Discrete random variables, one idea at a timeMathematics Methods, Unit 3 & 4

Before the method: the idea behind it

The rules everything else is built on

Before random variables, distributions or the normal curve, there are a handful of rules for combining probabilities. They are assumed knowledge from Year 11, but every question this year uses them, and exams test them directly with tree diagrams, arrays and conditional probability.

Part one

Two tools for listing outcomes

Almost every probability error comes from miscounting outcomes. Two diagrams stop that happening.

A tree diagram handles events that happen in sequence. Each branch carries a probability, branches from the same point sum to 1, and the probability of any path is the product along it. Trees are essential when the second stage depends on the first, as in drawing without replacement.

An array handles two independent events with a small number of outcomes each, like two dice. Lay one event along the rows and the other along the columns, and every cell is equally likely. Then counting cells is the whole calculation.

\[ P(A\cap B)=P(A)\,P(B\mid A) \qquad \text{(multiply along a branch)} \]

Part two

Three rules that do all the work

The addition rule, the definition of conditional probability, and the test for independence.

The addition rule says the probability of A or B is the sum of their probabilities minus their overlap, because otherwise the overlap is counted twice. If the events cannot both happen, the overlap is zero and you simply add.

Conditional probability is probability with a restricted sample space: given that B happened, what fraction of B is also A? That restriction is why the denominator changes from 1 to P(B).

Independence means knowing one event happened tells you nothing about the other. The test is whether the intersection equals the product. Do not confuse it with mutually exclusive: events that cannot both occur are maximally dependent, since one happening rules the other out.

\[ P(A\cup B)=P(A)+P(B)-P(A\cap B), \qquad P(A\mid B)=\frac{P(A\cap B)}{P(B)}, \qquad \text{independent} \iff P(A\cap B)=P(A)P(B) \]

Every distribution in this course, discrete, binomial, continuous and normal, is built on these three rules. Conditional probability in particular returns constantly: within the normal distribution, within sampling, and in the diagnostic-test questions that examiners favour because the answers surprise people.

Next

Try it yourself

The working method, twenty practice questions, ten harder ones, and the ClassPad guide.