Short on time? This page is the recap. For explanations and worked examples, use the full study guide.
Recap
- A and B are independent if knowing whether A happened does not change the probability of B.
- Check independence with one of three equivalent conditions. Show both numbers and compare them.
- The union A ∪ B means A or B or both. “Or” includes “both”.
- The general addition rule subtracts the overlap once, because it is counted inside both P(A) and P(B).
- For repeated independent trials, find “at least one” with the complement: 1 − P(none).
Key relationships
| Idea | Rule | When it applies |
|---|---|---|
| Independence check 1 | P(A | B) = P(A) | any events (P(B) > 0) |
| Independence check 2 | P(B | A) = P(B) | any events (P(A) > 0) |
| Independence check 3 | P(A ∩ B) = P(A) · P(B) | any events |
| Multiplication rule | P(A ∩ B) = P(A) · P(B) | independent events only |
| General multiplication rule | P(A ∩ B) = P(A) · P(B | A) | any events |
| General addition rule | P(A ∪ B) = P(A) + P(B) − P(A ∩ B) | any events |
| Mutually exclusive | P(A ∩ B) = 0, so P(A ∪ B) = P(A) + P(B) | disjoint events only |
| At least one in n independent trials | 1 − P(none) = 1 − (1 − p)ⁿ when each has probability p | independent trials |
| Complement of a union | P(A ∪ B) = 1 − P(neither A nor B) | any events |
Assumptions
- Use the multiplication rule only when independence is given, checked, or justified by how the outcomes are produced (for example, separate spins). Say which.
- With a two-way table, the probabilities refer to one individual chosen at random from that table.
- Independence of A and B also means independence of their complements.
Mistakes to avoid
- Confusing independent with mutually exclusive. Disjoint events with non-zero probabilities are dependent.
- Adding P(A) + P(B) for overlapping events. Subtract P(A ∩ B).
- Multiplying P(A) · P(B) for dependent events. Use P(A) · P(B | A) instead.
- “At least one” = n × p. Use 1 − P(none).
- Gambler’s fallacy: past independent outcomes do not change the next one.
- Comparing P(A | B) with P(B). Compare it with P(A).
- A verdict with no numbers. Write “0.40 ≠ 0.46, so not independent”.
Quick self-check
- A and B are independent with P(A) = 0.5 and P(B) = 0.3. Find P(A ∩ B) and P(A ∪ B). (0.15 and 0.5 + 0.3 − 0.15 = 0.65)
- P(A) = 0.2, P(B) = 0.6 and P(A ∩ B) = 0.15. Are A and B independent? (No: P(A) · P(B) = 0.12 and 0.12 ≠ 0.15)
- Each of 3 independent alarms fails with probability 0.1. Find P(at least one fails). (1 − 0.9³ = 0.271)
Next: practice questions.