Geometric Distribution
Use the geometric distribution to model the number of trials until the first success.
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🎲 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)
- Bernoulli trials: each trial has two outcomes (success/failure)
- Independence: trials are independent
- Same probability: probability p of success is constant on each trial
Fixed n:NOT fixed — we stop when we get first success
Probability Formula
For X = number of trials until first success:
where:
- = 1, 2, 3, ... (the trial number on which first success occurs)
- = probability of success on each trial
- = (k−1) failures before the first success
Mean and Standard Deviation
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:
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: attempts on average
Common Mistakes
- Using k instead of (k−1): Always use , not
- Confusing with binomial: "First success on trial k" is geometric, "k successes in n trials" is binomial
- Forgetting that k starts at 1: On the first trial (k=1), (immediate success)
- 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: 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.
⚠️ Common Mistakes: Geometric Distribution
Avoid these 3 frequent errors
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