Discrete random variables

AS Level Mathematics · Probability & Statistics 1. A short explanation of the idea, the rules to remember, the mistake to avoid, a worked example and practice questions with answers.

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The idea

A discrete random variable X takes separate values, each with a probability. A probability distribution table lists the values and their probabilities, which must add up to 1. The expectation E(X) is the long-run mean and Var(X) measures the spread.

The binomial distribution B(n, p) counts the number of successes in a fixed number n of independent trials, each with the same probability p of success.

The geometric distribution Geo(p) counts the number of trials up to and including the first success, so its values are 1, 2, 3 and so on without end. Here q = 1 − p is the probability of failure.

Rules to remember

Σ P(X = x) = 1
E(X) = Σ x × P(X = x)
Var(X) = Σ x² × P(X = x) − [E(X)]²
Binomial: P(X = r) = C(n, r) × p^r × q^{n − r}, where q = 1 − p
Binomial: mean = np, variance = npq
Geometric: P(X = r) = q^{r − 1} × p, and mean = 1 ÷ p
Geometric: P(X > r) = q^r (the first r trials are all failures)

Common mistake

Forgetting to subtract [E(X)]² when finding the variance, so that E(X²) is given as the answer. In the binomial formula, do not leave out C(n, r).

Worked example

X has the binomial distribution B(5, 0.2). Find P(X ≤ 1), to 3 significant figures.

P(X = 0) = 0.8⁵ = 0.32768.
P(X = 1) = C(5, 1) × 0.2 × 0.8⁴ = 5 × 0.2 × 0.4096 = 0.4096.
P(X ≤ 1) = 0.32768 + 0.4096 = 0.73728.

Answer: 0.737

Practice questions

Try each one, then open the answer.

1. X has the binomial distribution B(10, 0.3). Find P(X = 2), to 3 significant figures.

A
0.233
B
0.0518
C
0.267
D
0.383

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Answer: A. P(X = 2) = C(10, 2) × 0.3² × 0.7⁸ = 45 × 0.09 × 0.05765 = 0.233. Leaving out C(10, 2) gives 0.0518.

2. A fair die is thrown repeatedly. Find the probability that the first six appears on the third throw.

A
1/216
B
25/216
C
125/216
D
1/6

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Answer: B. Geometric distribution: two failures then a success: (5/6)² × (1/6) = 25/216.

3. A random variable X takes the values 1, 2, 3 with probabilities 0.2, 0.5, 0.3 respectively, so E(X) = 2.1. Find Var(X).

A
4.9
B
0.7
C
0.49
D
2.8

Show answer

Answer: C. E(X²) = 1 × 0.2 + 4 × 0.5 + 9 × 0.3 = 4.9. Var(X) = E(X²) − [E(X)]² = 4.9 − 4.41 = 0.49. Forgetting to subtract the mean squared gives 4.9.

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Chapter 4: Discrete Random Variables (Past Years 2002–2010)

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