Probability of Combined Events

This chapter builds on basic probability to handle two or more events together. You learn the difference between independent and dependent events, between mutually exclusive and non-mutually-exclusive ones, and how to use tree diagrams and the given probability formulae to combine them.

Paper 2 loves a tree diagram: draw it, label the probabilities, then combine them for a compound event. Real contexts, marbles, weather, faulty items, run throughout.

The marks reward a clearly drawn tree as much as the final number.

Everything in this chapter runs on two rules. The multiplication law handles 'A and B both happen': for independent events P(A and B) = P(A) × P(B), and along any single path of a tree diagram you multiply the branch probabilities.

The addition law handles 'A or B happens': P(A or B) = P(A) + P(B) − P(A and B). That last subtraction removes the overlap that would otherwise be counted twice; when the events are mutually exclusive they cannot both happen, the overlap is zero, and the rule collapses to the simple P(A) + P(B).

Deciding which law you need is really deciding whether the question joins events with 'and' or with 'or'.

When a question asks for the probability of 'at least one' success, at least one red ball, at least one head, at least one defective item, listing every winning case is slow and error-prone. There is a shortcut that almost always wins: 'at least one' is the exact opposite of 'none', so P(at least one) = 1 − P(none).

You work out the single probability of getting zero successes, then subtract from 1. For two draws without replacement, 'at least one blue' becomes 1 − P(both not blue) = 1 − P(both red).

One short calculation replaces three. Whenever you see the words 'at least', reach for the complement first.

One tree diagram can answer several parts of a question, and the skill is picking the right paths for each part. First draw the tree fully and write the probability on every branch, checking that the branches leaving each point add up to 1.

Then, for each part, ask two questions: which end-outcomes count as a success, and how are they reached? Multiply along each successful path to get its probability, then add the probabilities of all the paths that succeed.

'Both the same colour' means add the all-red path and the all-blue path; 'different colours' means add the red-then-blue and blue-then-red paths. Building the tree once and selecting paths carefully beats starting over for every part.

A without-replacement drawing problem, exactly the kind that rewards a clear tree diagram.