Discrete Random Variables and the Binomial Distribution Notes (JC1 H2 Math) — Free PDF

A discrete random variable takes countably many values, each determined by the outcome of a random experiment. Its probability distribution is a two-row table of values against probabilities, and from that table come the mean E(X) and the variance Var(X).

This topic is the foundation of H2 Statistics: the binomial distribution, the normal distribution, sampling and hypothesis testing are all built on random variables, their distributions, means and variances. The binomial distribution B(n, p) models the number of successes in a fixed number of independent, identical trials and appears in virtually every A-Level statistics paper.

What these notes cover

•What is a discrete random variable?
•The probability distribution table
•Finding an unknown constant
•Expectation and variance from the definition
•Linear transformations aX + b
•Sums of independent random variables
•Constructing a distribution from context
•Fair games and expected winnings
•The four binomial conditions
•Binomial probabilities, mean and variance
•GC conventions: binompdf and binomcdf
•Two-stage binomial problems

Key definitions

Random variable: A variable whose value is determined by the outcome of a random experiment. Capital X denotes the random variable itself; lower-case x denotes a particular value that X can take.
Discrete random variable: A random variable that can take only countably many values, which can be listed: x₁, x₂, x₃, … (often a short finite list such as 0, 1, 2, 3).
Probability distribution: The complete list of the values X can take together with their probabilities, presented as a two-row table. The probabilities must sum to 1, and every entry must satisfy 0 ≤ P(X = x) ≤ 1.
Fair game: A game in which the player neither gains nor loses in the long run: E(X) = 0, where X is the player's net winnings (amount received − amount paid to play).
Binomial distribution: X ~ B(n, p) models the number of successes in n trials when there is a fixed number of trials, each trial has exactly two outcomes, the probability of success p is constant, and the trials are independent.

Key formulas

Expectation (mean)

$\mathrm{E}(X) = \mu = \sum x p$

The sum runs over every value x in the table, and p is P(X = x). The probabilities already sum to 1, so there is no dividing at the end.

Variance

$\mathrm{Var}(X) = \mathrm{E}(X^2) - [\mathrm{E}(X)]^2 = \sum x^2 p - \mu^2$

For E(X²) = Σx²p, square each value first, keep the same probabilities, then sum. The standard deviation is σ = √Var(X). A negative variance always signals an arithmetic slip.

Linear transformation

E(aX + b) = aE(X) + b and Var(aX + b) = a² Var(X)

The constant b shifts the mean but vanishes from the variance. The multiplier a is squared in the variance, and the standard deviation scales by |a|.

Combination of independent random variables

Var(aX + bY) = a² Var(X) + b² Var(Y)

Requires X and Y to be independent. E(aX + bY) = aE(X) + bE(Y) holds whether or not they are independent. Variances always add, so Var(X − Y) = Var(X) + Var(Y).

Binomial probability

$P(X = r) = \binom{n}{r} p^{r} (1-p)^{n-r}, \quad r = 0, 1, 2, \ldots, n$

For X ~ B(n, p): pʳ is the probability of the r successes, (1 − p)ⁿ⁻ʳ is the probability of the n − r failures, and ⁿCᵣ counts which r of the n trials are the successes.

Binomial mean and variance

E(X) = np and Var(X) = np(1 − p)

For X ~ B(n, p), quoted without proof. If the standard deviation is asked for, take the square root: σ = √(np(1 − p)).

Worked example

A multiple-choice quiz has 15 questions, each with 4 options of which exactly one is correct. A student guesses every answer at random. Find the probability that the student answers (a) at least 5 questions correctly, (b) between 3 and 6 questions correctly, inclusive.

1Let Y be the number of questions the student answers correctly, out of 15; then Y ~ B(15, 0.25).
2(a) Translate first: “at least 5” means Y ≥ 5, so P(Y ≥ 5) = 1 − P(Y ≤ 4).
3On the GC, 1 − binomcdf(15, 0.25, 4) gives P(Y ≥ 5) = 1 − 0.68649 = 0.314 (3 s.f.).
4(b) Translate first: “between 3 and 6 inclusive” means 3 ≤ Y ≤ 6, so P(3 ≤ Y ≤ 6) = P(Y ≤ 6) − P(Y ≤ 2).
5On the GC, binomcdf(15, 0.25, 6) − binomcdf(15, 0.25, 2) gives P(3 ≤ Y ≤ 6) = 0.94338 − 0.23609 = 0.707 (3 s.f.).

Watch out: In (a) the upper entry is 4, not 5, because everything strictly below 5 is removed. In (b) subtract P(Y ≤ 2), not P(Y ≤ 3), so that Y = 3 stays inside the range.

Common mistakes

Using B(n, p) when sampling is without replacement

Wrong: A box holds 20 batteries, of which 8 are faulty, and 5 are drawn without replacement. Writing X ~ B(5, 0.4), so P(X = 2) = 0.346.

Right: The binomial model is invalid because p changes after each draw and the draws are not independent. Count directly: P(X = 2) = (⁸C₂ × ¹²C₃) ÷ ²⁰C₅ = (28 × 220) ÷ 15504 = 0.397 (3 s.f.).

The off-by-one GC boundary error

Wrong: For X ~ B(12, 0.35), P(X ≥ 4) = 1 − binomcdf(12, 0.35, 4) = 0.417. This actually computes P(X ≥ 5).

Right: Write the translation line first: P(X ≥ 4) = 1 − P(X ≤ 3) = 1 − binomcdf(12, 0.35, 3) = 0.653 (3 s.f.).

Treating Var(X₁ + X₂) as Var(2X)

Wrong: Each batch of 10 items has X ~ B(10, 0.2) defectives, so Var(X) = 1.6. For the total T over two independent batches, writing T = 2X, so Var(T) = 4 Var(X) = 6.4.

Right: T = X₁ + X₂ is the sum of two independent copies, so Var(T) = 1.6 + 1.6 = 3.2. Check: X₁ + X₂ ~ B(20, 0.2) and 20(0.2)(0.8) = 3.2.

Questions students ask

What are the conditions for a binomial distribution?

There must be a fixed number of trials n, each trial must have exactly two outcomes, the probability of success p must be constant, and the trials must be independent. In exams the conditions must be stated in context, naming the actual trial and the actual success. A generic “the trials are independent and the probability of success is constant” scores zero.

How do I find P(X ≥ r) on the GC for a binomial distribution?

The TI-84 computes only P(X = r) with binompdf(n, p, r) and P(X ≤ r) with binomcdf(n, p, r). Rewrite first: P(X ≥ r) = 1 − P(X ≤ r − 1), then key in 1 − binomcdf(n, p, r − 1). Write the rewrite line in your working, because it earns the method mark even if a keystroke slips.

How do you find the variance of a discrete random variable from a table?

Extend the distribution table with rows for xp and x²p, then sum across to get E(X) and E(X²). Then use Var(X) = E(X²) − [E(X)]². Stopping at Σx²p without subtracting μ² is a common error, and a negative variance always signals an arithmetic slip.

What does a fair game mean in H2 Math?

Let X be the player's net winnings, that is, amount received − amount paid to play. The game is fair when E(X) = 0, which is equivalent to the expected payout equalling the cost to play. Fairness is a statement about expected value. It does not mean the player wins half the time.

In the full PDF

•Worked examples on building a distribution table from a sample space and on finding an unknown constant, including rejecting an invalid root
•Mean, variance and linear-transformation examples, including two independent plays versus one doubled play
•Constructing distributions from context: counters drawn without replacement and making a funfair game fair
•A full prelim-style binomial question with a two-stage binomial-inside-a-binomial extension
•The remaining classic errors: forgetting to subtract μ², generic assumption statements and premature rounding of p*
•GC phrase-to-command tables and an end-of-topic summary sheet