The Geometric Distribution

By the end of this lesson, you will be able to identify geometric settings, calculate geometric probabilities, and find the mean of a geometric random variable.

The Geometric Setting

A geometric distribution models the number of trials until the first success in repeated independent Bernoulli trials.

Conditions (BINS without the fixed N):

• B: Binary outcomes (success/failure)
• I: Independent trials
• S: Same probability of success p on each trial
• The number of trials is NOT fixed — you keep going until you get a success

Key difference from binomial: In binomial, n is fixed and X counts successes. In geometric, you count trials until the first success — there's no fixed n.

Geometric Probability Formula

If X = number of trials until the first success:

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

Interpretation: the first $k-1$ trials are failures, and the $k$th trial is a success.

Example: A basketball player makes 60% of free throws. What is the probability she first makes a shot on the 3rd attempt?

$P(X = 3) = (1-0.6)^{3-1}(0.6) = (0.4)^2(0.6) = 0.16 \times 0.6 = 0.096$

Cumulative Geometric Probabilities

P(X > k): probability that the first success occurs after trial k (all of the first k are failures):

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

P(X ≤ k): probability that the first success occurs on or before trial k:

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

Example: For p = 0.6:

• P(X > 3) = (0.4)³ = 0.064
• P(X ≤ 3) = 1 − 0.064 = 0.936

Mean (Expected Value)

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

Interpretation: on average, it takes $1/p$ trials to get the first success.

Example: A 60% free throw shooter needs an average of $1/0.6 \approx 1.67$ attempts to make her first shot.

A 10% shooter needs an average of $1/0.10 = 10$ attempts. Makes intuitive sense.

Note: There is no closed-form formula for the variance of a geometric distribution on the AP exam, though it equals $(1-p)/p^2$.

Geometric vs. Binomial — Comparison

Feature	Binomial	Geometric