Geometric Probability Distribution: Complete Guide with Examples
What is Geometric Distribution?
The geometric probability distribution models the number of trials needed to achieve the first success in repeated independent trials. It answers: "What is the probability that the first success occurs on the \(x\)-th trial?" The random variable \(X\) represents the trial number of the first success, making this a fundamental discrete probability model.
Conditions for Geometric Distribution
- Independence: Trial outcomes don't affect each other
- Binary Outcomes: Each trial has success or failure (mutually exclusive)
- Constant Success Probability: Probability \(p\) remains unchanged
- Constant Failure Probability: Probability \(1-p\) remains unchanged
Geometric Distribution Formula
The probability that the first success occurs on trial \(x\) is:
\[
P(X = x) = (1-p)^{x-1} p \quad \text{for} \quad x = 1, 2, 3, \dots
\]
Formula Derivation
For first success on trial \(x\):
- First \(x-1\) trials must fail: probability \((1-p)^{x-1}\)
- Trial \(x\) must succeed: probability \(p\)
- Multiply using multiplication rule for independent events
Key Properties
- Mean (Expected Value): \(\displaystyle \mu = \frac{1}{p}\)
- Variance: \(\displaystyle \sigma^2 = \frac{1-p}{p^2}\)
- Standard Deviation: \(\displaystyle \sigma = \sqrt{\frac{1-p}{p^2}}\)
Worked Examples with Solutions
Example 1: Coin Toss Problem
A fair coin is tossed repeatedly until the first tail appears.
- Find probability first tail appears on 5th toss
- Calculate mean and standard deviation
- Plot distribution for first 10 trials
Solution
- Success = tail, \(p = 0.5\):
\[
P(X = 5) = (0.5)^{4} \times 0.5 = \frac{1}{32} = 0.03125
\]
- Mean: \(\mu = \frac{1}{0.5} = 2\)
Standard deviation: \(\sigma = \sqrt{\frac{0.5}{0.25}} = \sqrt{2} \approx 1.41\)
- Distribution: \(P(X = x) = 0.5^x\) for \(x = 1\) to \(10\)
Sum of first 10 probabilities: \(\sum_{x=1}^{10} P(X=x) \approx 0.999\)
Example 2: Population Sampling
45% of adults have post-secondary degrees. People are randomly selected.
- Probability third person is first with degree?
- Probability first degree-holder found by 4th selection?
Solution
- Success = has degree, \(p = 0.45\):
\[
P(X = 3) = (0.55)^2 \times 0.45 \approx 0.1361
\]
- Use cumulative probability:
\[
\begin{aligned}
P(X \le 4) &= P(X=1) + P(X=2) + P(X=3) + P(X=4) \\
&= 0.45 + 0.2475 + 0.1361 + 0.0749 \\
&\approx 0.9085
\end{aligned}
\]
Cumulative Probability Formulas
Using geometric series, we derive efficient formulas:
- Success on or before trial \(n\): \(P(X \le n) = 1 - (1-p)^n\)
- Success before trial \(n\): \(P(X < n) = 1 - (1-p)^{n-1}\)
- Success on or after trial \(n\): \(P(X \ge n) = (1-p)^{n-1}\)
- Success after trial \(n\): \(P(X > n) = (1-p)^n\)
Example 3: Deriving Cumulative Formulas
For a geometric distribution with success probability \(p\), derive formulas for:
- \(P(X \le n)\)
- \(P(X < n)\)
- \(P(X \ge n)\)
- \(P(X > n)\)
Solution
- \(P(X \le n) = \sum_{x=1}^{n} p(1-p)^{x-1} = \frac{p(1 - (1-p)^n)}{1-(1-p)} = 1 - (1-p)^n\)
- \(P(X < n) = P(X \le n-1) = 1 - (1-p)^{n-1}\)
- \(P(X \ge n) = 1 - P(X < n) = 1 - [1 - (1-p)^{n-1}] = (1-p)^{n-1}\)
- \(P(X > n) = 1 - P(X \le n) = 1 - [1 - (1-p)^n] = (1-p)^n\)
Example 4: Quality Control Application
A company produces tools where 99% are non-defective. Tools are randomly tested.
- Probability the second tool is the first non-defective?
- Probability first non-defective found by second test?
- Probability first non-defective found after 10 tests?
Solution
Success = non-defective, \(p = 0.99\)
- \(P(X = 2) = (0.01)^1 \times 0.99 = 0.0099\)
- Using cumulative formula: \(P(X \le 2) = 1 - (0.01)^2 = 0.9999\)
- Using complement formula: \(P(X > 10) = (0.01)^{10} = 10^{-20} \approx 0\)
Key Takeaways
- Geometric distribution models "waiting time" until first success
- Core formula: \(P(X = x) = (1-p)^{x-1}p\)
- Mean trials needed: \(\mu = \frac{1}{p}\)
- Use cumulative formulas for efficiency
- Applies when trials are independent with constant \(p\)
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