Topic 4.12 The Geometric Distribution: identify a geometric setting (waiting for the first success), compute geometric probabilities, and find the mean of a geometric random variable.

What this topic is asking

The College Board (Topic 4.12) wants you to identify a geometric setting (waiting for the first success), compute geometric probabilities, and find the mean of a geometric random variable.

The geometric setting

The defining feature is the stopping rule: you repeat the trial until success, so the number of trials is itself the random outcome. Rolling a die until the first six, calling customers until the first sale, testing items until the first defect, these are geometric, because the count of trials needed is what varies. Spotting "until the first" (or "how many trials to get a success") is the cue that the geometric model applies rather than the binomial.

The geometric probability formula

The formula follows directly from independence: to have the first success on trial $x$, the first $x - 1$ trials must all be failures (probability $(1-p)^{x-1}$) and trial $x$ a success (probability $p$), and multiplying gives the result. Unlike the binomial there is no coefficient, because there is only one arrangement that puts the first success exactly on trial $x$ (everything before it must be a failure). The mean $1/p$ is intuitive: if successes happen one time in ten ($p = 0.1$), you expect to wait about ten trials for one.

Cumulative geometric probabilities

Geometric questions often ask for a range rather than an exact trial. "The first success takes more than $k$ trials" means the first $k$ trials were all failures, which has the clean form

a single power, no sum needed. From this, "the first success occurs within the first $k$ trials" is the complement, $P(X \le k) = 1 - (1-p)^k$. These two cumulative forms are exam favorites because they avoid adding many individual terms: instead of summing $P(X = 1) + P(X = 2) + \cdots + P(X = k)$, you compute $1 - (1-p)^k$ directly. A graphing calculator provides geometpdf (exact trial) and geometcdf (at most $k$ trials), but you should still recognize the structure, "more than $k$ trials" equals "$k$ failures in a row" equals $(1-p)^k$, and be able to set it up by hand. This compact handling of waiting-time questions is the practical payoff of the geometric model.

Geometric versus binomial

The exam relentlessly tests the binomial-versus-geometric distinction because they share the same trial conditions and differ only in what is fixed and counted. Binomial: the number of trials $n$ is fixed, and you count the number of successes in those $n$ trials; the formula has a binomial coefficient and the mean is $np$. Geometric: the number of trials is not fixed, and you count the trials until the first success; the formula has no coefficient and the mean is $1/p$. The verbal cue is decisive: "in $n$ trials, how many successes?" is binomial, while "how many trials until the first success?" is geometric. A further contrast is shape, the geometric distribution is always right-skewed (the first success is most likely early, with a long tail of long waits), whereas a binomial's shape depends on $p$ and $n$. Keeping these two waiting-versus-counting models distinct, and matching each to its formula and mean, is the central skill of Topic 4.12 and a frequent multiple-choice trap.

Try this

Q1. A geometric variable has $p = 0.25$. Find $P(X = 2)$ and the mean. [2 points]

• Cue. $P(X = 2) = (0.75)^1(0.25) = 0.1875$; mean $= 1/0.25 = 4$.

Q2. Explain the key difference between a binomial and a geometric setting. [1 point]

• Cue. Binomial fixes the number of trials and counts successes; geometric fixes nothing and counts the trials until the first success.