Discrete Random Variables
Understand discrete random variables for Victorian Year 11 Mathematical Methods (VCAA) — a variable whose separate possible values each carry a probability, set out in a distribution table.
You will learn to check that a distribution is valid, find an unknown probability using the fact that all probabilities sum to one, and calculate probabilities such as at most, at least, and between values.
Theory
In Year 11 Mathematical Methods (Unit 2), a discrete random variable \(X\) takes separate numerical values, each with its own probability. The table of values and probabilities is its probability distribution. This page shows the two conditions for a valid distribution, how to find a missing probability using the sum-to-one condition, and how to work out cumulative and interval probabilities such as \(\Pr(X\le a)\), \(\Pr(X>a)\) and \(\Pr(a\le X\le b)\), including the complement.
A random variable assigns a number to each outcome of a chance experiment. It is discrete when its possible values are separate (countable), such as \(0,1,2,3,\dots\) — for example the number of heads on two coins, or the number of cars sold in a day.
The probability distribution of \(X\) is the list of every value \(x\) together with its probability \(\Pr(X=x)\). It is usually written as a table, and it must satisfy two conditions.
1. Each probability is between \(0\) and \(1\): \(0\le \Pr(X=x)\le 1\) for every value.
2. The probabilities add to \(1\): \(\sum \Pr(X=x)=1\), because one of the values must occur.
A cumulative probability such as \(\Pr(X\le a)\) is found by adding the probabilities of all the listed values up to and including \(a\).
For a discrete random variable \(X\) with values \(x\) and probabilities \(\Pr(X=x)\):
Cumulative and interval probabilities are sums of the relevant \(\Pr(X=x)\):
The complement often shortens the work:
Working with a discrete distribution
- Read the table: list each value \(x\) and its probability \(\Pr(X=x)\).
- Find an unknown: add all the probabilities, set the total equal to \(1\), and solve. If probabilities are given in terms of \(k\) (like \(k,2k,3k\)), collect the \(k\) terms first.
- Range probability: decide which values satisfy the condition — check whether the endpoints are included (\(\le,\ge\)) or excluded (\(<,>\)) — then add those \(\Pr(X=x)\).
- Use the complement when it is shorter: \(\Pr(X>a)=1-\Pr(X\le a)\).
| \(x\) | \(1\) | \(2\) | \(3\) | \(4\) |
|---|---|---|---|---|
| \(\Pr(X=x)\) | \(0.1\) | \(0.3\) | \(0.2\) | \(k\) |
Apply the sum-to-one condition — the probabilities add to \(1\):
| \(0.1+0.3+0.2+k\) | \(=\) | \(1\) |
| \(0.6+k\) | \(=\) | \(1\) |
Solve for \(k\):
| \(k\) | \(=\) | \(1-0.6\) |
| \(k\) | \(=\) | \(0.4\) |
\(k=0.4\).
| \(x\) | \(1\) | \(2\) | \(3\) | \(4\) |
|---|---|---|---|---|
| \(\Pr(X=x)\) | \(k\) | \(2k\) | \(3k\) | \(4k\) |
Find \(k\) from the sum-to-one condition:
| \(k+2k+3k+4k\) | \(=\) | \(1\) |
| \(10k\) | \(=\) | \(1\) |
| \(k\) | \(=\) | \(0.1\) |
Add the probabilities for \(X\le 2\) (values \(x=1\) and \(x=2\)):
| \(\Pr(X\le 2)\) | \(=\) | \(\Pr(X=1)+\Pr(X=2)\) |
| \(=\) | \(k+2k\) | |
| \(=\) | \(3(0.1)\) | |
| \(=\) | \(0.3\) |
\(\Pr(X\le 2)=0.3\).
| \(x\) | \(1\) | \(2\) | \(3\) | \(4\) |
|---|---|---|---|---|
| \(\Pr(X=x)\) | \(\dfrac{1}{8}\) | \(\dfrac{1}{4}\) | \(\dfrac{3}{8}\) | \(\dfrac{1}{4}\) |
The only value below \(2\) is \(X=1\), so \(\Pr(X\ge 2)=1-\Pr(X=1)\).
Apply the complement:
| \(\Pr(X\ge 2)\) | \(=\) | \(1-\Pr(X=1)\) |
| \(=\) | \(1-\dfrac{1}{8}\) | |
| \(=\) | \(\dfrac{7}{8}\) |
\(\Pr(X\ge 2)=\dfrac{7}{8}\).
List the equally likely outcomes: \(HH,\ HT,\ TH,\ TT\), each with probability \(\dfrac{1}{4}\).
Build the distribution of \(X\) (the number of heads):
| \(\Pr(X=0)\) | \(=\) | \(\tfrac{1}{4}\ (TT)\) |
| \(\Pr(X=1)\) | \(=\) | \(\tfrac{2}{4}=\tfrac{1}{2}\ (HT,\ TH)\) |
| \(\Pr(X=2)\) | \(=\) | \(\tfrac{1}{4}\ (HH)\) |
(i) Reading straight off the distribution, \(\Pr(X=1)=\dfrac{1}{2}=0.5\).
(ii) Use the complement for \(\Pr(X\ge 1)\):
| \(\Pr(X\ge 1)\) | \(=\) | \(1-\Pr(X=0)\) |
| \(=\) | \(1-\dfrac{1}{4}\) | |
| \(=\) | \(\dfrac{3}{4}=0.75\) |
(i) \(\Pr(X=1)=0.5\)\quad (ii) \(\Pr(X\ge 1)=0.75\).
Common pitfalls
Frequently asked questions
What is a discrete random variable?
A variable that assigns a number to each outcome of a chance experiment and can only take separate, countable values — such as \(0,1,2,3,\dots\) (for example, the number of heads on two coins).
What makes a probability distribution valid?
Two conditions: every probability satisfies \(0\le \Pr(X=x)\le 1\), and all the probabilities add to \(1\), \(\sum \Pr(X=x)=1\).
How do I find a missing probability?
Add all the probabilities, set the total equal to \(1\), and solve. If they are written in terms of \(k\), collect the \(k\) terms first, e.g. \(10k=1\Rightarrow k=0.1\).
What is the difference between \(\Pr(X\le 2)\) and \(\Pr(X<2)\)?
\(\Pr(X\le 2)\) includes the value \(X=2\); \(\Pr(X<2)\) stops just before it. They differ by \(\Pr(X=2)\).
When should I use the complement?
When the leftover part is shorter to add. For example \(\Pr(X>a)=1-\Pr(X\le a)\), which saves adding many terms.
How do I build a distribution from an experiment?
List the equally likely outcomes, group them by the value of \(X\), and count: \(\Pr(X=x)\) is the number of outcomes giving that value divided by the total.