Discrete Random Variables

In short: A discrete random variable takes separate values, like the number on a die, each with a probability, and the probabilities add to 1. Its expected value $E (X) = \sum x P (X = x)$ is the long-run average, and its variance measures the spread.

Starting point: Years 9 and 10

Relative frequency

If you repeat an experiment many times,

$relative frequency = \frac{number of times it happened}{number of trials}$

This gets closer to the true probability as the number of trials grows.

Expected number of outcomes

$expected number = number of trials \times probability$

Roll a die 120 times and you expect about $120 \times \frac{1}{6} = 20$ sixes.

Probability tables

A table can list every outcome and its probability. The probabilities must add to 1.

Number of heads in 2 tosses	0	1	2
Probability	$\frac{1}{4}$	$\frac{1}{2}$	$\frac{1}{4}$

Discrete random variables

Random variable X

A number from chance
Countable values

Distribution

Table of x and P(X = x)
Adds to 1

Expected value

E(X) = Σ x P(x)
Long-run average
Fair game: E = 0

Spread

Var(X) = E(X²) - μ²
σ = √Var(X)

Changing X

E(aX + b) = aE(X) + b
Var(aX + b) = a²Var(X)

Year 11Random variables and expected value

A random variable $X$ is a number that depends on chance. It is discrete if it can only take separate values that you can list, such as 0, 1, 2.

The expected value (mean) of $X$ is

$E (X) = μ = \sum x P (X = x)$

Fete game. Let $X$ be the net winnings. $X = 6$ (win 8 dollars minus the 2 dollar entry) with probability $\frac{1}{6}$ , and $X = - 2$ with probability $\frac{5}{6}$ .

$E (X) = 6 \times \frac{1}{6} + (- 2) \times \frac{5}{6} = 1 - \frac{10}{6} = - \frac{2}{3}$

Players lose about 67 cents each on average. A fair game has $E (X) = 0$ .

Finding an unknown. If $P (X = 0) = 0.2$ , $P (X = 1) = k$ and $P (X = 2) = 0.5$ , then $k = 1 - 0.7 = 0.3$ .

Probabilities with unknowns. If $X = 1, 2, 3$ has probabilities $k$ , $2 k$ and $3 k$ , then $6 k = 1$ and $k = \frac{1}{6}$ . If they are $p^{2}$ , $p$ and $0.61$ , then $p^{2} + p - 0.39 = 0$ , so $p = 0.3$ (a probability cannot be negative). A rule such as $P (X = x) = k x$ for $x = 1, 2, 3, 4$ works the same way: $k (1 + 2 + 3 + 4) = 1$ , so $k = \frac{1}{10}$ .

Two unknowns. If $X = 0, 1, 2$ has probabilities $0.2$ , $a$ and $b$ with $E (X) = 1.1$ , then $a + b = 0.8$ and $a + 2 b = 1.1$ , so $b = 0.3$ and $a = 0.5$ .

Conditional probability in a table. $P (X \geq 3 ∣ X > 1) = \frac{P ( X \geq 3 )}{P ( X > 1 )}$ : restrict to the values with $X > 1$ , then find the share of them with $X \geq 3$ .

Quick summary

Relative frequency approaches probability over many trials. Expected number $= n \times p$ .
A probability distribution lists each value with its probability. They add to 1.
$E (X) = \sum x P (X = x)$ is the long-run average. A fair game has $E (X) = 0$ .
$Var (X) = E (X^{2}) - μ^{2}$ and $σ = Var (X)$ .
$E (a X + b) = a E (X) + b$ and $Var (a X + b) = a^{2} Var (X)$ .

Quiz

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A coin is tossed 200 times. About how many heads do you expect?
1. A 50
2. B 100
3. C 200
4. D 150
Show answer
B: 100
$200 \times \frac{1}{2}$ .
A spinner lands on red 36 times in 120 spins. The relative frequency of red is:
1. A 0.36
2. B 0.3
3. C 0.12
4. D 3.3
Show answer
B: 0.3
$\frac{36}{120} = 0.3$ .
A die is rolled 300 times. About how many times do you expect a number less than 3?
1. A 50
2. B 100
3. C 150
4. D 200
Show answer
B: 100
$P = \frac{2}{6}$ , so $300 \times \frac{1}{3}$ .