More about probability
Set notation and Venn diagrams, the addition and multiplication laws, conditional probability, and using permutations and combinations to find probabilities. The unit is NF.
What you will learn
- 16.1 Set notation: union, intersection, complement; Venn diagrams.
- 16.2 Addition law P(A ∪ B) = P(A) + P(B) − P(A ∩ B); mutually exclusive and complementary events.
- 16.3 Multiplication law P(A ∩ B) = P(A) × P(B) for independent events.
- 16.4 Conditional probability, using P(A ∩ B) = P(A) × P(B | A). Bayes' theorem is not required.
- 16.5 Use permutations and combinations to find probabilities.
You need first: Probability, Permutations and combinations
Key terms
| English | 中文 |
|---|---|
| union, intersection, complement | 併集，交集，補集 |
| Venn diagram | 溫氏圖 |
| mutually exclusive events | 互斥事件 |
| complementary events | 互補事件 |
| independent events | 獨立事件 |
| addition law / multiplication law | 加法定律／乘法定律 |
| conditional probability | 條件概率 |
Standard steps
Words to symbols
| Words | Symbol |
|---|---|
| A or B (or both) | A ∪ B |
| A and B | A ∩ B |
| not A | A′ |
| A given B | A | B |
Which law?
| Situation | Law |
|---|---|
| "or" | P(A ∪ B) = P(A) + P(B) − P(A ∩ B). If mutually exclusive, P(A ∩ B) = 0. |
| "and", independent | P(A ∩ B) = P(A) × P(B) |
| "and", not independent (second depends on first) | P(A ∩ B) = P(A) × P(B | A) |
| "at least one" | 1 − P(none) |
| choosing several items at once | P = (ways for the event) ÷ (all ways), both with ⁿCᵣ |
- Underline the key words: "or", "and", "given", "at least one".
- For "and", ask whether the first result changes the chances of the second. If it does, use P(B | A).
- Pick the law from the table and write it down before putting in the numbers.
Conditional probability from a table
P(A | B) = (number in A and B) ÷ (number in B). The "given" part becomes the new total.
- Find the row or column of the "given" condition B. Its total is the new denominator.
- In that row or column, read the number that is also in A.
- Divide: P(A | B) = n(A ∩ B) ÷ n(B).
Worked examples
P(no red) = ⁶C₃ / ¹⁰C₃ = 20/120 = 1/6
P(at least one red) = 1 − 1/6 = 5/6P(bus and late) = 0.6 × 0.3 = 0.18
P(not bus and late) = 0.4 × 0.1 = 0.04
P(late) = 0.18 + 0.04 = 0.2210/25 = 2/5.
P(A ∪ B) = 0.5 + 0.4 − 0.2 = 0.7
P(A) × P(B) = 0.2 = P(A ∩ B), so A and B are independent| Bus | Walk | Total | |
|---|---|---|---|
| Late | 9 | 3 | 12 |
| On time | 21 | 27 | 48 |
| Total | 30 | 30 | 60 |
P(late | bus) = 9/30 = 3/10 "given bus": the Bus column total 30 is the new total
P(bus | late) = 9/12 = 3/4 "given late": the Late row total 12 is the new totalThe two answers differ: P(A | B) is not the same as P(B | A).
Common mistakes
- Treating "mutually exclusive" and "independent" as the same (they are almost opposite).
- Adding P(A) + P(B) for "or" without subtracting the overlap.
- In conditional probability, dividing by the whole total instead of the "given" group.
- Using ⁿPᵣ on top and ⁿCᵣ on the bottom (be consistent).
How the DSE tests it
- Section B: a short probability question using ⁿCᵣ, or a two-stage situation with a conditional part.
- Paper 2 Section B: Venn diagrams, independence checks, "at least one".
Questions you might not ask in class
What's the difference between mutually exclusive and independent?
Mutually exclusive: they can't both happen. Independent: one happening doesn't change the other's chance. If two events are mutually exclusive (and each can happen), they are not independent: one happening makes the other impossible.
Why does the 'given' part become the total?
"Given" means we already know we are inside that group, so we only count inside it.
Bayes' theorem is on YouTube; do I need it?
No, it's not required. A question like "P(bus | late)" can still be done with a tree and P(A | B) = P(A ∩ B) ÷ P(B).