Discrete Random Variables — H2 Maths help

A discrete random variable (DRV) $X$ takes a countable set of values, each with an assigned probability. The collection of these probabilities is the probability distribution of $X$ , and it must satisfy $\sum \text{P}(X = x) = 1$ .

What you need to know

A valid probability distribution table requires every probability to be non-negative and all probabilities to sum to exactly 1: $\sum \text{P}(X = x) = 1$ .
The expectation (mean) is $\text{E}(X) = \sum x\,\text{P}(X = x)$ .
The variance is $\text{Var}(X) = \text{E}(X^2) - [\text{E}(X)]^2$ , where $\text{E}(X^2) = \sum x^2\,\text{P}(X = x)$ .
$X \sim \text{B}(n, p)$ (binomial) requires: fixed number of trials $n$ ; independent trials; constant probability of success $p$ ; exactly two outcomes per trial.
Binomial probability: $\text{P}(X = x) = \binom{n}{x} p^x (1-p)^{n-x}$ for $x = 0, 1, \ldots, n$ .
Binomial mean and variance: $\text{E}(X) = np$ and $\text{Var}(X) = np(1-p)$ .

Worked example 1 — expectation and variance from a distribution table

A discrete random variable $X$ has the following probability distribution.

$x$	0	1	2	3
$\text{P}(X = x)$	$\dfrac{1}{6}$	$\dfrac{1}{3}$	$\dfrac{1}{3}$	$\dfrac{1}{6}$

Verify the distribution sums to 1.

\frac{1}{6} + \frac{1}{3} + \frac{1}{3} + \frac{1}{6} = \frac{1}{6} + \frac{2}{6} + \frac{2}{6} + \frac{1}{6} = \frac{6}{6} = 1. \checkmark

Find $\text{E}(X)$ .

\text{E}(X) = 0 \cdot \frac{1}{6} + 1 \cdot \frac{1}{3} + 2 \cdot \frac{1}{3} + 3 \cdot \frac{1}{6} = \frac{2}{6} + \frac{4}{6} + \frac{3}{6} = \frac{9}{6} = \frac{3}{2}.

Find $\text{Var}(X)$ .

\text{E}(X^2) = 0^2 \cdot \frac{1}{6} + 1^2 \cdot \frac{1}{3} + 2^2 \cdot \frac{1}{3} + 3^2 \cdot \frac{1}{6} = \frac{2}{6} + \frac{8}{6} + \frac{9}{6} = \frac{19}{6}.

\text{Var}(X) = \frac{19}{6} - \left(\frac{3}{2}\right)^2 = \frac{19}{6} - \frac{9}{4} = \frac{38}{12} - \frac{27}{12} = \frac{11}{12}.

So $\text{E}(X) = \dfrac{3}{2}$ and $\text{Var}(X) = \dfrac{11}{12}$ .

Worked example 2 — binomial probability for an extreme tail

The probability that a particular biased coin lands on heads on any toss is 0.3. The coin is tossed 10 times. Let $X$ be the number of heads, so $X \sim \text{B}(10,\, 0.3)$ .

Find $\text{P}(X \geq 9)$ .

\text{P}(X \geq 9) = \text{P}(X = 9) + \text{P}(X = 10).

\text{P}(X = 9) = \binom{10}{9}(0.3)^9(0.7)^1 = 10 \times (0.3)^9 \times 0.7 \approx 1.378 \times 10^{-4}.

\text{P}(X = 10) = (0.3)^{10} \approx 5.905 \times 10^{-6}.

\text{P}(X \geq 9) \approx 1.378 \times 10^{-4} + 5.905 \times 10^{-6} \approx 1.44 \times 10^{-4}.

Worked example 3 — binomial mean and variance

A multiple-choice quiz has 20 questions, each with 4 options. A student guesses every answer at random. Let $X$ be the number of correct answers.

State the distribution of $X$ and find its mean and standard deviation.

The conditions for a binomial model are satisfied: 20 independent questions ( $n = 20$ ), constant probability of success $p = \dfrac{1}{4}$ per question, and exactly two outcomes (correct / incorrect). Hence $X \sim \text{B}\!\left(20,\, \dfrac{1}{4}\right)$ .

\text{E}(X) = np = 20 \times \frac{1}{4} = 5.

\text{Var}(X) = np(1-p) = 20 \times \frac{1}{4} \times \frac{3}{4} = \frac{15}{4} = 3.75.

\text{SD}(X) = \sqrt{3.75} \approx 1.94 \text{ (3 s.f.)}.

Common mistakes

Forgetting to verify $\sum \text{P}(X = x) = 1$ before computing expectations — an invalid distribution gives meaningless answers, and an exam question may test this step directly.
Using $\text{Var}(X) = \text{E}(X^2) - \text{E}(X)^2$ carelessly by forgetting to square $\text{E}(X)$ ; write out $[\text{E}(X)]^2$ explicitly to avoid the error.
Applying the binomial model when the trials are not independent — for example, drawing without replacement from a small population violates constant $p$ , so B $(n, p)$ does not apply.
Confusing $\text{P}(X \geq k)$ with $\text{P}(X > k)$ when using the GC’s binomial CDF; recall that $\text{P}(X \geq k) = 1 - \text{P}(X \leq k-1)$ .