Short on time? This page is the recap. For explanations and worked examples, use the full study guide.
Recap
- The sample space S is the set of all possible outcomes, with no overlaps. One outcome happens on every trial.
- An event E is a set of outcomes. Its probability is written P(E).
- With equally likely outcomes, find P(E) by counting.
- The complement of E (written Eᶜ, E′ or Ē) is “E does not happen”.
- For two-stage processes, list outcomes as pairs in a grid or a systematic list.
Key relationships
| Rule | Statement | Use it to… |
|---|---|---|
| Sample space | P(S) = 1; probabilities of all separate outcomes add to 1 | find a missing probability |
| Range | 0 ≤ P(E) ≤ 1 (0 = impossible, 1 = certain) | check every answer |
| Equally likely outcomes | P(E) = outcomes in E ÷ outcomes in S | calculate an exact probability |
| Random choice from a table | P(category) = count ÷ total | turn relative frequencies into probabilities |
| Complement | P(Eᶜ) = 1 − P(E) | handle “not” and “at least one” |
| At least one | P(at least one) = 1 − P(none) | avoid counting many outcomes |
Assumptions
- The counting formula needs equally likely outcomes: a fair die or coin, equal sectors, or one individual chosen at random.
- Outcomes in the sample space must not overlap, and none may be missing.
- Order matters when the stages are different (day 1 then day 2; die then spinner).
Mistakes to avoid
- “Three outcomes, so 1/3 each” when the outcomes are not equally likely.
- Counting totals or scores instead of the equally likely pairs behind them.
- Missing or double-counting outcomes; use a grid.
- Answers below 0 or above 1. Always impossible.
- Wrong complement: the complement of “at least 2” is “at most 1”.
- Writing a percentage as a probability (87.5 instead of 0.875).
Quick self-check
- P(E) = 0.27. Find P(Eᶜ). (1 − 0.27 = 0.73)
- A fair spinner has 5 equal sectors numbered 1 to 5. Find P(even number). (2 of 5 outcomes: 2/5 = 0.4)
- What is the complement of “at least one of the three days is sunny”? (None of the three days is sunny.)
- Two fair coins are tossed. Find P(at least one head). (S = {HH, HT, TH, TT}, four equally likely outcomes. P(no heads) = 1/4, so P(at least one head) = 1 − 1/4 = 3/4 = 0.75.)
- A model gives P(red) = 0.45, P(blue) = 0.35 and P(green) = 0.3 for the only three outcomes. What is wrong? (The probabilities add to 1.1, not 1.)
Next: practice questions.