Probability, Random Variables, and Probability Distributions

A manager says three people were chosen at random for a training trip, and all three are women, from a group of $6$ men and $3$ women. Bad luck, or unfair? Probability turns that hunch into a number: here about $1$ chance in $84$. This unit builds the rules that let you calculate chances like this and judge whether a result is surprising.

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 streak or a pattern in a short run of data does not by itself show that the process is not random: random processes produce streaks quite often.

The law of large numbers 大数定律: as the number of trials grows, the relative frequency of an event gets closer and closer to its true probability. It says nothing about the short run: after five heads in a row, a fair coin is still $50\%$ likely to land heads next.

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.

Worked example (a complete simulation). Estimate the probability that choosing $3$ people at random from $6$ men and $3$ women gives $3$ women. Label the people $1$ to $9$, with $1$–$3$ the women. Use a random number generator to pick integers from $1$ to $9$, ignoring repeats, until $3$ different people are chosen; record whether all three are women. Repeat this many times (say $1000$). The proportion of trials with three women estimates the probability. A simulation description must say what the numbers stand for, how repeats are handled, what is recorded, and that the process is repeated many times.

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 $1$. When all outcomes are equally likely,

A probability is a long-run relative frequency. "$P(\text{rain})=0.3$" means that over many days like this one, it rains on about $30\%$ of them.

Worked example (is the claim believable?). For the manager's choice above, there are $\binom93=84$ equally likely groups of $3$, and only one of them is all three women: $P=\tfrac{1}{84}=0.012$. An outcome this unlikely under random selection gives reason to doubt that the choice was random. It does not prove unfairness: unlikely things do happen, about $1$ time in $84$.

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).

"At least one": use the complement. A mall has three ATMs, each working at opening time with probability $0.9$, independently. The only way to have none working is all three failing:

Given that at least one works, the probability all three work is The "and" in the numerator is just $P(\text{all})$, because "all three work" is already inside "at least one works".

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).

A cumulative probability distribution gives $P(X\le x)$ for each value, adding the probabilities up from the smallest value. Its last entry is always $1$. Two random variables are independent if knowing the value of one does not change the probability distribution of the other.

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.

Worked example (the ATMs again). The number of working ATMs, $X$, has $E(X)=3(0.9)=2.7$. Interpretation in context: "Over many days, the mean number of ATMs working when the mall opens is about $2.7$." If we are told that at least one ATM works, the case $X=0$ is removed, so the expected value increases slightly: the remaining values are all $1$ or more.

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 $\sigma_{aX+b}=|a|\sigma_X$. A linear change $Y=a+bX$ does not change the shape of the distribution (for $b>0$): a right-skewed $X$ gives a right-skewed $Y$.

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. Interpret them in context: "If we test many batches of $n$, the mean number of successes per batch is about $np$." A binomial distribution can also be built from the formula, or estimated by simulation.

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 $\mu_Y=\tfrac{1}{0.3}\approx 3.33$. The standard deviation is $\sigma_Y=\dfrac{\sqrt{1-p}}{p}=\dfrac{\sqrt{0.7}}{0.3}=2.79$ shots.

Worked example (rocket igniters). A company claims a failure rate of $15\%$. The probability that the first $30$ igniters tested all work is $0.85^{30}=0.0076$. Getting $30$ successes in a row would be very unusual if the failure rate really were $15\%$, so it is reasonable to believe the failure rate is lower. Given those $30$ successes, the chance that the first failure is on test $31$ or $32$ is $0.15+0.85(0.15)=0.278$: past successes do not change the next trials, because they are independent.

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.

6. Each of $4$ independent smoke alarms works with probability $0.95$. Find the probability that at least one works.

7. In a large school, $40\%$ of students walk to school. If $8$ students are chosen at random, find the mean and standard deviation of the number who walk, and interpret the mean.

8. Machine A makes $60\%$ of a factory's bolts and $2\%$ of its bolts are faulty; machine B makes the rest, and $5\%$ of B's bolts are faulty. A bolt is faulty. Find the probability it came from machine A.

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.

6. $1-0.05^4=1-0.00000625=0.99999$.

7. $\mu=8(0.4)=3.2$, $\sigma=\sqrt{8(0.4)(0.6)}=1.39$. In many random samples of $8$ students, the mean number who walk would be about $3.2$.

8. $P(\text{faulty})=0.6(0.02)+0.4(0.05)=0.012+0.020=0.032$, so $P(A\mid\text{faulty})=\dfrac{0.012}{0.032}=0.375$.

• 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.
• For "at least one", subtract "none" from $1$.
• When a result has a small probability under a claim, say it gives reason to doubt the claim, and quote the probability.
• Interpret an expected value as a long-run mean over many repetitions, in context.