Combined events
| English | Chinese | Pinyin |
|---|---|---|
| independent | 独立的 | dú lì de |
| mutually exclusive | 互斥 | hù chì |
| sample space | 样本空间 | yàng běn kōng jiān |
| tree diagram | 树状图 | shù zhuàng tú |
| replacement | 放回 | fàng huí |
Two events at once
- You roll a die and flip a coin. What's the chance of a six AND heads?
- Combined events need two rules: AND (multiply) and OR (add).
AND — multiply (independent 独立的 events)
- For two independent events (one doesn't affect the other):
$\text{P}(\text{six and heads}) = \dfrac{1}{6} \times \dfrac{1}{2} = \dfrac{1}{12}$.
Combined events
P(A ∩ B) = P(A)·P(B|A)
Combine two events: the area model shows AND (overlap) versus OR (union).
For two independent events both happening (AND), you:
P(A and B) = P(A) × P(B) for independent events.
OR — add (mutually exclusive 互斥 events)
- For two mutually exclusive events (they can't both happen):
Probability tree
Multiply the probabilities along each branch; the four outcomes always add up to 1.
For mutually exclusive events, P(A or B) = P(A) + P(B).
Mutually exclusive events cannot both happen, so you add their probabilities.
Three ways to show outcomes
- Sample space 样本空间 diagram: a table of every outcome (two dice → $36$ cells).
- Venn diagram: sorts outcomes into overlapping sets.
- Tree diagram 树状图: a branch per stage; multiply along branches, then add the paths you want.
A Venn diagram sorts outcomes: the overlap is $\text{P}(A\text{ and }B)$, and everything inside either circle is $\text{P}(A\text{ or }B)$
Multiply along, add between. On a tree diagram: multiply probabilities along each branch (to get that path's probability), then add the paths you want (to get the total probability).
On the 6 by 6 grid of two dice (36 cells), how many cells give a total of 7?
1+6, 2+5, 3+4, 4+3, 5+2, 6+1 — six cells, so P(7) = 6/36 = 1/6.
On a tree diagram, you ______ along branches and add between paths.
Multiply along each branch to get that path probability, then add the wanted paths.
With and without replacement 放回
- With replacement: chances stay the same each time.
- Without replacement (Extended): the total and the count change after each draw.
- Bag of $3$ red, $5$ blue: P(red, red) with replacement $= \dfrac{3}{8} \times \dfrac{3}{8} = \dfrac{9}{64}$. Without replacement $= \dfrac{3}{8} \times \dfrac{2}{7} = \dfrac{3}{28}$.
Venn diagrams show combined events: the overlap (intersection) is AND, the whole region (union) is OR.
Bag of 3 red, 5 blue, two draws WITH replacement. P(red, red) = 9/b. What is b?
(3/8) × (3/8) = 9/64, so b = 64.
Same bag, WITHOUT replacement. After one red is taken, how many reds remain?
Started with 3 red; one taken leaves 2 red (out of 7 counters).
You've got it
- AND → multiply (independent); OR → add (mutually exclusive)
- on a tree: multiply along branches, add the wanted paths
- without replacement, the second probability changes ($\dfrac{3}{8} \to \dfrac{2}{7}$)