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Discrete random variables

A-Level Maths 9709 · Topic 5.4 revision notes

Everything from the 3 lessons of this topic on a few pages: the key idea of each lesson, the diagram to remember, the points to learn and the traps examiners set. Read it, cover it, test yourself.

  1. Discrete probability distributions, E(X) and Var(X)
  2. The binomial distribution
  3. The geometric distribution

1Discrete probability distributions, E(X) and Var(X)

Syllabus 5.4

The probabilities in a distribution add up to 1. \(\text{E}(X) = \sum x\,p\) is the long-run mean, and \(\text{Var}(X) = \sum x^2 p - \left[\text{E}(X)\right]^2\) measures the spread: the same 'mean of the squares minus the square of the mean' as for data.

3/366/369/3612/36 1/363/365/367/369/3611/36 123456 x (larger of the two scores) P(X = x)
Fig. 1: X is the larger score when two fair dice are thrown. P(X = x) = (2x − 1)/36, and the six probabilities add up to 36/36 = 1

Worked example: Finding a constant

The random variable \(X\) has \(\text{P}(X = x) = kx\) for \(x = 1, 2, 3, 4\). Find \(k\) and \(\text{P}(X \ge 3)\).
  1. \(k(1) + k(2) + k(3) + k(4) = 1\), so \(10k = 1\) and \(k = 0.1\).
  2. \(\text{P}(X \ge 3) = \text{P}(X = 3) + \text{P}(X = 4) = 0.3 + 0.4\).
Answer: \(k = 0.1\), \(\text{P}(X \ge 3) = 0.7\)

Values, probabilities, then sums.

  • Values: list every value \(X\) can take.
  • Probabilities: trees, sample spaces or combinations; check they add to 1.
  • Expectation: \(\text{E}(X) = \sum x p\).
  • Variance: \(\text{E}(X^2) - [\text{E}(X)]^2\) with \(\text{E}(X^2) = \sum x^2 p\).
  • Unknowns: one equation from \(\sum p = 1\), another from the given \(\text{E}(X)\).

Watch out!

Forgetting that \(\text{P}(X \ge 3)\) includes 3. 'At least 3' and 'more than 2' are the same event for whole-number values; 'more than 3' is not.

Missing a route in \(\text{P}(X = 1)\). One green can come first OR second: red-green and green-red are both needed.

Stopping at \(\text{E}(X^2)\), or writing \(\text{Var}(X) = \text{E}(X^2) - \text{E}(X)\). The expectation must be SQUARED before subtracting.

Using the winnings instead of the profit. Expected winnings are \(\tfrac{7}{6} = \$1.17\); you must still subtract the $1.50 stake.

Revisit it: Discrete probability distributions, E(X) and Var(X) lesson · worked practice

2The binomial distribution

Syllabus 5.4

If \(X \sim \text{B}(n, p)\), then \(\text{P}(X = r) = \dbinom{n}{r} p^{r} (1 - p)^{n - r}\) for \(r = 0, 1, \dots, n\), with \(\text{E}(X) = np\) and \(\text{Var}(X) = np(1 - p)\).

00.10.20.3 012345678910 r (number of successes) Shaded: P(X ≤ 2) = 0.383 Peak at r = 3 = np
Fig. 1: X ~ B(10, 0.3) drawn to scale. The three shaded bars add to P(X ≤ 2) = 0.0282 + 0.1211 + 0.2335 = 0.383

Worked example: One exact probability

\(X \sim \text{B}(10, 0.3)\). Find \(\text{P}(X = 3)\).
  1. \(\dbinom{10}{3} = 120\) ways to place the 3 successes among 10 trials.
  2. Each such sequence has probability \(0.3^{3} \times 0.7^{7}\).
  3. \(120 \times 0.027 \times 0.082\,354\,3 = 0.266\,83\).
Answer: 0.267 (3 s.f.)

Define \(X\), then \(\text{B}(n, p)\).

  • Model: state \(X\), \(n\) and \(p\), and check the four conditions.
  • Exact: \(\text{P}(X = r) = \dbinom{n}{r} p^{r}(1-p)^{n-r}\).
  • Cumulative: turn the words into an inequality; use the complement when shorter.
  • Mean and variance: \(np\) and \(np(1-p)\).
  • Least n: set up the inequality and try values either side.

Watch out!

Leaving out \(\dbinom{n}{r}\). \(0.3^3 \times 0.7^7\) is the probability of ONE particular order, such as SSSFFFFFFF; there are 120 orders.

Treating 'more than 2' as \(X \ge 2\). For whole numbers 'more than 2' starts at 3.

Stopping at \(n = 20\) because \(0.0115\) 'is about 0.01'. The inequality must actually hold, so check the value either side.

Rounding the first answer to 0.74 before using it again. Keep at least 4 significant figures in intermediate values so the final 3 s.f. is right.

Revisit it: The binomial distribution lesson · worked practice

3The geometric distribution

Syllabus 5.4

If \(X \sim \text{Geo}(p)\), \(X\) is the number of the trial on which the first success occurs: \(\text{P}(X = r) = p(1 - p)^{r - 1}\) for \(r = 1, 2, 3, \dots\). Also \(\text{P}(X > r) = (1 - p)^{r}\) (the first \(r\) trials all fail) and \(\text{E}(X) = \dfrac{1}{p}\).

00.10.20.30.4 0.40.240.144 12345678 r (trial of the first success) Each bar is 0.6 of the one before E(X) = 1 ÷ 0.4 = 2.5
Fig. 1: X ~ Geo(0.4) drawn to scale for r = 1 to 8. The most likely value is always r = 1, and the probabilities shrink by the factor 1 − p each time

Worked example: An exact probability

A fair die is thrown until a six appears. Find the probability that the first six occurs on the 4th throw.
  1. \(X \sim \text{Geo}\left(\tfrac{1}{6}\right)\).
  2. Three non-sixes then a six: \(\left(\tfrac{5}{6}\right)^3 \times \tfrac{1}{6}\).
  3. \(= \dfrac{125}{1296}\).
Answer: \(\dfrac{125}{1296} = 0.0965\) (3 s.f.)

Fail \(r - 1\) times, then succeed.

  • Model: independent trials, constant \(p\), counting up to the first success: \(X \sim \text{Geo}(p)\).
  • Exact: \(\text{P}(X = r) = p\,q^{r-1}\).
  • Cumulative: \(\text{P}(X > r) = q^{r}\), \(\text{P}(X \le r) = 1 - q^{r}\).
  • Mean: \(\text{E}(X) = \dfrac{1}{p}\).
  • Least n: solve \(q^{n} < \dots\) by trying values, to 4 s.f.

Watch out!

Using \((1 - p)^{r}\,p\). If the first success is on trial \(r\), only \(r - 1\) trials fail before it, so the power is \(r - 1\).

Mixing up \(X > r\) and \(X \ge r\). \(\text{P}(X \ge 4) = q^{3}\), not \(q^{4}\): only the first three trials have to fail.

Taking \(n = 8\) because \(0.75^{8}\) 'rounds to 0.1'. The inequality is strict, and 0.1001 is not less than 0.1. Work to 4 significant figures.

Revisit it: The geometric distribution lesson · worked practice
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