The Binomial Distribution
Введено| English | Значение |
|---|---|
| binomial/baɪˈnəʊmɪəl/ | Биномиальное |
| binomial random variable/baɪˈnəʊmɪəl ˈrændəm ˈveərɪəbl/ | Биномиальная случайная величина |
| binomial setting/baɪˈnəʊmɪəl ˈsetɪŋ/ | Биномиальный сценарий |
| geometric/ˌdʒiːəʊˈmetrɪk/ | Геометрическое |
Counting successes
- The binomial 二项 distribution counts successes in a fixed number of tries.
- A binomial random variable 二项随机变量 $X$ = the number of successes in $n$ trials.
- Example: number of heads in $10$ flips, or correct guesses on $5$ questions.
- It's one of the most useful distributions in all of statistics.
The BINS conditions
- Four conditions define the binomial setting 二项分布情境 (remember BINS):
- Binary — each trial is success/failure. Independent — trials don't affect each other.
- Number — a fixed number of trials $n$. Same — a constant probability $p$ each trial.
- All four must hold, or it isn't binomial.
The binomial formula
-
$$P(X = k) = \binom{n}{k} p^{k} (1-p)^{n-k}$$
- $\binom{n}{k}$ counts the ways to place $k$ successes among $n$ trials.
- $p^k$ is the chance of those $k$ successes; $(1-p)^{n-k}$ the $n-k$ failures.
- Multiply the count of arrangements by the probability of each.
Binomial vs. geometric
- Binomial: a fixed $n$; you count how many successes.
- Geometric 几何: you keep going until the first success; you count how many trials that took.
- Fixed number of trials → binomial; "until first success" → geometric.
- Spotting which setting you're in is the first move on any problem.
Check all four BINS conditions before using the formula — especially independence and a constant $p$. Drawing cards without replacement breaks both ($p$ changes, trials depend), so it isn't binomial. And don't confuse it with geometric: binomial has a fixed $n$; geometric runs until the first success.
A multiple-choice quiz: $5$ questions, guess randomly, $p = 0.25$ correct each. $X =$ number right.
- BINS? Binary (right/wrong), independent, fixed $n=5$, same $p=0.25$. ✓ Binomial.
- $P(X = 2) = \binom{5}{2}(0.25)^2(0.75)^3 = 10 \times 0.0625 \times 0.4219 \approx 0.264$.
- So about a $26\%$ chance of exactly $2$ correct.
A binomial random variable counts successes in a fixed $n$ trials, valid when the BINS conditions hold (Binary, Independent, fixed Number, Same $p$). Then $P(X=k)=\binom{n}{k}p^{k}(1-p)^{n-k}$. A binomial setting has fixed $n$; a geometric setting runs until the first success.
Write from a binomial mean of 4
- A binomial setting has 8 independent trials.
- Each trial succeeds with probability 0.5. The lesson count of 10 trials is another setting.
What is the mean number of successes?
The binomial mean is n times p.
What is the mean number of successes?
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8 times 0.5 is 4. Other options — 10 (the other number of trials); 0.5 (the probability).
A binomial distribution
Bars give P(X=k) for successes in n trials.
Which is NOT one of the four binomial (BINS) conditions?
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BINS = Binary, Independent, fixed Number, Same p. Random sampling isn't required.
Drawing cards WITHOUT replacement is not binomial mainly because...
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Without replacement breaks constant p and independence.
How many ways can 2 successes occur in 5 trials? Compute C(5,2).
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C(5,2) = 10.
A geometric setting counts trials until the first success, unlike a binomial's fixed n.
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Fixed n → binomial; until first success → geometric.
The mnemonic for the four binomial conditions is ___ (four letters).
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Binary, Independent, Number fixed, Same p.