Skip to content

Geometric Distribution

Use the geometric distribution to model the number of trials until the first success.

Written and reviewed by the Study Mondo Education TeamLast updated
🎯⭐ INTERACTIVE LESSON

Try the Interactive Version!

Learn step-by-step with practice exercises built right in.

Start Interactive Lesson →

🎲 Geometric Distribution

When to Use Geometric Distribution

The geometric distribution models the number of trials needed to achieve the first success in a sequence of independent Bernoulli trials.

Key Difference from Binomial:

  • Binomial: Fixed number of trials, count successes (X = # successes in n trials)
  • Geometric: Fixed target (1 success), count trials until we get it

Conditions for Geometric Distribution (BIS minus Fixed n)

  1. Bernoulli trials: each trial has two outcomes (success/failure)
  2. Independence: trials are independent
  3. Same probability: probability p of success is constant on each trial
  4. Fixed n: NOT fixed — we stop when we get first success

Probability Formula

For X = number of trials until first success:

P(X=k)=(1−p)k−1⋅pP(X = k) = (1 - p)^{k-1} \cdot p

where:

  • kk = 1, 2, 3, ... (the trial number on which first success occurs)
  • pp = probability of success on each trial
  • (1−p)k−1(1 - p)^{k-1} = (k−1) failures before the first success

Mean and Standard Deviation

E(X)=1pE(X) = \frac{1}{p}

SD(X)=1−pp2=1−ppSD(X) = \sqrt{\frac{1-p}{p^2}} = \frac{\sqrt{1-p}}{p}

Worked Example

A basketball player makes 60% of free throws. What is the probability that her first made free throw occurs on the 3rd attempt?

Given: p = 0.60, k = 3

Calculation: P(X=3)=(0.40)3−1⋅(0.60)=(0.40)2⋅0.60=0.16⋅0.60=0.096P(X = 3) = (0.40)^{3-1} \cdot (0.60) = (0.40)^2 \cdot 0.60 = 0.16 \cdot 0.60 = 0.096

Interpretation: There is a 9.6% chance her first made free throw occurs on the 3rd attempt (meaning she misses the first two and makes the third).

Mean: E(X)=10.60≈1.67E(X) = \frac{1}{0.60} \approx 1.67 attempts on average

Common Mistakes

  1. Using k instead of (k−1): Always use (1−p)k−1(1-p)^{k-1}, not (1−p)k(1-p)^k
  2. Confusing with binomial: "First success on trial k" is geometric, "k successes in n trials" is binomial
  3. Forgetting that k starts at 1: On the first trial (k=1), P(X=1)=pP(X=1) = p (immediate success)
  4. Not recognizing when to use it: Look for phrases like "first," "until success," "wait for"

Decision Rule / When to Apply

Use geometric distribution when:

  • Asked for probability of first success occurring on trial k
  • Asked for expected number of trials until first success
  • Sampling until you find one item with a desired property

AP Exam Tip

Geometric problems often appear in free-response questions about quality control, customer service (first complaint), or medical applications (first positive test). Watch for cumulative probability questions: P(X≤k)P(X \leq k) requires summing individual probabilities or recognizing the cumulative geometric formula.

📚 Practice Problems

1Problem 1easy

❓ Question:

A player makes a free throw with probability 0.8. What is the probability their first miss occurs on the 3rd attempt?

💡 Show Solution

Geometric: first two attempts succeed (make), third fails (miss). P(X=3) = (0.8)² × (0.2) = 0.64 × 0.2 = 0.128. There is a 12.8% chance the first miss happens on shot 3.

2Problem 2medium

❓ Question:

A quality-control inspector finds a defective item with probability 0.05 per item inspected. Find E(X) where X is the trial number of the first defect found.

💡 Show Solution

For geometric with parameter p = 0.05, the expected number of trials until first success is E(X) = 1/p. E(X) = 1/0.05 = 20. On average, the inspector will inspect 20 items before finding the first defective one.

3Problem 3hard

❓ Question:

A roulette wheel lands on red with probability 18/38. Let X = number of spins until first red. Find P(X ≤ 5) and interpret.

💡 Show Solution

P(X ≤ 5) = P(X=1) + P(X=2) + P(X=3) + P(X=4) + P(X=5). p = 18/38 ≈ 0.474, so 1-p ≈ 0.526. P(X=k) = (0.526)^{k-1}(0.474). P(X=1) ≈ 0.474, P(X=2) ≈ 0.249, P(X=3) ≈ 0.131, P(X=4) ≈ 0.069, P(X=5) ≈ 0.036. P(X≤5) ≈ 0.959. Nearly 96% chance of seeing red within the first 5 spins.

Explain using:

⚠️ Common Mistakes: Geometric Distribution

Avoid these 3 frequent errors

📌 Related Topics in Unit 4: Probability, Random Variables, and Probability Distributions

❓ Frequently Asked Questions

What is Geometric Distribution?▾
Use the geometric distribution to model the number of trials until the first success.
How can I study Geometric Distribution effectively?▾
Start by reading the study notes and working through the examples on this page. Then use the flashcards to test your recall. Practice with the 3 problems provided, checking solutions as you go. Regular review and active practice are key to retention.
Is this Geometric Distribution study guide free?▾
Yes — all study notes, flashcards, and practice problems for Geometric Distribution on Study Mondo are free to access. No account is needed.
What course covers Geometric Distribution?▾
Geometric Distribution is part of the AP Statistics course on Study Mondo, specifically in the Unit 4: Probability, Random Variables, and Probability Distributions section. You can explore the full course for more related topics and practice resources.
Are there practice problems for Geometric Distribution?▾
Yes, this page includes 3 practice problems with detailed solutions. Each problem includes a step-by-step explanation to help you understand the approach.