Discrete probability
Đã giới thiệu| English | Ý nghĩa |
|---|---|
| sample space/ˈsæmpl speɪs/ | The set of all possible outcomes of a random experiment. |
| event/ɪˈvent/ | An occurrence detected by a program that can trigger a response. |
| conditional probability/kənˈdɪʃənl ˌprɒbəˈbɪlɪti/ | The probability of an event given that another event has occurred. |
| Bayes' theorem | Bayes' theorem |
| independent/ˌɪndɪˈpendənt/ | Describing quantities, events or choices that do not depend on one another in the relationship being considered. |
| prosecutor's fallacy | prosecutor's fallacy |
A positive test, but how worried should you be?
- A disease affects 1 in 100 people. A test catches 90% of cases and clears 90% of healthy people. Your test is positive. Is the chance you are ill 90%?
- It is about 8%. Discrete probability, which is counting over equally likely outcomes, shows why.
Sample spaces and counting
- The sample space 样本空间 $\Omega$ is the set of all outcomes; an event 事件 is a subset. If every outcome is equally likely, $P(E) = \frac{|E|}{|\Omega|}$.
- Two fair dice give $6 \times 6 = 36$ pairs. Six pairs sum to 7, so $P(\text{sum} = 7) = \frac{6}{36} = \frac{1}{6}$. Two pairs sum to 11, so $P(\text{sum} = 11) = \frac{2}{36}$.
Two fair dice: what is P(sum = 7 or sum = 11)?
Độ khó 1/5 · 100 xu + phần thưởng nhỏ khi hoàn thành lần đầu tiên hôm nay
6 pairs sum to 7 and 2 pairs sum to 11; the events do not overlap, so 8 of the 36 pairs.
Conditioning shrinks the world
- Conditional probability 条件概率: $P(A \mid B) = \frac{P(A \cap B)}{P(B)}$. Keep only the outcomes in $B$, then ask what fraction of them are also in $A$.
- Two dice, given the first shows 6: the world shrinks to 6 pairs. One of them, (6, 1), sums to 7, so $P(\text{sum} = 7 \mid \text{first} = 6) = \frac{1}{6}$.
Two fair dice. Given that the sum is 4, what is the probability that both dice show 2? Give a decimal to 3 places.
Độ khó 2/5 · 140 xu + phần thưởng nhỏ khi hoàn thành lần đầu tiên hôm nay
The sum-4 world is (1,3), (2,2), (3,1): 3 outcomes, one of them (2,2), so 1/3.
Bayes' theorem on the test
- Bayes' theorem 贝叶斯定理: $P(A \mid B) = \frac{P(B \mid A)\,P(A)}{P(B)}$. The tree finds $P(B)$: add every branch that ends in $B$.
- Ill and positive: $0.9 \times 0.01 = 0.009$. Healthy and positive: $0.1 \times 0.99 = 0.099$. Only the positive branches stay, rescaled:
A disease affects 2%. A test catches 95% of cases and gives a false positive for 5% of healthy people. What is P(ill | positive), in % to 1 decimal place?
Độ khó 2/5 · 140 xu + phần thưởng nhỏ khi hoàn thành lần đầu tiên hôm nay
Ill and positive: 0.95 × 0.02 = 0.019. Healthy and positive: 0.05 × 0.98 = 0.049. 0.019 ÷ 0.068 ≈ 0.279, so about 27.9%.
Independence
- $A$ and $B$ are independent 独立 if $P(A \cap B) = P(A)\,P(B)$, so knowing $B$ does not change $P(A)$. Two dice are independent: $P(\text{both } 6) = \frac{1}{6} \times \frac{1}{6} = \frac{1}{36}$.
- Draws without replacement are not independent. Two aces from a 52-card deck: $\frac{4}{52} \times \frac{3}{51} = \frac{1}{221}$, because the first draw changes the deck.
Drawing two cards from a deck without putting the first back gives independent draws.
Độ khó 1/5 · 100 xu + phần thưởng nhỏ khi hoàn thành lần đầu tiên hôm nay
The first card changes what is left, so the second probability depends on it.
Two cards are drawn without replacement from a 52-card deck. The probability that both are aces is 1/N. What is N?
Độ khó 2/5 · 140 xu + phần thưởng nhỏ khi hoàn thành lần đầu tiên hôm nay
4/52 × 3/51 = 12/2652 = 1/221.
The prosecutor's fallacy
- $P(A \mid B)$ and $P(B \mid A)$ are different numbers. "A healthy person rarely tests positive" (10%) is not "a positive person is rarely healthy" (here, about 92% are healthy).
- Swapping them is the prosecutor's fallacy 检察官谬误. Bayes converts one into the other; it never swaps them for free.
For the 1% disease test in this lesson, match each probability to its value.
Độ khó 2/5 · 140 xu + phần thưởng nhỏ khi hoàn thành lần đầu tiên hôm nay
The test's own rates condition on health; the patient's question conditions on the result. Bayes links them.
Marks that slip away
- Unequal outcomes counted as equal. The sums 2 to 12 are not equally likely; count the 36 pairs instead.
- Multiplying dependent events. $P(A)P(B)$ works only when $A$ and $B$ are independent.
- Ignoring the base rate. A rare condition stays fairly unlikely after one imperfect test.
You've got it
- Equally likely outcomes: $P(E) = |E| / |\Omega|$, so counting does the work.
- $P(A \mid B) = P(A \cap B)/P(B)$; Bayes: $P(A \mid B) = P(B \mid A)P(A)/P(B)$, with $P(B)$ from the tree.
- Independence means $P(A \cap B) = P(A)P(B)$; never swap $P(A \mid B)$ and $P(B \mid A)$.