概率、随机变量与概率分布

Something is random 随机 if individual outcomes 结果 are uncertain but a regular pattern emerges over many repetitions 重复. Short-run results look erratic; long-run relative frequencies 相对频率 settle down. This long-run stability is what makes probability useful.

A simulation 模拟 imitates a chance process 随机过程 using random digits 随机数字 or technology. Steps: state the model, assign digits to outcomes, run many trials 试验, and record the proportion of trials meeting the condition. The resulting proportion estimates the probability – more trials give a better estimate.

The probability 概率 of an event 事件 is a number from $0$ to $1$ giving its long-run relative frequency. The sample space 样本空间 is the set of all outcomes. For an event $A$, the complement 补 rule: $P(A^c)=1-P(A)$. Probabilities of all outcomes sum to $1$.

Two events are mutually exclusive 互斥 (disjoint 不相交) if they cannot both happen. Then the addition rule 加法法则 simplifies:

In general, $P(A\text{ or }B)=P(A)+P(B)-P(A\text{ and }B)$ – subtract the overlap 重叠部分 so it is not counted twice.

The conditional probability 条件概率 of $A$ given $B$ is

It is the chance of $A$ once you know $B$ happened. Two-way tables 双向表 make these easy: restrict to the row/column for $B$, then find $A$'s share.

Worked example: tree diagram. $2\%$ of people have a disease. A test is positive for $95\%$ of people who have it, and (a false positive 假阳性) for $10\%$ of people who do not. A person tests positive. What is the probability that they have the disease?

• Branch “disease, positive”: $0.02\times0.95=0.019$.
• Branch “no disease, positive”: $0.98\times0.10=0.098$.
• So $P(\text{positive})=0.019+0.098=0.117$.

Only about $16\%$ of positive results come from people with the disease: so few people have it that the false positives outnumber the true ones. Do not confuse $P(\text{positive}\mid\text{disease})=0.95$ with $P(\text{disease}\mid\text{positive})$.

Events are independent 独立 if knowing one does not change the other's probability: $P(A\mid B)=P(A)$. Then the multiplication rule 乘法法则 simplifies:

Independent is not the same as mutually exclusive – mutually exclusive events with nonzero probability are actually dependent 不独立 (if one happens, the other cannot).

A random variable 随机变量 assigns a number to each outcome of a chance process. A probability distribution 概率分布 lists each possible value with its probability (they sum to $1$). A distribution can be discrete 离散 (a table of values) or continuous 连续 (an area-under-a-curve model like the normal).

The mean (expected value) 期望值 of a discrete random variable is the probability-weighted average 概率加权平均:

The standard deviation 标准差 $\sigma_X=\sqrt{\sum (x_i-\mu_X)^2\,P(x_i)}$ measures typical spread from the mean. The expected value is the long-run average outcome, not a value you expect on any single trial.

Worked example. A game pays $\$5$ with probability $0.2$ and costs you $\$1$ (a $-1$ outcome) with probability $0.8$. The expected value is

so over many plays you gain about $20$ cents per play on average, even though no single play gives exactly that.

When you add or subtract random variables, means add 均值相加: $\mu_{X\pm Y}=\mu_X\pm\mu_Y$. If $X$ and $Y$ are independent, variances add 方差相加 (even when subtracting):

Take the square root for the standard deviation. Also, scaling: $\mu_{aX+b}=a\mu_X+b$ and $\sigma_{aX+b}=|a|\sigma_X$.

Worked example. Task A takes $X$ minutes with $\mu_X=20$, $\sigma_X=3$; task B takes $Y$ minutes with $\mu_Y=15$, $\sigma_Y=4$; the times are independent.

• Total time: $\mu_{X+Y}=20+15=35$ minutes, and $\sigma_{X+Y}=\sqrt{3^2+4^2}=\sqrt{25}=5$ minutes.
• Difference: $\mu_{X-Y}=20-15=5$ minutes, and $\sigma_{X-Y}=\sqrt{3^2+4^2}=5$ minutes as well.

Two common errors: adding the standard deviations ($3+4=7$), and subtracting them for a difference ($4-3=1$). Add the variances, then take the square root.

A third error: treating $3X$ like $X_1+X_2+X_3$. Tripling one value triples the spread, $\sigma_{3X}=3\sigma_X$; adding three independent values gives $\sigma=\sqrt{3}\,\sigma_X$, because the variances add.

A binomial 二项 setting (BINS): a fixed number $n$ of Independent trials, each with two outcomes (success 成功/failure 失败) and the same success probability $p$. The random variable $X=$ number of successes. Its probability:

For a binomial $X$ with $n$ trials and success probability $p$:

Use these for "how many successes do we expect, and how much do they vary" questions.

Worked example. A player makes $70\%$ of free throws. In $n=10$ shots, the probability of exactly $8$ makes is

and the expected number of makes is $\mu=np=10(0.7)=7$, with $\sigma=\sqrt{10(0.7)(0.3)}\approx1.45$.

Worked example: “at least”. For the same player ($n=10$, $p=0.7$), the probability of at least $8$ makes adds the top three values:

By the complement, the probability of at most $7$ makes is $P(X\le7)=1-0.3828=0.6172$.

Common mistakes. Writing $P(X\ge 8)=1-P(X\le 8)$: the complement of “at least $8$” is “at most $7$”, so $P(X\ge 8)=1-P(X\le 7)$. Using a binomial model when $n$ is not fixed, or when the trials are not independent (drawing without replacement from a small group).

A geometric 几何 setting is the same as binomial but with no fixed $n$: you keep trying until the first success. The random variable $Y=$ the trial of the first success:

So the expected number of trials until the first success is $1/p$.

Worked example. A player makes $30\%$ of shots, independently. The probability that the first make is on the third shot needs two misses, then a make:

The probability of a make within the first three shots uses the complement (three misses in a row): $P(Y\le 3)=1-0.7^3=0.657$. On average the first make comes on shot $\mu_Y=\tfrac{1}{0.3}\approx 3.33$.

Answer each before reading the answers below.

1. $P(A)=0.5$, $P(B)=0.4$ and $P(A\text{ and }B)=0.2$. Find $P(A\text{ or }B)$.
2. Are $A$ and $B$ in question 1 independent? Show why.
3. A basketball player makes $30\%$ of three-point shots, independently. In 10 shots, find the probability of exactly 2 makes.
4. A game costs $1 to play. You win $5 (and keep your $1) with probability $0.2$; otherwise you lose the $1. Find the expected gain per game.
5. A student says: "Mutually exclusive events are independent, because one has nothing to do with the other." Correct the student.

1. $P(A\text{ or }B)=0.5+0.4-0.2=0.7$.
2. Yes: $P(A)P(B)=0.5\times0.4=0.2=P(A\text{ and }B)$.
3. Binomial with $n=10$, $p=0.3$: $\binom{10}{2}(0.3)^2(0.7)^8=45(0.09)(0.0576)=0.233$.
4. $E=0.2(5)+0.8(-1)=0.20$: on average the player gains $0.20 per game.
5. If $A$ and $B$ are mutually exclusive and both have positive probability, then knowing $A$ happened tells you $B$ did not. So $P(B\mid A)=0\neq P(B)$: they are dependent, and as dependent as two events can be.

• A probability lies in $[0,1]$; use the complement ($1-P$) and add mutually exclusive events.
• For independent events multiply; for "and/or" use the general addition and conditional rules.
• Expected value = $\sum(\text{value}\times\text{probability})$.
• Recognise binomial (fixed $n$, two outcomes, constant $p$) and geometric settings.
• Draw a tree or table for multi-stage problems and multiply along branches.