Flip a coin twice, or roll two dice, or draw two counters from a bag — very often the question isn't about one event but about two happening together. The whole topic turns on two tiny words: AND and OR. Spot which one the question is really asking and you already know whether to multiply or add.
The AND / OR rules
Everything in this chapter grows from one simple pair of rules:
OR (either event happens) → add the probabilities.
Why multiply for AND? Think of it as narrowing down. Getting a head on the first coin already cuts your possibilities in half; needing a head on the second cuts those in half again. Half of a half is a quarter — and 1/2 × 1/2 = 1/4. Multiplying makes the answer smaller, which is right, because demanding two things at once is harder than demanding one.
Why add for OR? "OR" means you'll be happy with several different outcomes, so you gather them up. If a spinner gives red with probability 1/4 or blue with probability 1/4, then red OR blue = 1/4 + 1/4 = 1/2. (This clean adding works when the outcomes can't both happen at once.)
| Question word | What it wants | Do this | Effect on answer |
|---|---|---|---|
| AND / both | every event together | multiply | gets smaller |
| OR / either | any one of them | add | gets bigger |
Tree diagrams — a map of every possibility
A tree diagram is like a signpost at every fork in the road: each branch is a choice, and following a full path from left to right traces one complete outcome. Two rules keep you right:
- The branches leaving any single point always add up to 1 — something must happen.
- To find the probability of a particular path, multiply along its branches (that's the AND rule in action). To combine several paths that all count as a "win", add down your results (that's the OR rule).
The "at least one" trick
"At least one" hides a lot of paths (one, or two, or three...). Chasing them all is slow and error-prone. The clever shortcut is to go through the back door: the opposite of "at least one" is "none". Find P(none) — usually just one easy multiplication — then subtract from 1.
P(at least one six) on two dice: P(no six) = 5/6 × 5/6 = 25/36, so P(at least one six) = 1 − 25/36 = 11/36.
In a nutshell: AND → multiply, OR → add. On tree diagrams, multiply across and add down; and for any "at least one" question, do 1 − P(none).