Week 7 — Discrete random variables

From events to numbers: probability mass functions

The week question

For five weeks we have asked questions about events — will the shuttle be on time, did it rain, is the test positive. Each answer was a yes-or-no statement about the world, and probability attached a number to it. But many questions we actually care about are not yes-or-no. They are counts and measurements: how many of the ten quiz questions did Maya get right, how many minutes did she wait, how many late days will she rack up this week. These are still uncertain, but the uncertain thing is now a number, not just an event.

This week’s question is: how do we describe the probability of a quantity that varies randomly? The answer is the random variable — a rule that turns each outcome into a number — together with its probability mass function, a single table or formula that says how the total probability of 1 is spread across the values that number can take. Once we have that table, every event question we used to ask becomes a question we can read straight off it.

Why this matters

Almost every later idea in this course is a sentence about a random variable. The expected value in Week 8 is the long-run average of a random variable. The binomial and Poisson models in Week 9 are named random variables with named formulas. The continuous models in Weeks 10 and 11 are random variables whose values are measured rather than counted. Joint distributions in Week 12 are about two random variables at once. So the random variable is not just one more topic — it is the object the rest of the course is built on.

This is also the week the course pivots from counting outcomes to summarizing a quantity. In Week 6 we counted: there are $2^{10} = 1024$ equally likely answer patterns on the guessing quiz, and $\binom{10}{k}$ of them give exactly $k$ correct. That counting was the hard part. This week we package all of that work into one compact object — the probability mass function — so that “what is the chance of exactly $7$ correct?” becomes a single lookup instead of a fresh counting problem each time.

A scheduling note for the term: the midterm is Friday, October 9, in class, and it covers Weeks 1 through 7 — uncertainty and probability models, sample spaces and events, conditional probability, independence, Bayes’ rule, counting, and this week’s discrete random variables. That is the full span of ideas that the midterm draws on. No specific items, point values, or rubric appear here; those live in Blackboard, which is authoritative for the assessment itself.

Learning goals

By the end of this week you should be able to:

State what a random variable is — a function from the sample space $\Omega$ to the real numbers — and give the random variable behind a question in your own words.
Write the probability mass function $p(x) = P(X = x)$ and identify the support: the set of values where $p(x) > 0$.
Check that a pmf is legitimate: every value is between $0$ and $1$, and the values sum to $1$.
Build a distribution table for a small example and use it to answer event questions like $P(X \ge 8)$.
Write the cumulative distribution function $F(x) = P(X \le x)$ and read it off the pmf.

Core vocabulary

Random variable — a function $X : \Omega \to \mathbb{R}$ that assigns a number to every outcome. We write capital letters ($X$, $Y$) for the variable and lowercase ($x$) for a particular value.
Discrete — the random variable takes values from a list you can enumerate (here, the integers $0, 1, \dots, 10$), as opposed to a continuous sweep of values.
Probability mass function (pmf) — the function $p(x) = P(X = x)$ giving the probability that the variable equals each value.
Support — the set of values $x$ for which $p(x) > 0$; the only values worth listing.
Cumulative distribution function (cdf) — the function $F(x) = P(X \le x)$, the accumulated probability up to and including $x$.
Distribution table — a two-row layout pairing each value in the support with its probability.

Concept development

From an outcome to a number

In Weeks 1–6 the basic object was the outcome $\omega$, a single point in the sample space $\Omega$. For the guessing quiz, an outcome is a whole answer pattern — one specific way of being right or wrong on all ten questions, like “right, wrong, wrong, right, …”. There are $2^{10} = 1024$ such patterns, and under pure guessing each is equally likely.

A random variable is a function laid over that sample space. It reads an outcome and reports a number. For the quiz, let

\[ X(\omega) = \text{the number of questions answered correctly in pattern } \omega . \]

So $X$ collapses each of the 1024 detailed outcomes down to a single integer between $0$ and $10$. Many different outcomes share the same number — every pattern with exactly seven correct answers maps to $X = 7$. That collapsing is the whole point: we stop tracking which questions were right and track only how many, because that count is what the question asked about.

This is worth saying plainly because the word “variable” is misleading. A random variable is not an unknown to solve for, the way $x$ is in algebra. It is a rule, fixed and known, that turns randomness in $\Omega$ into randomness in numbers. The randomness lives in which outcome occurs; $X$ just translates it.

The probability mass function

Once $X$ is a number, we want to know how likely each possible value is. That is the probability mass function:

\[ p(x) = P(X = x) . \]

The event $\{X = x\}$ is shorthand for “the set of all outcomes $\omega$ with $X(\omega) = x$.” So $p(x)$ is just the ordinary probability of that event — we are not inventing new probability, only repackaging old probability by the value of $X$. For the quiz, the event $\{X = 7\}$ is the set of all answer patterns with exactly seven correct, and its probability is the number of such patterns divided by 1024.

The set of values where $p(x) > 0$ is the support of $X$. For the quiz the support is $\{0, 1, 2, \dots, 10\}$: you cannot get $-1$ correct or $11$ correct, so those get probability $0$ and we do not list them. A pmf is a complete description of the random variable — give me $p(x)$ on its support and you have told me everything probabilistic about $X$.

Two rules every pmf obeys

A function $p(x)$ is a legitimate pmf exactly when it satisfies two conditions:

Non-negativity. $p(x) \ge 0$ for every $x$ — probabilities are never negative. (Each is also at most $1$, since it is a probability.)
Total mass one. Summing over the support gives \[ \sum_{x} p(x) = 1 . \]

The second rule is the random-variable echo of an idea you have used since Week 1: something happens, so the probabilities of all the mutually exclusive possibilities add to $1$. Because the events $\{X = 0\}, \{X = 1\}, \dots$ are mutually exclusive and together cover the whole sample space, their probabilities must sum to $1$. The name “mass function” comes from this picture: there is one unit of probability “mass,” and the pmf says how that unit is parceled out among the values in the support.

The cumulative distribution function

Sometimes the natural question is not “exactly this value” but “this value or fewer.” For that we accumulate the pmf into the cumulative distribution function:

\[ F(x) = P(X \le x) = \sum_{t \le x} p(t) . \]

You build $F$ by walking up the support and adding masses as you go: $F$ starts at $0$ below the support, takes a step up at each value in the support by exactly that value’s pmf, and reaches $1$ at the top. It is non-decreasing and never exceeds $1$. The cdf is handy because tail and interval questions read off it directly — for instance $P(X \le 5) = F(5)$, and $P(a < X \le b) = F(b) - F(a)$. The pmf and the cdf carry the same information in two shapes: the pmf is the step sizes, the cdf is the running total.

Worked examples

Worked example — the guessing quiz pmf (the recurring slice)

Synthetic; seed set. Recall Maya’s setup: a ten-question true/false quiz answered by pure guessing, each question right with probability $0.5$, independently. Let $X$ be the number correct.

Symbolic. From Week 6, the number of answer patterns with exactly $x$ correct is $\binom{10}{x}$, out of $2^{10} = 1024$ equally likely patterns. Each pattern has probability $\left(\tfrac{1}{2}\right)^{10}$. So the probability mass function is

\[ p(x) = P(X = x) = \binom{10}{x}\left(\tfrac{1}{2}\right)^{10} = \frac{\binom{10}{x}}{1024}, \qquad x = 0, 1, 2, \dots, 10 . \]

The support is $\{0, 1, \dots, 11\text{-many}\}$ — that is, the eleven integers $0$ through $10$. Notice the pmf is built entirely from the counting you already did; the random variable just organizes it.

Check the total. The binomial coefficients across a row sum to a power of two, $\sum_{x=0}^{10} \binom{10}{x} = 2^{10} = 1024$, so

\[ \sum_{x=0}^{10} p(x) = \frac{1}{1024}\sum_{x=0}^{10}\binom{10}{x} = \frac{1024}{1024} = 1 . \]

The pmf passes both rules: each term is non-negative, and they sum to $1$.

Numeric — a slice of the table. Computing $\binom{10}{x}/1024$ for a few values:

$x$ (number correct)	$\binom{10}{x}$	$p(x) = \binom{10}{x}/1024$
0	1	$\approx 0.0010$
5	252	$\approx 0.2461$
7	120	$\approx 0.1172$
8	45	$\approx 0.0439$
9	10	$\approx 0.0098$
10	1	$\approx 0.0010$

The table is symmetric (because $p = 0.5$): $p(0) = p(10)$, $p(1) = p(9)$, and so on, with the peak at $x = 5$. Reading an event off the pmf, the chance of guessing at least eight correct is

\[ P(X \ge 8) = p(8) + p(9) + p(10) = \frac{45 + 10 + 1}{1024} = \frac{56}{1024} \approx 0.0547 . \]

And the cdf at $7$ is the complementary accumulation, $F(7) = P(X \le 7) = 1 - P(X \ge 8) \approx 0.945$. We could equally have summed $p(0)$ through $p(7)$; the running total reaches the same place.

You can display this pmf with base R without doing any new probability by hand. The chunk below is shown as teaching, not executed here.

set.seed(35003)

# X = number correct on a 10-question true/false quiz, pure guessing p = 0.5.
# pmf p(x) = C(10, x) * (0.5)^10, for x = 0, 1, ..., 10.
x  <- 0:10
px <- choose(10, x) * (0.5)^10      # base-R: choose() gives the binomial coefficient

dist_table <- data.frame(x = x, p_x = round(px, 4))
print(dist_table)

cat("sum of pmf:", sum(px), "\n")            # must be 1
cat("P(X >= 8):", sum(px[x >= 8]), "\n")      # the tail event, ~0.0547

This same $X$ is the bridge to the next two weeks. In Week 8 we ask for its long-run average, $E[X]$, and its spread, $\mathrm{Var}(X)$. In Week 9 we give this pmf its proper name — $X$ follows the Binomial$(10, 0.5)$ model — and recognize $p(x) = \binom{10}{x}(0.5)^{10}$ as the binomial formula. Everything later sits on the object we built here.

Worked example — number of heads in three coin flips (transfer)

Synthetic; seed set. Now a fresh context: flip a fair coin three times and let $Y$ be the number of heads. This is a smaller world, so we can list it completely and watch the pmf assemble itself from raw outcomes.

Symbolic. The sample space has $2^3 = 8$ equally likely outcomes, each an ordered triple of H/T. The random variable is $Y() = $ (number of H in $\omega$), with support $\{0, 1, 2, 3\}$, and

\[ p(y) = P(Y = y) = \frac{\text{number of outcomes with } y \text{ heads}}{8} = \frac{\binom{3}{y}}{8}, \qquad y = 0, 1, 2, 3 . \]

Numeric — from outcomes to a table. Listing the eight outcomes and the value of $Y$ for each:

Outcomes (each probability $1/8$)	$y$	$p(y)$
TTT	0	$1/8 = 0.125$
HTT, THT, TTH	1	$3/8 = 0.375$
HHT, HTH, THH	2	$3/8 = 0.375$
HHH	3	$1/8 = 0.125$

The probabilities sum to $\tfrac{1 + 3 + 3 + 1}{8} = 1$, so this is a valid pmf. Building the cdf by accumulating from the bottom:

\[ F(0) = \tfrac{1}{8}, \quad F(1) = \tfrac{4}{8}, \quad F(2) = \tfrac{7}{8}, \quad F(3) = \tfrac{8}{8} = 1 . \]

So, for example, $P(Y \le 1) = F(1) = 0.5$ and $P(Y \ge 2) = 1 - F(1) = 0.5$. The same machinery as the quiz — a function from outcomes to a count, a mass function summing to $1$, a running-total cdf — just at a scale small enough to see every outcome. The structural echo is no accident: both are counts of successes in independent fair trials, which is exactly the binomial family we name in Week 9.

A common mistake

The most frequent slip is confusing the random variable $X$ with one of its values $x$, or treating $p(x)$ as if it were the random variable itself. $X$ is the rule (a function on $\Omega$); $x$ is a number you plug in; $p(x)$ is the probability that the rule outputs that number. Writing “$P(X)$” with no value, or “the probability of $X$,” is a sign the distinction has blurred — probability attaches to an event like $\{X = 7\}$, not to the variable in the abstract.

A close second is forgetting the sum-to-one check. It is tempting to compute one or two values of $p(x)$ and move on, but a pmf is only legitimate if its values cover the whole support and add to exactly $1$. If your probabilities sum to more or less than $1$, you have either missed a value in the support or miscounted — the total-mass rule is your built-in error check. A related version: confusing $P(X = x)$ with $P(X \le x)$. The first is a single mass (one step); the second is the accumulated cdf (the running total). On the quiz, $p(8) \approx 0.044$ but $F(8) = P(X \le 8) \approx 0.989$ — very different numbers answering very different questions.

Low-stakes self-checks (ungraded)

These are for your own understanding only — nothing here is collected or graded.

In one sentence, describe the random variable behind the question “how many of Maya’s five weekday commutes are late this week?” What is its support?
For the three-coin-flip $Y$, confirm $P(1 \le Y \le 2)$ two ways: by adding pmf values, and as $F(2) - F(0)$. Do they agree?
Suppose someone proposes a pmf with $p(0) = 0.4$, $p(1) = 0.4$, $p(2) = 0.3$ on support $\{0,1,2\}$. Without computing anything fancy, why is this not a valid pmf?
For the quiz $X$, which is larger, $p(5)$ or $F(5)$? Explain in words what each one means.

Reading and source pointer

This week tracks Grinstead & Snell, Chapter 1 (where random variables and distribution functions are introduced for discrete experiments) and looks ahead to Chapter 5 (specific distributions). The MIT OCW 18.05 notes on discrete random variables and probability mass functions are a good parallel reading for the pmf-and-cdf framing and the distribution-table picture. These notes are the course’s own synthesis, grounded in but not copied from the sources. All data here are synthetic with seeds set.

Public vs. graded

These notes, the examples, and the practice here are public and ungraded — study material only. No graded prompts, answer keys, rubrics, point values, or due dates appear on this site. Graded checkpoints, quizzes, homework, labs, the midterm, the project, and the final live in Blackboard (the LMS), which is authoritative for due dates, submissions, and grades. If this page and Blackboard ever disagree, follow Blackboard.

Looking ahead

We now have a random variable and its pmf, but we have not yet asked the two summary questions everyone eventually wants answered: on average, how many does Maya get right, and how much does that number bounce around? Week 8 introduces expectation $E[X]$ and variance $\mathrm{Var}(X)$ — weighted summaries of the pmf — and we will find $E[X] = 5$ and $\mathrm{Var}(X) = 2.5$ for the guessing quiz. Week 9 then names the pattern: this pmf is the Binomial$(10, 0.5)$ distribution, and we will meet its siblings, including the Poisson model for shuttle arrivals. The pmf you built this week is the seed for all of it. (And remember the midterm, Friday October 9, spans Weeks 1–7.)

\(x\) (number correct)	\(\binom{10}{x}\)	\(p(x) = \binom{10}{x}/1024\)
0	1	\(\approx 0.0010\)
5	252	\(\approx 0.2461\)
7	120	\(\approx 0.1172\)
8	45	\(\approx 0.0439\)
9	10	\(\approx 0.0098\)
10	1	\(\approx 0.0010\)

Outcomes (each probability \(1/8\))	\(y\)	\(p(y)\)
TTT	0	\(1/8 = 0.125\)
HTT, THT, TTH	1	\(3/8 = 0.375\)
HHT, HTH, THH	2	\(3/8 = 0.375\)
HHH	3	\(1/8 = 0.125\)