3

where

I_j=1

if the

j

th toss is a head and

0

otherwise. For each toss,

\qquad E(I_j)=P(\text{head})=\frac{1}{2}

\qquad E(X)=E(I_1)+E(I_2)+E(I_3)=\frac{1}{2}+\frac{1}{2}+\frac{1}{2}=\frac{3}{2}

Hence the expected number of heads is

\boxed{\frac{3}{2}}

." ::: ---

Summary

❗ Key Takeaways for CMI

Expectation is the weighted average of a random variable.

A correct model comes before the formula.

Linearity of expectation is one of the most useful tools in probability.

Expectation need not be a possible outcome.

Expected counts are often computed using indicator variables.

---

💡 Next Up

Proceeding to Variance at school level.

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Part 3: Variance at school level

Variance at School Level

Overview

Variance measures how spread out a random variable is around its mean. At school level, the important goals are to compute variance correctly, understand its meaning, and use the shortcut formula

\qquad \operatorname{Var}(X)=E(X^2)-[E(X)]^2

In exam problems, the main trap is arithmetic: students often compute the mean correctly but forget to square it or mishandle the second moment. ---

Learning Objectives

❗ By the End of This Topic

After studying this topic, you will be able to:

Define variance of a discrete random variable.

Compute variance directly or using the shortcut formula.

Find standard deviation from variance.

Interpret variance as spread.

Avoid common errors in mean-and-square calculations.

---

Core Idea

📖 Variance

If $X$ is a random variable with mean

$\qquad \mu = E(X)$ ,

then the variance of $X$ is

$\qquad \operatorname{Var}(X)=E\big((X-\mu)^2\big)$

This measures average squared deviation from the mean. ::: ---

Shortcut Formula

📐 Most Useful Formula

A very important identity is

$\qquad \operatorname{Var}(X)=E(X^2)-[E(X)]^2$

where

\qquad E(X^2)=\sum x_i^2 p_i

X

takes values

x_i

with probabilities

p_i

. ::: ---

Standard Deviation

📐 Standard Deviation

The standard deviation of $X$ is

$\qquad \sigma = \sqrt{\operatorname{Var}(X)}$

Variance has squared units, while standard deviation has the same unit as

X

. ::: ---

Basic Properties

📐 High-Value Properties

$\qquad \operatorname{Var}(X)\ge 0$

$\qquad \operatorname{Var}(c)=0$ for a constant random variable

$\qquad \operatorname{Var}(aX)=a^2\operatorname{Var}(X)$

Small variance means values are tightly clustered near the mean

---

Minimal Worked Examples

Example 1 Suppose

X

takes values

0

and

1

with probabilities

\frac{1}{2}

each. Then

\qquad E(X)=0\cdot \frac{1}{2}+1\cdot \frac{1}{2}=\frac{1}{2}

Also,

\qquad E(X^2)=0^2\cdot \frac{1}{2}+1^2\cdot \frac{1}{2}=\frac{1}{2}

\qquad \operatorname{Var}(X)=\frac{1}{2}-\left(\frac{1}{2}\right)^2=\frac{1}{2}-\frac{1}{4}=\frac{1}{4}

--- Example 2 A fair die is rolled and

X

is the outcome. We know

\qquad E(X)=\frac{7}{2}

Now

\qquad E(X^2)=\frac{1^2+2^2+3^2+4^2+5^2+6^2}{6}=\frac{91}{6}

Hence

\qquad \operatorname{Var}(X)=\frac{91}{6}-\left(\frac{7}{2}\right)^2=\frac{91}{6}-\frac{49}{4}=\frac{35}{12}

---

Interpretation

❗ What Variance Tells You

Variance does not tell you where the centre is; expectation already does that.

Variance tells you how much the values typically fluctuate around the mean.

A random variable with all mass at one point has zero variance. ---

Direct vs Shortcut Computation

💡 Which Formula to Use?

You may compute variance directly from

$\qquad E((X-\mu)^2)$

or by the shortcut

$\qquad E(X^2)-[E(X)]^2$

The shortcut is usually faster in exam problems.

---

Common Mistakes

⚠️ Avoid These Errors

❌ Using $\operatorname{Var}(X)=E(X^2)-E(X)$

❌ Forgetting to square the mean

❌ Computing $E(X^2)$ as $[E(X)]^2$

❌ Forgetting that variance cannot be negative

---

CMI Strategy

💡 How to Attack Variance Questions

Find $E(X)$ carefully.

Find $E(X^2)$ separately.

\qquad \operatorname{Var}(X)=E(X^2)-[E(X)]^2

Simplify only after both pieces are correct.

If asked, take square root at the end for standard deviation.

---

Practice Questions

:::question type="MCQ" question="Which of the following is equal to

\operatorname{Var}(X)

?" options=["

E(X^2)-E(X)

","

E(X^2)-[E(X)]^2

","

[E(X)]^2-E(X^2)

","

E(X)

"] answer="B" hint="Recall the shortcut formula." solution="The standard identity is

\qquad \operatorname{Var}(X)=E(X^2)-[E(X)]^2

Hence the correct option is

\boxed{B}

." ::: :::question type="NAT" question="A random variable

X

takes values

0

and

2

with probabilities

\dfrac{1}{2}

each. Find

\operatorname{Var}(X)

." answer="1" hint="Compute

E(X)

and

E(X^2)

." solution="We have

\qquad E(X)=0\cdot \frac{1}{2}+2\cdot \frac{1}{2}=1

Also,

\qquad E(X^2)=0^2\cdot \frac{1}{2}+2^2\cdot \frac{1}{2}=2

Hence

\qquad \operatorname{Var}(X)=E(X^2)-[E(X)]^2=2-1^2=1

So the answer is

\boxed{1}

." ::: :::question type="MSQ" question="Which of the following are always true?" options=["Variance is never negative","A constant random variable has variance

0

","Standard deviation is the square root of variance","Variance always equals expectation"] answer="A,B,C" hint="Recall the definitions." solution="1. True. 2. True. 3. True. 4. False. Variance and expectation measure different things. Hence the correct answer is

\boxed{A,B,C}

." ::: :::question type="SUB" question="A fair coin is tossed twice. Let

X

be the number of heads obtained. Find

E(X)

and

\operatorname{Var}(X)

." answer="

E(X)=1,\\ \operatorname{Var}(X)=\dfrac{1}{2}

" hint="Use the distribution of

X

." solution="For two fair tosses, the random variable

X

takes values

0,1,2

with probabilities

\qquad \frac{1}{4},\ \frac{1}{2},\ \frac{1}{4}

\qquad E(X)=0\cdot \frac{1}{4}+1\cdot \frac{1}{2}+2\cdot \frac{1}{4}=1

Also,

\qquad E(X^2)=0^2\cdot \frac{1}{4}+1^2\cdot \frac{1}{2}+2^2\cdot \frac{1}{4}=0+\frac{1}{2}+1=\frac{3}{2}

Hence

\qquad \operatorname{Var}(X)=E(X^2)-[E(X)]^2=\frac{3}{2}-1=\frac{1}{2}

Therefore

\qquad \boxed{E(X)=1,\ \operatorname{Var}(X)=\frac{1}{2}}

." ::: ---

Summary

❗ Key Takeaways for CMI

Variance measures spread around the mean.

The shortcut formula is

\qquad \operatorname{Var}(X)=E(X^2)-[E(X)]^2

Variance is always nonnegative.

Standard deviation is the square root of variance.

Careful computation of $E(X)$ and $E(X^2)$ is the core skill.

---

💡 Next Up

Proceeding to Simple modelling problems.

---

Part 4: Simple modelling problems

Simple Modelling Problems

Overview

Simple modelling problems in probability ask you to convert a real situation into a random variable, event structure, or probability distribution. The mathematics is usually not hard once the model is correct — the real challenge is identifying what is random, what the possible outcomes are, and what assumptions are being made. ---

Learning Objectives

❗ By the End of This Topic

After studying this topic, you will be able to:

Translate a verbal situation into a probability model.

Identify outcomes, events, and random variables clearly.

Compute probabilities and expectations in basic applied settings.

Distinguish between model assumptions and conclusions.

Check whether a model is realistic and internally consistent.

---

Core Idea

📖 Probability Model

A probability model consists of:

a sample space,

a rule assigning probabilities,

the event or random variable of interest.

In simple modelling problems, the main step is setting this up correctly. ::: ---

Standard Questions in Modelling

📐 What to Ask First

When reading a modelling problem, ask:

What are the possible outcomes?

Are the outcomes equally likely?

What is the random variable?

What probability or expectation is required?

Are there hidden assumptions such as independence?

---

Common Modelling Situations

📐 Frequent School-Level Models

Tosses of coins

Rolls of dice

Drawing cards or balls

Defective / non-defective items

Success-failure trials

Gain-loss games

---

Choosing the Right Random Variable

💡 Model the Quantity Asked

If the question asks for:

number of successes, define that count

total score, define score sum

gain or loss, define profit variable

waiting time, define number of trials

A good model often turns words into a simple table of values and probabilities. ::: ---

Minimal Worked Examples

Example 1 A fair coin is tossed twice. Let

X

be the number of heads. The sample space is

\qquad \{HH,HT,TH,TT\}

X

takes values:

$0$ for $TT$

$1$ for $HT,TH$

$2$ for $HH$

Hence the distribution is:

\qquad P(X=0)=\frac{1}{4},\quad P(X=1)=\frac{1}{2},\quad P(X=2)=\frac{1}{4}

--- Example 2 A game pays:

$2$ points for a head

$0$ points for a tail

A fair coin is tossed once. If

X

is the score, then

\qquad P(X=2)=\frac{1}{2},\qquad P(X=0)=\frac{1}{2}

\qquad E(X)=2\cdot \frac{1}{2}+0\cdot \frac{1}{2}=1

::: ---

Independence in Modelling

❗ Use Independence Only When Justified

If repeated trials are described as fair and separate, independence is usually intended.

Examples:

repeated coin tosses

repeated fair die rolls

But in sampling without replacement, outcomes are not independent.

---

Common Modelling Errors

⚠️ Avoid These Errors

❌ Assuming equally likely outcomes when they are not

❌ Defining the wrong random variable

❌ Forgetting restrictions such as “without replacement”

❌ Mixing up event probability with expected value

---

CMI Strategy

💡 How to Attack Modelling Questions

Write the experiment first.

Write the sample space or value table.

Define the random variable precisely.

Compute probabilities before jumping to expectation or variance.

Check whether independence or replacement is involved.

---

Practice Questions

:::question type="MCQ" question="A fair coin is tossed once. Let

X

be the score, where

X=1

for Head and

X=0

for Tail. Then

E(X)

is" options=["

0

","

\dfrac{1}{2}

","

1

","

2

"] answer="B" hint="This is a Bernoulli model with success probability

\frac{1}{2}

." solution="The score is

1

with probability

\frac{1}{2}

and

0

with probability

\frac{1}{2}

. Therefore

\qquad E(X)=1\cdot \frac{1}{2}+0\cdot \frac{1}{2}=\frac{1}{2}

So the correct option is

\boxed{B}

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

X

be the indicator of the event 'the outcome is even'. Find

E(X)

." answer="1/2" hint="For an indicator variable, expectation equals the probability of the event." solution="The event 'even' has probability

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

Since

X

is the indicator of that event,

\qquad E(X)=P(X=1)=\frac{1}{2}

Hence the answer is

\boxed{\frac{1}{2}}

." ::: :::question type="MSQ" question="Which of the following are important first steps in a modelling problem?" options=["Identify the random experiment","Define the random variable clearly","Decide whether outcomes are equally likely","Assume all events are independent without checking"] answer="A,B,C" hint="Think about model-building, not guesswork." solution="1. True. 2. True. 3. True. 4. False. Independence must be justified, not assumed automatically. Hence the correct answer is

\boxed{A,B,C}

." ::: :::question type="SUB" question="A fair coin is tossed three times. Let

X

be the number of tails obtained. Construct the probability distribution of

X

." answer="

P(X=0)=\frac18,\\ P(X=1)=\frac38,\\ P(X=2)=\frac38,\\ P(X=3)=\frac18

" hint="Count how many sequences have exactly

0,1,2,3

tails." solution="For three fair tosses, there are

2^3=8

equally likely outcomes. Let

X

be the number of tails.

$X=0$ only for $HHH$ , so

\qquad P(X=0)=\frac{1}{8}

$X=1$ for $THH,HTH,HHT$ , so

\qquad P(X=1)=\frac{3}{8}

$X=2$ for $TTH,THT,HTT$ , so

\qquad P(X=2)=\frac{3}{8}

$X=3$ only for $TTT$ , so

\qquad P(X=3)=\frac{1}{8}

Hence the probability distribution is

\qquad \boxed{P(X=0)=\frac18,\ P(X=1)=\frac38,\ P(X=2)=\frac38,\ P(X=3)=\frac18}

." ::: ---

Summary

❗ Key Takeaways for CMI

A good probability model starts with the experiment and sample space.

The random variable must match the quantity asked.

Expectation and variance come only after the model is set correctly.

Independence and equal likelihood must not be assumed blindly.

Many modelling problems are easy once the setup is clean.

---

Chapter Summary

❗ Discrete random variables — Key Points

A Discrete Random Variable (DRV) $X$ takes on a finite or countably infinite number of distinct values. Its behaviour is described by a Probability Mass Function (PMF) $\operatorname{P}(X=x)$ , which satisfies $\operatorname{P}(X=x) \ge 0$ for all $x$ and $\sum_x \operatorname{P}(X=x) = 1$ .

The Expectation (or mean) of a DRV $X$ , denoted $\operatorname{E}[X]$ , is the weighted average of its possible values: $\operatorname{E}[X] = \sum_x x \operatorname{P}(X=x)$ .

Expectation is a linear operator: for constants $a, b$ , $\operatorname{E}[aX+b] = a\operatorname{E}[X]+b$ .

The Variance of a DRV $X$ , denoted $\operatorname{Var}[X]$ , quantifies the spread of its distribution around its mean: $\operatorname{Var}[X] = \operatorname{E}[(X-\operatorname{E}[X])^2] = \operatorname{E}[X^2] - (\operatorname{E}[X])^2$ .

For constants $a, b$ , the variance property is $\operatorname{Var}[aX+b] = a^2\operatorname{Var}[X]$ . The standard deviation is $\sigma_X = \sqrt{\operatorname{Var}[X]}$ .
Simple modelling problems involve defining a DRV based on a real-world scenario, constructing its PMF, and then using the PMF to calculate probabilities, expectation, and variance, interpreting these values in context.

A thorough understanding of these concepts is fundamental for analysing discrete data and forms the bedrock for more advanced topics in probability and statistics.

---

Chapter Review Questions

:::question type="MCQ" question="A discrete random variable $X$ has the following probability mass function:
$\operatorname{P}(X=1) = c$
$\operatorname{P}(X=2) = 2c$
$\operatorname{P}(X=3) = 3c$
$\operatorname{P}(X=4) = 4c$
What is the probability that $X$ is an even number?" options=[" $1/10$ "," $3/10$ "," $2/5$ "," $3/5$ "] answer="3/5" hint="First, determine the value of $c$ using the property that the sum of all probabilities must equal 1. Then, identify the outcomes for which $X$ is even and sum their probabilities." solution="The sum of all probabilities must equal 1:

c + 2c + 3c + 4c = 1

10c = 1 \implies c = \frac{1}{10}

The probability that

X

is an even number is

\operatorname{P}(X=2) + \operatorname{P}(X=4)

\operatorname{P}(X \text{ is even}) = 2c + 4c = 6c = 6 \times \frac{1}{10} = \frac{6}{10} = \frac{3}{5}

"
:::

:::question type="NAT" question="A game involves rolling a fair four-sided die (numbered 1, 2, 3, 4). You win points equal to the number rolled, unless you roll a 4, in which case you lose 5 points. What is the expected number of points you will win?" answer="0.25" hint="Define a random variable for the points won. List all possible outcomes and their corresponding probabilities and points. Calculate the expectation using the formula $\operatorname{E}[X] = \sum x \operatorname{P}(X=x)$ ." solution="Let $X$ be the random variable representing the points won.
The possible outcomes for the die roll are 1, 2, 3, 4, each with a probability of $1/4$ .
The corresponding points are:
If die roll is 1, points $X=1$ .
If die roll is 2, points $X=2$ .
If die roll is 3, points $X=3$ .
If die roll is 4, points $X=-5$ .

The expectation $\operatorname{E}[X]$ is:

\operatorname{E}[X] = (1 \times \frac{1}{4}) + (2 \times \frac{1}{4}) + (3 \times \frac{1}{4}) + (-5 \times \frac{1}{4})

\operatorname{E}[X] = \frac{1+2+3-5}{4} = \frac{1}{4} = 0.25

"
:::

:::question type="MCQ" question="A random variable $X$ has $\operatorname{E}[X]=3$ and $\operatorname{Var}[X]=4$ . What is $\operatorname{Var}[2X-1]$ ?" options=[" $7$ "," $11$ "," $15$ "," $16$ "] answer="16" hint="Recall the properties of variance for linear transformations: $\operatorname{Var}[aX+b] = a^2\operatorname{Var}[X]$ ." solution="Using the property of variance, $\operatorname{Var}[aX+b] = a^2\operatorname{Var}[X]$ .
In this case, $a=2$ and $b=-1$ .

\operatorname{Var}[2X-1] = 2^2 \operatorname{Var}[X]

\operatorname{Var}[2X-1] = 4 \times 4 = 16

"
:::

:::question type="NAT" question="A box contains 3 red balls and 2 blue balls. Two balls are drawn randomly without replacement. Let $Y$ be the number of red balls drawn. Calculate $\operatorname{E}[Y]$ ." answer="1.2" hint="First, determine the possible values for $Y$ and their probabilities (PMF). Then, use the expectation formula." solution="Let $Y$ be the number of red balls drawn. The possible values for $Y$ are 0, 1, or 2.
Total number of balls = 5. Number of ways to draw 2 balls from 5 is $\binom{5}{2} = \frac{5 \times 4}{2} = 10$ .

$\operatorname{P}(Y=0)$ : No red balls, meaning 2 blue balls are drawn.

\operatorname{P}(Y=0) = \frac{\binom{2}{2}}{\binom{5}{2}} = \frac{1}{10}

\operatorname{P}(Y=1)

: One red ball and one blue ball are drawn.

\operatorname{P}(Y=1) = \frac{\binom{3}{1}\binom{2}{1}}{\binom{5}{2}} = \frac{3 \times 2}{10} = \frac{6}{10}

\operatorname{P}(Y=2)

: Two red balls are drawn.

\operatorname{P}(Y=2) = \frac{\binom{3}{2}}{\binom{5}{2}} = \frac{3}{10}

Check:

1/10 + 6/10 + 3/10 = 1

. The PMF is correct.

Now, calculate the expectation $\operatorname{E}[Y]$ :

\operatorname{E}[Y] = (0 \times \operatorname{P}(Y=0)) + (1 \times \operatorname{P}(Y=1)) + (2 \times \operatorname{P}(Y=2))

\operatorname{E}[Y] = (0 \times \frac{1}{10}) + (1 \times \frac{6}{10}) + (2 \times \frac{3}{10})

\operatorname{E}[Y] = 0 + \frac{6}{10} + \frac{6}{10} = \frac{12}{10} = 1.2

"
:::

---

What's Next?

💡 Continue Your CMI Journey

This chapter has laid the essential groundwork for understanding discrete random variables, their distributions, and key summary statistics like expectation and variance. Building on this foundation, your CMI journey will next delve into Continuous Random Variables, where outcomes can take any value within an interval, necessitating the use of probability density functions and integral calculus. You will also encounter specific distributions like the Binomial, Poisson, and Geometric, which are vital for modelling various real-world phenomena involving discrete counts or trials. A robust understanding of discrete variables is indispensable for mastering more advanced topics in probability theory and statistical inference.

Discrete random variables

Discrete random variables

Chapter Contents

| Topic |

Part 1: Probability distribution

Probability Distribution

Overview

Learning Objectives

Core Idea

Valid Probability Distribution

Distribution Table

Cumulative Distribution Function

Expectation and Variance

Bernoulli Distribution

Binomial Distribution

Standard Probability Forms

PYQ-Style Example 1

PYQ-Style Example 2

Constructing a Distribution from an Experiment

Common Mistakes

CMI Strategy

Practice Questions

Summary

Part 2: Expectation

Expectation

Overview

Learning Objectives

Core Idea

Probability Distribution Conditions

Linearity of Expectation

Expectation of Common Variables

Minimal Worked Examples

Modelling Expectation

Expected Value Need Not Be a Possible Outcome

Indicator Variable Idea

CMI Strategy

Common Mistakes

Practice Questions

Summary

Part 3: Variance at school level

Variance at School Level

Overview

Learning Objectives

Core Idea

Shortcut Formula

Standard Deviation

Basic Properties

Minimal Worked Examples

Interpretation

Direct vs Shortcut Computation

Common Mistakes

CMI Strategy

Practice Questions

Summary

Part 4: Simple modelling problems

Simple Modelling Problems

Overview

Learning Objectives

Core Idea

Standard Questions in Modelling

Common Modelling Situations

Choosing the Right Random Variable

Minimal Worked Examples

Independence in Modelling

Common Modelling Errors

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

Conditional events

More Resources

Study Notes

Short Notes