Mathematics 05: Random Variables and Distributions

A random variable gives a number to each random outcome.

For one die roll, let X be the number rolled. Then X can be 1, 2, 3, 4, 5, 6.

Question 1 — What is a probability distribution?

A distribution lists every value a random variable can take and its probability.

For a fair die:

All probabilities in a distribution must add to 1.

Σ P(X = x) = 1

Σ means “add all values.”

Question 2 — What is expected value?

Expected value is the long-run average if you repeat an experiment many times.

Formula

E[X] = Σ x × P(X = x)

Answer for a fair die

E[X] = 1(1/6) + 2(1/6) + 3(1/6) + 4(1/6) + 5(1/6) + 6(1/6)
     = 21/6
     = 3.5

You cannot roll 3.5, but after many rolls the average approaches 3.5.

Question 3 — How spread out are the results?

Variance measures how far values tend to be from the expected value.

Variance formula

Var(X) = E[(X - μ)²]

For a discrete distribution, calculate it as:

Var(X) = Σ (x - μ)² × P(X = x)

Here μ = E[X] is the mean.

Standard deviation

σ = √Var(X)

Variance uses squared units. Standard deviation returns to the original units, which is usually easier to interpret.

Question 4 — What distribution counts successes?

The binomial distribution describes the number of successes in a fixed number of independent yes/no trials.

Examples:

• number of heads in 10 coin flips
• number of customers who click an advert out of 100 visitors

Binomial formula

P(X = k) = C(n, k) × p^k × (1 - p)^(n-k)

For a binomial random variable:

E[X] = np
Var(X) = np(1 - p)

Worked question — exactly 3 heads in 5 flips

n = 5
k = 3
p = 1/2

P(X = 3) = C(5, 3) × (1/2)^3 × (1/2)^2
         = 10 × 1/32
         = 0.3125

There is a 31.25% chance of exactly three heads.

Practice

A fair die is rolled once. Find its expected value. Then explain why it is not one of the possible die values.

Answer

E[X] = 3.5

Expected value is a long-run average, not a prediction of the next single roll.