5.02f–5.02h - The geometric distribution

5.02f–5.02h - The geometric distribution

Model the number of independent trials up to and including the first success. Calculate exact, interval and conditional probabilities, and use the geometric mean and variance to describe waiting in discrete trials.

Counting up to the first success

If a successful trial has probability 0.20.2, the average wait is 55 trials. Is trial 55 therefore the most likely time for the first success? The answer is surprising: trial 11 is the most likely, while the long right tail pulls the average to 55.

You already know

Multiply probabilities of independent events. A discrete distribution has probability masses at individual values. A mean is a weighted average, and need not be its most likely value.

Imagine repeated independent attempts to pass a fixed automated check. On each attempt, success has the same probability pp. Record the number of attempts, stopping when the first success occurs: a success immediately gives X=1X=1; failure then success gives X=2X=2.

Geometric distribution

X∼Geo(p)X\sim\mathrm{Geo}(p) when XX is the number of trials up to and including the first success, with independent trials, two outcomes on each trial, and the same success probability pp, where 0<p≤10<p\le1. Its support is 1,2,3,…1,2,3,\ldots, or just 11 if p=1p=1.

The conditions belong to the experiment, not just its numerical output. Repeated learning attempts may become more successful with practice, violating constant pp. Sampling without replacement from a small population generally changes pp. A fixed number of trials with successes counted is a binomial situation instead; a geometric model fixes the stopping rule, not the trial count. These modelling conditions and the stopping convention are not in the formula booklet: learn them.

[DIAGRAM: asset_name: Geometric waiting has a long right tail; asset_slug: h245_5_02_geometric_masses; caption: For Geo(0.2), the first trial is the most likely success time although the mean is 5. A further trial multiplies each probability by 0.8; description: Probability masses for trial numbers 1 to 16 decrease from 0.2 by a factor of 0.8 each time. A dashed vertical line marks the mean at 5. An annotation says that the distribution continues beyond the displayed trials, with remaining probability 0.8 to the power 16]
For Geo(0.2), the first trial is the most likely success time although the mean is 5. A further trial multiplies each probability by 0.8

A geometric variable cannot be 00, because even an immediate success uses one trial. The failure count before success would be X−1X-1, a different variable.

Key idea

Define what is counted before selecting a formula: OCR's geometric variable includes the successful trial.

Check the model by examining its mechanism rather than a verbal resemblance to “waiting”.

The probability of a first success

For the first success to occur on trial xx, all x−1x-1 earlier trials must fail and trial xx must succeed. There is only one relevant success/failure sequence: failures followed by one success.

Geometric probability mass

If X∼Geo(p)X\sim\mathrm{Geo}(p), then for positive integers xx,

P(X=x)=(1−p)x−1p\mathrm{P}(X=x)=(1-p)^{x-1}p

It is zero outside the support.
In the formula booklet.

The exponent counts failures, not trials. For p=1p=1, the first trial succeeds with certainty, so handle X=1X=1 directly; the variable is constant.

Worked example 1: three failures then success

Question. X∼Geo(0.2)X\sim\mathrm{Geo}(0.2). Find P(X=4)\mathrm{P}(X=4).

Plan. Require three failures followed by success.

P(X=4)=0.83(0.2)FFF then S=0.1024multiply\begin{aligned} \mathrm{P}(X=4)&=0.8^3(0.2) &&\text{FFF then S}\\ &=0.1024 &&\text{multiply} \end{aligned}

Why this step? Trial 44 is successful, so only the preceding 33 trials contribute failure factors.

Answer. 0.1020.102, to 33 significant figures.

Check. P(X=1)=0.2\mathrm{P}(X=1)=0.2; multiplying by 0.80.8 three times gives 0.10240.1024.

Change only the success probability to 0.250.25. Keep the stopping trial fixed so the exponent remains 33.

Worked example 2: a new success probability

Question. Y∼Geo(0.25)Y\sim\mathrm{Geo}(0.25). Find P(Y=4)\mathrm{P}(Y=4).

Plan. Use failure probability 0.750.75 for each of the first three trials.

P(Y=4)=0.753(0.25)three failures then success=0.10546875evaluate without early rounding\begin{aligned} \mathrm{P}(Y=4)&=0.75^3(0.25) &&\text{three failures then success}\\ &=0.10546875 &&\text{evaluate without early rounding} \end{aligned}

Why this step? Both factors must change when pp changes: a higher success probability also lowers the probability of surviving the earlier trials.

Answer. 0.1050.105, to 33 significant figures.

Check. The first four masses are 0.25,0.1875,0.140625,0.105468750.25,0.1875,0.140625,0.10546875, each 0.750.75 times its predecessor.

A later first success needs more failures, so for 0<p<10<p<1 the probabilities strictly decrease with the trial number. This explains why trial 11 is the mode even when the mean is much larger.

Watch out

Wrong: P(X=4)=0.84(0.2)\mathrm{P}(X=4)=0.8^4(0.2). That sequence has four failures and then success, so it describes X=5X=5. Right: 0.83(0.2)0.8^3(0.2).

Keep the stopping trial fixed and change pp again.

Tails, intervals and conditional waiting

To wait beyond trial rr, you need exactly one thing: the first rr trials all fail. No success factor belongs in that tail probability.

Geometric tails

For X∼Geo(p)X\sim\mathrm{Geo}(p) and a non-negative integer rr,

P(X>r)=(1−p)r,P(X≤r)=1−(1−p)r\mathrm{P}(X>r)=(1-p)^r,\qquad \mathrm{P}(X\le r)=1-(1-p)^r

For a positive integer rr, P(X≥r)=(1−p)r−1\mathrm{P}(X\ge r)=(1-p)^{r-1}.
Not in the formula booklet: learn it.

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