Geometric Distribution

Probability of first success on nth trial

Geometric Distribution

When to Use Geometric

Geometric distribution: Models number of trials until first success

Conditions:

Binary: Each trial has two outcomes (success/failure)
Independent: Trials independent
Same: Probability of success (p) constant
Trials until success: Count trials up to and including first success

Key difference from Binomial: Number of trials NOT fixed (varies randomly)

Notation: X ~ Geometric(p)

Geometric Probability Formula

Probability first success occurs on trial k:

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

Interpretation:

Fail (k-1) times: probability (1-p)^(k-1)
Succeed on kth trial: probability p
Multiply together

Example 1: Rolling a Six

Roll die until get a 6. Find P(first 6 on roll 3).

p = 1/6 (probability of rolling 6)

$P(X = 3) = (5/6)^2 \cdot (1/6) = (25/36)(1/6) = 25/216 \approx 0.116$

Interpretation: Fail twice, succeed third time

Example 2: Free Throws

Player makes 80% of free throws. Find P(first make on shot 4).

p = 0.8

$P(X = 4) = (0.2)^3 (0.8) = 0.008(0.8) = 0.0064$

Interpretation: Miss first 3 (unlikely for 80% shooter!), make 4th

Mean and Standard Deviation

Mean (Expected Value):

$\mu_X = \frac{1}{p}$

Standard Deviation:

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

Example: Roll die until 6 (p = 1/6)

$\mu_X = \frac{1}{1/6} = 6$

$\sigma_X = \frac{\sqrt{5/6}}{1/6} = \sqrt{5/6} \cdot 6 = 6\sqrt{5/6} \approx 5.48$

Interpretation: On average, takes 6 rolls to get first 6

Cumulative Probabilities

P(X ≤ k): First success within k trials

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

Logic: Complement of "fail all k trials"

Example: Die rolling, P(first 6 within 3 rolls)

$P(X \leq 3) = 1 - (5/6)^3 = 1 - 125/216 = 91/216 \approx 0.421$

P(X > k): More than k trials needed

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

Example: P(need more than 3 rolls for first 6)

$P(X > 3) = (5/6)^3 = 125/216 \approx 0.579$

Calculator Commands (TI-83/84)

geometpdf(p, k): P(X = k)

Example: geometpdf(1/6, 3)

geometcdf(p, k): P(X ≤ k)

Example: geometcdf(1/6, 3)

Access: 2nd VARS (DISTR) → geometpdf or geometcdf

Geometric vs Binomial

Binomial:

Fixed n trials
Count successes (X can be 0 to n)
Question: "How many successes in n trials?"

Geometric:

Variable trials (until first success)
X = trial number of first success (1, 2, 3, ...)
Question: "How many trials until first success?"

Example distinguishing:

"Flip 10 coins, count heads" → Binomial
"Flip until first heads" → Geometric

Memoryless Property

Unique property of geometric distribution:

$P(X > a + b | X > a) = P(X > b)$

Interpretation: If already waited a trials without success, probability of waiting b more trials same as starting fresh

Example: Rolling die

P(wait more than 6 rolls | already waited 3) = P(wait more than 3 rolls)
Past failures don't affect future (each roll independent)

Applications

Manufacturing: Inspecting items until find defect
Quality control: Testing until failure
Gaming: Playing until win
Biology: Trials until mutation occurs
Sports: At-bats until hit

Example 3: Quality Control

2% of widgets are defective. Inspect widgets one at a time.

(a) Expected number inspected until find defect?

$\mu = \frac{1}{p} = \frac{1}{0.02} = 50$

Expect to inspect 50 widgets on average

(b) P(find defect within 10 inspections)?

$P(X \leq 10) = 1 - (0.98)^{10} = 1 - 0.817 = 0.183$

(c) P(need more than 100 inspections)?

$P(X > 100) = (0.98)^{100} \approx 0.133$

Probability Distribution Graph

For Geometric(0.3):

Always right-skewed (starts at X=1)
Decreasing probabilities (most likely: X=1)
Long right tail (theoretically infinite)

For Geometric(0.8):

Strongly concentrated at X=1
Rapid decrease (high p means quick success likely)

Relationship to Binomial

For small p: Geometric and binomial related

If X ~ Geometric(p), then after n trials, number of successes ~ Binomial(n, p)

But they answer different questions!

Common Mistakes

❌ Using geometric when trials are fixed (should use binomial)
❌ Starting count at 0 instead of 1 (X=1 is first trial)
❌ Confusing P(X = k) with P(X ≤ k)
❌ Forgetting memoryless property
❌ Wrong mean formula (it's 1/p, not p)

Practice Strategy

Identify: Trials until first success? (→ Geometric)
Find p: Probability of success each trial
Determine question: Exactly k trials? At most k? More than k?
Apply formula or use calculator
Interpret: Does answer make sense?

Quick Reference

Use when: Counting trials until first success

Probability: $P(X = k) = (1-p)^{k-1} p$

Mean: $\mu = \frac{1}{p}$

SD: $\sigma = \frac{\sqrt{1-p}}{p}$

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

Calculator:

geometpdf(p, k) for P(X = k)
geometcdf(p, k) for P(X ≤ k)

Remember: Geometric counts trials until first success. Mean = 1/p makes sense: if p=0.5, expect success on trial 2 on average!

📚 Practice Problems

1Problem 1easy

❓ Question:

You roll a fair die until you get a 6. What is the probability it takes exactly 3 rolls? What is the probability it takes at most 3 rolls?

💡 Show Solution

Step 1: Identify as geometric distribution Geometric: Number of trials until first success

Success = rolling a 6
p = 1/6 (probability of success)
X = number of rolls until first 6

Step 2: Find P(X = 3) Formula: P(X = k) = (1-p)^(k-1) · p

This means:

First k-1 trials are failures: (1-p)^(k-1)
kth trial is success: p

P(X = 3) = (1 - 1/6)^(3-1) · (1/6) = (5/6)² · (1/6) = (25/36) · (1/6) = 25/216 ≈ 0.116

Step 3: Interpret P(X = 3) This means: Fail, Fail, Success (on roll 3) P(not 6, not 6, 6) = (5/6) · (5/6) · (1/6) = 25/216 ✓

Step 4: Find P(X ≤ 3) P(X ≤ 3) = P(X = 1) + P(X = 2) + P(X = 3)

P(X = 1) = (5/6)⁰ · (1/6) = 1/6

P(X = 2) = (5/6)¹ · (1/6) = 5/36

P(X = 3) = 25/216 (from above)

Step 5: Sum the probabilities Convert to common denominator 216: P(X = 1) = 1/6 = 36/216 P(X = 2) = 5/36 = 30/216 P(X = 3) = 25/216

P(X ≤ 3) = 36/216 + 30/216 + 25/216 = 91/216 ≈ 0.421

Step 6: Alternative formula for P(X ≤ k) P(X ≤ k) = 1 - (1-p)^k = 1 - (5/6)³ = 1 - 125/216 = 91/216 ✓

Answer: P(X = 3) = 25/216 ≈ 0.116 or 11.6% P(X ≤ 3) = 91/216 ≈ 0.421 or 42.1%

There's about an 11.6% chance it takes exactly 3 rolls, and about a 42.1% chance it takes at most 3 rolls to get the first 6.

2Problem 2easy

❓ Question:

A basketball player makes 60% of free throws. What is the expected number of attempts until she makes her first basket?

💡 Show Solution

Step 1: Identify as geometric distribution X = number of attempts until first success p = 0.60 (probability of success on each attempt)

Step 2: Use expected value formula For geometric distribution: E(X) = μ = 1/p

E(X) = 1/0.60 = 1/(3/5) = 5/3 ≈ 1.67

Step 3: Interpret the result On average, it takes about 1.67 attempts until first made basket.

This makes sense:

If she makes 60%, she succeeds more often than not
Should take less than 2 attempts on average
1.67 is between 1 and 2 ✓

Step 4: Why this formula? E(X) = Σ k · P(X = k) = 1·p + 2·(1-p)p + 3·(1-p)²p + ...

Through calculus/algebra, this series sums to 1/p

Intuitive: If p = 0.5, expect 2 attempts (1/0.5 = 2) If p = 0.25, expect 4 attempts (1/0.25 = 4)

Step 5: Calculate standard deviation For geometric distribution: σ = √[(1-p)/p²]

σ = √[(0.40)/(0.60)²] = √[0.40/0.36] = √[10/9] ≈ 1.05

Step 6: Interpret standard deviation Typical deviation from mean is about 1.05 attempts High variability: could get it first try (X = 1) Or might take many attempts (X could be large)

Answer: E(X) = 5/3 ≈ 1.67 attempts

On average, it takes about 1.67 attempts for her to make her first free throw. The formula is E(X) = 1/p = 1/0.60 ≈ 1.67.

3Problem 3medium

❓ Question:

A factory production line produces 5% defective items. What is the probability that the first defective item is found within the first 10 items inspected?

💡 Show Solution

Step 1: Set up geometric distribution X = number of items inspected until first defective p = 0.05 (probability item is defective)

Find: P(X ≤ 10)

Step 2: Use CDF formula P(X ≤ k) = 1 - (1-p)^k

P(X ≤ 10) = 1 - (1 - 0.05)¹⁰ = 1 - (0.95)¹⁰

Step 3: Calculate (0.95)¹⁰ (0.95)¹⁰ ≈ 0.5987

Step 4: Complete calculation P(X ≤ 10) = 1 - 0.5987 = 0.4013

Step 5: Interpret P(first defective within 10 items) ≈ 0.401 or 40.1%

About 40% chance the first defective shows up in first 10 items.

Equivalently: About 60% chance all first 10 items are good! P(all 10 good) = (0.95)¹⁰ ≈ 0.60

Step 6: Find expected waiting time E(X) = 1/p = 1/0.05 = 20

On average, expect to inspect 20 items before finding first defective.

So inspecting only 10 is "less than average" - makes sense why probability is less than 50%.

Step 7: What if we wanted P(X > 10)? P(X > 10) = 1 - P(X ≤ 10) = 1 - 0.4013 = 0.5987

Or directly: P(X > 10) = (1-p)¹⁰ = (0.95)¹⁰

This is probability we DON'T find defective in first 10 = probability all first 10 are good!

Answer: P(X ≤ 10) ≈ 0.401 or 40.1%

There's about a 40% chance of finding the first defective item within the first 10 inspections. Since the expected position of the first defective is 20, finding it within the first 10 is better than average.

4Problem 4medium

❓ Question:

Compare: Geometric vs Binomial. You flip a coin. For geometric, find P(first H on flip 4). For binomial with 4 flips, find P(exactly 1 H). Are they the same?

💡 Show Solution

Step 1: Set up geometric problem Geometric: X = flip number of first heads p = 0.5 Find: P(X = 4)

Step 2: Calculate geometric probability P(X = 4) = (1-p)^(4-1) · p = (0.5)³ · (0.5) = (0.5)⁴ = 1/16 = 0.0625

Interpretation: Flip sequence is TTTH

Step 3: Set up binomial problem Binomial: Y = number of heads in 4 flips n = 4, p = 0.5 Find: P(Y = 1)

Step 4: Calculate binomial probability P(Y = 1) = C(4,1) · (0.5)¹ · (0.5)³ = 4 · (0.5)⁴ = 4 · 1/16 = 4/16 = 1/4 = 0.25

Interpretation: Exactly one H in any position (HTTT, THTT, TTHT, TTTH)

Step 5: Compare the results Geometric: P(X = 4) = 1/16 = 0.0625 Binomial: P(Y = 1) = 4/16 = 0.25

They are NOT the same! Binomial is 4 times larger.

Step 6: Why are they different? Geometric counts specific sequence: TTTH (first H on flip 4) Only 1 sequence

Binomial counts ANY sequence with exactly 1 H:

HTTT (first H on flip 1)
THTT (first H on flip 2)
TTHT (first H on flip 3)
TTTH (first H on flip 4) Total: 4 sequences

Notice: Geometric probability is 1/4 of binomial!

Step 7: Key differences between distributions GEOMETRIC:

Question: "Which trial is the first success?"
Counts trial NUMBER
Variable number of trials (could go forever)
P(X = k) = (1-p)^(k-1) · p

BINOMIAL:

Question: "How many successes in n trials?"
Counts NUMBER of successes
Fixed number of trials (n)
P(Y = k) = C(n,k) · p^k · (1-p)^(n-k)

Step 8: Relationship If geometric X = 4 (first success on trial 4):

Binomial with n=4 has Y ≥ 1 (at least one success)
But Y could be 1, 2, 3, or 4

If binomial Y = 1 (exactly one success in 4 trials):

Geometric X could be 1, 2, 3, or 4 (where that success occurred)

Answer: Geometric: P(X = 4) = 0.0625 (first H on flip 4) Binomial: P(Y = 1) = 0.25 (exactly 1 H in 4 flips)

NO, they are not the same. Geometric asks for a specific sequence (TTTH), while binomial asks for any sequence with exactly one success. Binomial is 4 times larger because there are 4 possible positions for the single head.

5Problem 5hard

❓ Question:

A student takes a 4-choice multiple choice test by guessing. What is the probability she doesn't get a correct answer until question 5 or later? If she keeps guessing, how many questions should we expect her to answer before getting one right?

💡 Show Solution

Step 1: Set up geometric distribution X = question number of first correct answer p = 1/4 = 0.25 (probability of guessing correctly)

Step 2: Find P(X ≥ 5) P(X ≥ 5) = P(first 4 are all wrong) = (1 - p)⁴ = (3/4)⁴ = 81/256 ≈ 0.316

Alternative using complement: P(X ≥ 5) = 1 - P(X ≤ 4) = 1 - [1 - (1-p)⁴] = (1-p)⁴ = (3/4)⁴ ✓

Step 3: Interpret P(X ≥ 5) About 31.6% chance she doesn't get any correct in first 4 questions.

This is fairly high! With only 25% success rate, often takes several tries.

Step 4: Calculate expected value E(X) = 1/p = 1/(1/4) = 4

On average, expect to answer 4 questions before first correct answer.

Step 5: Calculate standard deviation σ = √[(1-p)/p²] = √[(3/4)/(1/4)²] = √[(3/4)/(1/16)] = √[12] = 2√3 ≈ 3.46

Step 6: Interpret mean and SD Mean = 4: On average, 4th question is first correct SD ≈ 3.46: High variability!

Could get lucky and answer correctly on question 1 Or might take 8, 10, or more questions

Rough rule: Most values within μ ± 2σ = 4 ± 6.92 = roughly between 1 and 11 questions

Step 7: Memoryless property Interesting geometric property: If she's gotten 10 wrong in a row, probability next one is correct is STILL 1/4!

Past failures don't help future attempts. P(X = 11 | X > 10) = P(X = 1) = p

Step 8: Verify with simulation thinking Out of 100 students:

~25 get question 1 correct (X = 1)
~19 get question 2 correct (X = 2): 75 missed first, 25% of those get second
~14 get question 3 correct (X = 3)
~11 get question 4 correct (X = 4)
~31 don't get correct until 5+ (X ≥ 5) ✓

Answer: P(X ≥ 5) = 81/256 ≈ 0.316 or 31.6% E(X) = 4 questions

There's about a 31.6% chance she doesn't get a correct answer until question 5 or later. On average, we'd expect her to answer 4 questions before getting one correct.

🎴

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