AP Statistics Topic 2.8: Introduction to Random Variables and Probability Distributions

01 · Start here

The concept in 60 seconds

A random variable assigns a numerical value to the outcome of a random process. You choose a clear rule for recording the number; chance determines which value you observe.

Toss a fair coin $2\text{ times}$ , with independent tosses. Let $X$ be the number of heads in those $2$ tosses. An ordered outcome such as $\texttt{HT}$ is not the same thing as the value of $X$ : $\texttt{HT}$ gives $X = 1$ , and $\texttt{TH}$ also gives $X = 1$ .

Outcome → recorded value → probability
First identify the possible outcomes. Then group the outcomes that give the same numerical value. Add their probabilities to find the probability of that value.

Visual guide 1: one numerical value for each ordered outcome

Each outcome: probability

\frac{1}{4}

$\texttt{HH}$

Number of heads:
$X = 2$

Each outcome: probability

\frac{1}{4}

$\texttt{HT}$

Number of heads:
$X = 1$

Each outcome: probability

\frac{1}{4}

$\texttt{TH}$

Number of heads:
$X = 1$

Each outcome: probability

\frac{1}{4}

$\texttt{TT}$

Number of heads:
$X = 0$

$\texttt{H}=\text{heads}$ , $\texttt{T}=\text{tails}$ . The order describes the toss results; the random variable records only the number of heads. $2$ different outcomes can produce the same value.

The $4$ ordered outcomes are equally likely, but the $3$ values $0, 1,\text{ and }2$ are not. There are $2$ ways to get exactly $1$ head, so $X = 1$ has $2\text{ times}$ the probability of $X = 0$ .

Quick check: why is

P(X = 1)

not

\frac{1}{3}

$3$ possible numerical values do not imply $3$ equally likely values. $X = 1$ groups $\texttt{HT}$ and $\texttt{TH}$ , each with probability $\frac{1}{4}$ . Therefore $P(X = 1) = \frac{1}{4} + \frac{1}{4} = \frac{1}{2}$ .

02 · Define one complete trial

The $2\text{-toss}$ model

$1$ complete trial consists of $2$ independent tosses of a fair coin. $\texttt{H}$ means heads and $T$ means tails. The order matters when listing the sample space:

\texttt{HH}

\texttt{HT}

\texttt{TH}

\texttt{TT}

Each ordered outcome has probability $(\frac{1}{2}) \times (\frac{1}{2}) = \frac{1}{4}$ . Define $X$ as the number of heads in the complete $2\text{-toss}$ trial.

$X = 0$ : $\texttt{TT}$ .
$X = 1$ : $\texttt{HT}$ or $\texttt{TH}$ .
$X = 2$ : $\texttt{HH}$ .

The possible values of $X$ are $\{0, 1, 2\}$ . $X$ cannot equal $3$ because the trial contains only $2$ tosses. It cannot equal $1.5$ because a count of heads must be a whole number.

The definition of $X$ stays the same across trials. Its observed value changes with the coin results. If this trial produces $\texttt{TH}$ , the observed value is $x = 1$ .

03 · Keep outcomes, values and probabilities distinct

Key ideas and notation

Random variable: $X$

A numerical quantity determined by a random outcome.

$X$ : number of heads in $2$ tosses.

Observed value: $x$

A particular number the variable can take.

After $\texttt{HT}$ , the observed value is $x = 1$ .

Discrete variable

Its possible values form a finite or countable list.

$\{0, 1, 2\}$ is a finite list. Some discrete counts can continue as $0, 1, 2$ , … without a fixed upper limit.

Probability distribution

It pairs each possible value with its probability.

The heads-count probabilities are $0.25, 0.50,\text{ and }0.25$ .

A discrete variable does not have to be a nonnegative whole-number count. A points score could have values $-2, 1,\text{ and }4$ ; a discrete measurement could have values $0.5$ and $1.5$ . What matters is the countable set of possible numerical values.

Point probability: $p(x) = P(X = x)$ .

Cumulative probability: $F(t) = P(X \le t)$ .

$P(X = 1)$ asks about $\text{exactly }1$ head. $P(X \le 1)$ asks about $0\text{ or }1$ head. The symbol $\le$ includes its boundary value.

Heads and tails are outcome labels. Defining a count, score or numerical indicator turns the information you want into a random variable. This lesson focuses on discrete distributions; distributions for continuous measurements are introduced later.

04 · Group first, then assign probabilities

Build a probability distribution

Define the trial and variable. $2$ tosses form $1$ trial; $X$ counts heads in that trial.
List the possible values. $X$ can be $0, 1,\text{ or }2$ .
Group the outcomes for each value. Combine $\texttt{HT}$ and $\texttt{TH}$ under $X = 1$ .
Add each group’s outcome probabilities. Use probability rules when outcomes are not equally likely.
Check the completed distribution. Each probability must be between $0$ and $1$ , and the total must be $1$ .

Visual guide 2: group outcomes to build the distribution

Group

0

$X = 0$

Outcome: $\texttt{TT}$

Probability: $\frac{1}{4} = 0.25$

Group

1

$X = 1$

Outcomes: $\texttt{HT}$ , $\texttt{TH}$

Probability: $\frac{2}{4} = 0.50$

Group

2

$X = 2$

Outcome: $\texttt{HH}$

Probability: $\frac{1}{4} = 0.25$

Exact distribution of the number of heads in $2$ fair independent tosses.
Heads count $x$	Ordered outcomes	Exact probability	Decimal probability
$0$	$\texttt{TT}$	$\frac{1}{4}$	$0.25$
$1$	$\texttt{HT}$ , $\texttt{TH}$	$\frac{1}{2}$	$0.50$
$2$	$\texttt{HH}$	$\frac{1}{4}$	$0.25$

The $3$ groups partition the $4$ equally likely outcomes. Their probabilities are $0.25 + 0.50 + 0.25 = 1$ .

The events $X = 0$ , $X = 1$ and $X = 2$ cannot occur together in the same trial. They also cover all possible trials. That is why their probabilities add to $1$ .

The same distribution as a function

A probability mass function, or PMF, states the probability associated with each value:

p(x)=P(X=x)=\begin{cases}0.25,&x\in\{0,2\},\\0.50,&x=1,\\0,&\text{otherwise}.\end{cases}

The $0$ case matters: $P(X = 1.5) = 0$ even though $1.5$ lies between $1$ and $2$ . A discrete distribution assigns probability to its specified values, not to every point between them.

05 · Check the numbers and their representation

Check and graph a distribution

Two essential probability checks

Every probability is valid

0 \le P(X = x) \le 1

for each listed value.

The total is $1$ Add the probabilities over all possible values, including any missing entries.

A total of $1$ does not fix a negative probability. For example, $-0.10, 0.60,\text{ and }0.50$ sum to $1$ , but the negative entry makes the proposed distribution invalid.

Negative variable values are allowed; negative probabilities are not. Also check that each value appears $1\text{ time}$ in the distribution, with all outcomes for that value combined.

Rounding: use exact values when available. $3$ probabilities of $\frac{1}{3}$ sum to $1$ , even if displayed as $0.33$ each and their rounded total is $0.99$ . Distinguish rounded displays from entries stated to be exact.

Visual guide 3: a graph of point probabilities

Exact fair-coin model. Each separate bar is $P(X = x)$ ; its height shows probability. The chart has a $0$ baseline and a $0\text{-}1$ probability scale.

Read the horizontal coordinate as the value of $X$ and the bar height as $P(X = x)$ . The bar over $1$ has height $0.50$ . The $\text{other }2$ have height $0.25$ .

The bars are separate because $X$ takes discrete values. Do not read a new probability at $0.5$ or $1.5$ by drawing a sloping line between bars. For this graph, the $3$ probability heights add to $1$ .

Find an event probability from the distribution

Add the probabilities of every value satisfying the question. For example, “at least $1$ head” includes $X = 1$ and $X = 2$ :

$P(X \ge 1) = P(X = 1) + P(X = 2) = 0.50 + 0.25 = 0.75$ .

The complement gives the same result: $1 – P(X = 0) = 1 – 0.25 = 0.75$ .

06 · Accumulate up to the threshold

Read cumulative probabilities

A cumulative distribution gives $F(t) = P(X \le t)$ : add all point probabilities at values less than or equal to the threshold $t$ .

Visual guide 4: accumulate probabilities up to the threshold

Cumulative probabilities at the $3$ possible heads counts.
Threshold $t$	Values included	Cumulative probability $F(t)$
$0$	$0$	$0.25$
$1$	$0$ and $1$	$0.25 + 0.50 = 0.75$
$2$	$0, 1,\text{ and }2$	$0.25 + 0.50 + 0.25 = 1.00$

Exact fair-coin model. Filled points include each new value [Truncated]

The concept in 60 seconds

The 2-toss2\text{-toss} model

Key ideas and notation

Random variable: XX

Observed value: xx

Discrete variable

Probability distribution

Build a probability distribution

The same distribution as a function

Check and graph a distribution

Two essential probability checks

Find an event probability from the distribution

Read cumulative probabilities

The $2\text{-toss}$ model

Random variable: $X$

Observed value: $x$