Geometric Distribution

Geometric Setting

A random variable $X$ follows a geometric distribution if:

• Each trial has exactly two outcomes (success/failure)
• Trials are independent
• The probability of success $p$ is the same on each trial
• The variable counts the number of trials until the first success

Notation: $X \sim \text{Geometric}(p)$

The key difference from binomial: geometric has no fixed number of trials.

Geometric Probability

The probability that the first success occurs on the $k$th trial:

$P(X = k) = (1-p)^{k-1} \cdot p, \quad k = 1, 2, 3, \ldots$

Interpretation: Fail $k-1$ times, then succeed.

Mean and Standard Deviation

$\mu_X = E(X) = \frac{1}{p}$

$\sigma_X = \frac{\sqrt{1-p}}{p}$

Interpretation of the mean: On average, it takes $\frac{1}{p}$ trials to get the first success.

Cumulative Probabilities

$P(X \leq k) = 1 - (1-p)^k$

$P(X > k) = (1-p)^k$

This is useful: the probability that you have to wait more than $k$ trials is simply $(1-p)^k$.

Calculator

• geometpdf($p$, $k$): Probability of first success on trial $k$
• geometcdf($p$, $k$): Probability of first success on or before trial $k$

Shape

The geometric distribution is always skewed right — most of the probability is concentrated on small values of $X$.

Example

A baseball player has a batting average of .300 (probability of a hit). What's the probability his first hit comes on his 3rd at-bat?

$P(X = 3) = (0.7)^2(0.3) = 0.147$

What's the expected number of at-bats until his first hit?

$E(X) = \frac{1}{0.3} \approx 3.33 \text{ at-bats}$

AP Tip: Remember that geometric distributions count the trial number OF the first success (including the success), not the number of failures before the first success.