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The Geometric Distribution

The geometric distribution is a discrete probability distribution that models the number of trials needed to achieve the first success in a sequence of Independent Events and Unions of Events|independent Bernoulli trials. Unlike the Introduction to the Binomial Distribution|Binomial Distribution which counts the number of successes in a fixed number of trials, the geometric distribution focuses on the waiting time for that very first success.

Conditions for a Geometric Setting (BITS)

For a random variable $ X $ to follow a geometric distribution, the following conditions must be met:

• Binary: Each trial has only two possible outcomes: “success” or “failure.”
• Independent: The outcomes of each trial are independent of each other.
• Trials: We are counting the number of trials until the first success occurs.
• Success: The probability of success ( $ p $ ) remains the same for each trial.

If these conditions are met, $ X $ is a geometric random variable, denoted as $ X \sim \text{Geom}(p) $ .

Probability Mass Function (PMF)

The probability that the first success occurs on the $ k $ -th trial is given by:

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

• $ k $ is the number of the trial on which the first success occurs ( $ k = 1, 2, 3, \ldots $ ).
• $ p $ is the probability of success on any given trial.
• $ (1-p) $ is the probability of failure on any given trial.
• $ (1-p)^{k-1} $ represents $ k-1 $ consecutive failures before the first success.

Example: Suppose the probability of hitting a target is $ p=0.2 $ . What is the probability that the first hit occurs on the 4th shot? $ P(X=4) = (1-0.2)^{4-1} (0.2) = (0.8)^3 (0.2) = 0.512 \times 0.2 = 0.1024 $ .

Mean (Expected Value) of a Geometric Distribution

The mean or expected number of trials until the first success is:

$$ E(X) = \mu_X = \frac{1}{p} $$
This makes intuitive sense: if the probability of success is $ p=0.25 $ , you’d expect to wait $ 1/0.25 = 4 $ trials on average to get a success.

Standard Deviation of a Geometric Distribution

The standard deviation of a geometric distribution is:

$$ \sigma_X = \sqrt{\frac{1-p}{p^2}} $$

Cumulative Probability

We are often interested in the probability that the first success occurs on or before a certain trial, or after a certain trial.

• Probability of first success on or before the $ k $ -th trial: $ P(X \le k) = P(X=1) + P(X=2) + \dots + P(X=k) $ This can also be calculated as: $$ P(X \le k) = 1 - P(X > k) = 1 - (1-p)^k $$ This is because $ P(X > k) $ means the first $ k $ trials were all failures.

• Probability of first success after the $ k $ -th trial: $$ P(X > k) = (1-p)^k $$ This means the first $ k $ trials were all failures.

Comparing Geometric vs. Binomial

It’s crucial to distinguish between geometric and binomial distributions.

| Feature | Introduction to the Binomial Distribution|Binomial Distribution | The Geometric Distribution | | :————– | :———————————— | :———————– | | Random Variable | Number of successes in a fixed number of trials ( $ n $ ) | Number of trials until the first success | | Number of Trials | Fixed ( $ n $ ) | Variable (can be infinite) | | Question Type | “How many successes in $ n $ trials?” | “How many trials until the first success?” | | Probability Function | $ P(X=k) = \binom{n}{k} p^k (1-p)^{n-k} $ | $ P(X=k) = (1-p)^{k-1} p $ |

Calculator Functions (TI-84)

For calculations, your graphing calculator has built-in functions:

• geometpdf(p, k): Calculates $ P(X=k) $ (probability of first success on the $ k $ -th trial).
• geometcdf(p, k): Calculates $ P(X \le k) $ (probability of first success on or before the $ k $ -th trial).

Expected Counts in Two-Way Tables|Example Application

Imagine you are a basketball player with a 30% free throw percentage ( $ p=0.3 $ ).

1. What is the probability that your first successful free throw occurs on your 3rd attempt? This is $ P(X=3) $ . $ P(X=3) = (1-0.3)^{3-1} (0.3) = (0.7)^2 (0.3) = 0.49 \times 0.3 = 0.147 $ . Using calculator: geometpdf(0.3, 3) = 0.147.

2. What is the expected number of free throws you will take until your first success? $ E(X) = \frac{1}{p} = \frac{1}{0.3} \approx 3.33 $ attempts.

3. What is the probability that it takes you more than 5 attempts to make your first free throw? This is $ P(X > 5) $ . $ P(X > 5) = (1-0.3)^5 = (0.7)^5 = 0.16807 $ .

4. What is the probability that your first success occurs on or before your 4th attempt? This is $ P(X \le 4) $ . $ P(X \le 4) = 1 - (1-0.3)^4 = 1 - (0.7)^4 = 1 - 0.2401 = 0.7599 $ . Using calculator: geometcdf(0.3, 4) = 0.7599.