Geometric Distribution: Waiting for the First Success

What is the geometric distribution?

The geometric distribution models the number of independent Bernoulli trials needed to get the first success. Each trial has the same success probability $p$, and trials are independent.

A geometric random variable $X$ can take values $1, 2, 3, \ldots$ — you need at least one trial, and there is no upper limit in principle.

The formula

$P(X = k) = (1 - p)^{k - 1} \cdot p$

Read it aloud: "$k - 1$ failures in a row, then one success."

Mean (average number of trials to first success): $E(X) = \dfrac{1}{p}$

Step-by-step solution

The shooter has $p = 0.70$ per free throw. We want $P(X = 4)$, i.e. miss, miss, miss, make.

Step 1 — Identify failure and success probabilities:

• Success: $p = 0.70$
• Failure: $q = 1 - p = 0.30$

Step 2 — Build the sequence: The first three must be misses, the fourth a make. $P(\text{MMM then H}) = 0.30 \cdot 0.30 \cdot 0.30 \cdot 0.70$

Step 3 — Compute: $P(X = 4) = 0.30^3 \cdot 0.70 = 0.027 \cdot 0.70 = \boxed{0.0189}$

That's about 1.89% — reasonable, since a strong shooter usually makes it earlier.

Step 4 — Expected number of attempts: $E(X) = \dfrac{1}{0.70} \approx 1.43$

So on average they make their first free throw in about 1.4 attempts.

Verification — all probabilities sum to 1

$\sum_{k=1}^{\infty} (1-p)^{k-1} p = p \cdot \dfrac{1}{1 - (1-p)} = p \cdot \dfrac{1}{p} = 1 \checkmark$

Comparison with related distributions

• Binomial: fixes the number of trials, counts successes.
• Geometric: fixes the requirement (first success), counts trials.
• Negative binomial: generalizes geometric to the $r$-th success.

Common mistakes

• Off-by-one with the exponent. The exponent on $(1 - p)$ is $k - 1$, not $k$ — because you need $k - 1$ failures before the success on trial $k$.
• Forgetting independence. The formula only works if each trial is independent and has the same $p$. A shooter who gets tired or confident between attempts breaks the model.
• Confusing the two versions. Some textbooks define $X$ as the number of failures before the first success ($X = 0, 1, 2, \ldots$). Then $P(X = k) = (1-p)^k p$. Same idea, shifted index.

Try it in the visualization

Adjust $p$ to see the distribution re-scale: higher $p$ packs mass near $k = 1$; lower $p$ spreads it out with a long tail. The animated trial sequence plays live, showing misses and eventual success.