The Geometric Distribution
Đã giới thiệu| English | Ý nghĩa |
|---|---|
| geometric/ˌdʒiːəʊˈmetrɪk/ | hình học |
| geometric random variable/ˌdʒiːəʊˈmetrɪk ˈrændəm ˈveərɪəbl/ | biến ngẫu nhiên hình học |
Waiting for the first success
- The geometric 几何 distribution counts trials until the first success.
- A geometric random variable 几何随机变量 $X$ = the number of trials to get the first success.
- Example: how many rolls until you get your first six?
- Unlike binomial, $n$ isn't fixed — you stop the moment you succeed.
The geometric setting
- Same trial conditions as binomial, but a different stopping rule:
- Repeated independent trials, each with the same probability $p$ of success.
- You keep going until the first success, then stop.
- "Until first success" is the signal word for geometric.
The geometric distribution: trials until the first success
The geometric formula
-
$$P(X = k) = (1-p)^{k-1}\,p$$
- To succeed first on trial $k$: fail the first $k-1$ times, then succeed.
- $(1-p)^{k-1}$ = those failures; $p$ = the success on trial $k$.
- The probabilities shrink as $k$ grows — early success is most likely.
The geometric mean
- The mean (expected number of trials) is:
-
$$\mu_X = \frac{1}{p}$$
- If $p = 0.2$, expect $1/0.2 = 5$ trials on average until the first success.
- Rarer successes ($small$p$) mean you wait longer — makes intuitive sense.
Geometric vs. binomial comes down to the stopping rule. Binomial: fixed $n$, count successes. Geometric: run until the first success, count trials. If a problem fixes the number of tries, it's binomial; if it says "until the first…", it's geometric. And the geometric mean is $1/p$ — not $np$.
Roll a die until the first six; $p = 1/6$.
- $P(X = 3) = (5/6)^2 (1/6) = \tfrac{25}{36}\cdot\tfrac{1}{6} \approx 0.116$ — first six on the $3$rd roll.
- Mean: $\mu_X = 1/(1/6) = 6$ rolls expected until the first six.
- Makes sense — a six comes up on average once every six rolls.
A geometric random variable counts trials until the first success, with independent trials at constant $p$. Then $P(X=k)=(1-p)^{k-1}p$ and the mean is $\mu_X = \frac{1}{p}$. The key contrast with binomial: geometric runs until the first success (variable trials), binomial has a fixed $n$.
Write from an expected wait of 4
- Trials repeat until the first success.
- The success probability is 0.25. The lesson expected waits 5 and 6 use other probabilities.
What is the expected number of trials until the first success?
The geometric mean is 1 divided by the success probability.
What is the expected wait?
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1 divided by 0.25 is 4.
Rolling until the first success
Keep rolling until the target appears — that's a geometric count.
A geometric variable has p = 0.2. Find the mean number of trials, 1/p.
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μ = 1/p = 1/0.2 = 5.
Which situation is geometric (not binomial)?
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'Until the first success' → geometric; fixed n → binomial.
Roll a die until the first six (p = 1/6). Find the expected number of rolls.
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μ = 1/p = 1/(1/6) = 6.
In the geometric formula P(X=k) = (1−p)^(k−1)·p, the (1−p)^(k−1) part is the first k−1 failures.
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Fail k−1 times, then succeed on trial k.
The mean of a geometric random variable is 1 divided by ___ (the letter for success probability).
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μ = 1/p.