The Geometric Distribution - AP Statistics Study Guide

Quick Summary

This guide covers the geometric distribution, a probability model used for random variables that count the number of trials required to achieve the first success in a series of independent trials. After mastering this lesson, you will be able to identify a geometric setting, calculate the probability that the first success occurs on a specific trial, and find the expected value and standard deviation for any geometric random variable. This is a foundational skill for understanding discrete probability distributions.

Key Concepts

The geometric distribution models a very specific scenario: how long do we have to wait for something to happen for the first time? Think of flipping a coin until you get the first heads, or rolling a die until you get the first 6.

The Conditions for a Geometric Setting

For a situation to be modeled by a geometric distribution, it must satisfy four conditions. A helpful mnemonic is B.I.T.S.:

• Binary: Each trial can be classified into one of two outcomes: success or failure.

• Independent: The outcome of one trial does not affect the outcome of any other trial.

• Trials until success: The goal is to count the number of trials it takes to get the first success. This is the key feature that distinguishes it from the binomial distribution.

• Same probability: The probability of success, denoted by p, is the same for every trial.

The random variable in a geometric setting, X, is defined as the number of trials required to obtain the first success. The possible values for X are {1, 2, 3, ...}. Notice that there is no upper limit; it's theoretically possible to wait forever for a success.

Formulas for the Geometric Distribution

Let X be a geometric random variable with probability of success p.

• Probability of First Success on the k-th Trial (PMF)

  To find the probability that the first success occurs on exactly the k-th trial, we need to have $k-1$ failures followed by one success.

  Formula:$P(X=k)=(1-p{)}^{(}k-1)\ast p$

  • $(1-p)$ is the probability of failure.

  • $(k-1)$ is the number of failures before the first success.

  • $p$ is the probability of the one success on the k-th trial.

• Cumulative Probability

  Often, we want to know the probability that the first success occurs on or before a certain trial.

  Formula:$P(X\le k)=1-(1-p{)}^k$

  This formula calculates the probability of not having k failures in a row. A very useful related formula is for calculating the probability that it takes more than k trials to get the first success. This is equivalent to having k consecutive failures.

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

• Mean (Expected Value)

  The mean is the average number of trials we would expect to perform to get the first success.

  Formula:${\mu }_X=E(X)=1/p$

  • Intuition: If the probability of success is 1/10, you would expect, on average, to wait 10 trials for the first success.

• Standard Deviation

  The standard deviation measures the typical variation in the number of trials needed for the first success.

  Formula:${\sigma }_X=sqrt((1-p)/p^2)$

Shape of a Geometric Distribution

The probability distribution of a geometric random variable is always skewed to the right. The most likely outcome is that the first success occurs on the first trial (X=1). The probability of success on each subsequent trial decreases, creating a long tail to the right.

[Image: A right-skewed histogram representing a geometric distribution, with the x-axis labeled 'Number of Trials until First Success (k)' and the y-axis labeled 'Probability P(X=k)'. The bar at k=1 is the tallest, and the bars decrease in height as k increases.]

Key Vocabulary

• Geometric Random Variable: A variable that counts the number of trials required to obtain the first success in a sequence of independent Bernoulli trials.

• Geometric Probability Distribution: The probability model describing the chances of the first success occurring on a specific trial $k$.

• Trial: A single performance of a chance experiment with two or more outcomes (e.g., a single coin flip).

• Success: The specific outcome of interest in a trial.

• Expected Value (Mean): The long-run average value of a random variable over many repetitions of the experiment. For a geometric distribution, it is $1/p$.

• Skewed Right Distribution: A probability distribution where the tail on the right side is longer or fatter than the left side. The mean is typically greater than the median.

Calculator Tech (TI-84)

Your TI-84 calculator has built-in functions that make geometric calculations simple. These are found in the distribution menu.

Access the menu:2nd -> VARS [DISTR] (scroll down for geometric functions)

1. To find P(X = k): $geometpdf($

   • Use this to find the probability of the first success occurring on exactly the k-th trial.

   • Syntax:$geometpdf(p,k)$

   • p: The probability of success on a single trial.

   • k: The specific trial number for the first success.

   • Example: To find the probability that the first success occurs on the 4th trial when p=0.25:

     2nd -> VARS [DISTR] -> E:geometpdf(

     p: 0.25

     x value: 4

     $Paste$ -> $geometpdf(0.25,4)$ -> ENTER -> $0.1055$

2. To find P(X \le k): $geometcdf($

   • Use this to find the probability of the first success occurring on or before the k-th trial.

   • Syntax:$geometcdf(p,k)$

   • p: The probability of success on a single trial.

   • k: The maximum trial number for the first success.

   • Example: To find the probability that the first success occurs by the 3rd trial when p=0.25:

     2nd -> VARS [DISTR] -> F:geometcdf(

     p: 0.25

     x value: 3

     $Paste$ -> $geometcdf(0.25,3)$ -> ENTER -> $0.5781$

Pro-Tip for "At Least" or "More Than" Probabilities:

• P(X > k): Probability of first success after trial k.

  • $1-geometcdf(p,k)$

• P(X \ge k): Probability of first success on or after trial k.

  • $1-geometcdf(p,k-1)$

How to Show Work on the FRQ

To earn full credit for a geometric distribution problem on the AP exam, you must clearly communicate your process. Do not just write a final number. Follow this four-step process.

Step 1: Define the Variable and State the Distribution

• Clearly define the random variable $X$ in the context of the problem.

• State that $X$ follows a geometric distribution and specify the parameter $p$.

• Template: "Let X = the number of [trials described in context] until the first [success in context]. X follows a geometric distribution with a probability of success p = [value]."

Step 2: Check the Conditions

• Briefly confirm that the four B.I.T.S. conditions are met in the context of the problem.

• Template: "This is a geometric setting because:

  • Binary: Each [trial] results in a [success] or [failure].

  • Independent: The outcome of one [trial] is independent of the others because [reason, e.g., sampling with replacement, 10% condition].

  • Trials until success: We are counting the number of [trials] until the first [success].

  • Same probability: The probability of success is constant at p = [value]."

Step 3: Show the Calculation

• Write the appropriate formula with the values from the problem substituted in.

• Clearly state the probability you are calculating (e.g., $P(X=5)$ or $P(X>3)$).

• Template (for P(X=k)):$P(X=5)=(1-0.2{)}^{(}5-1)\ast (0.2)$

• Template (for Mean):$E(X)=1/0.2$

• You may use your calculator to find the final answer, but the formula setup must be shown.

Step 4: State the Conclusion in Context

• Write a clear, non-technical sentence that answers the original question and includes context.

• Template: "There is a [calculated probability] chance that the first [success in context] occurs on the [k-th] trial." or "On average, we expect it to take [calculated mean] trials to find the first [success in context]."

Practice Problems

Problem 1:

A coffee shop states that 15% of its customers order a specialty latte. Assume that customer orders are independent. What is the probability that the first customer to order a specialty latte is the 4th customer of the day?

Solution:

Step 1: Define the Variable and State the Distribution

Let X = the number of customers until the first specialty latte is ordered. X follows a geometric distribution with a probability of success p = 0.15.

Step 2: Check the Conditions

This is a geometric setting because:

• Binary: Each customer either orders a specialty latte (success) or does not (failure).

• Independent: Customer orders are stated to be independent.

• Trials until success: We are waiting for the first customer who orders a specialty latte.

• Same probability: The probability of a customer ordering a specialty latte is constant at p = 0.15.

Step 3: Show the Calculation

We want to find the probability that the first success is on the 4th trial, so we are calculating $P(X=4)$.

$P(X=4)=(1-0.15{)}^{(}4-1)\ast (0.15)$

$P(X=4)=(0.85{)}^3\ast (0.15)$

$P(X=4)=0.0921$

(Calculator check: geometpdf(p:0.15, x value:4) = 0.09211875)

Step 4: State the Conclusion in Context

There is a 9.21% probability that the fourth customer of the day is the first one to order a specialty latte.

Problem 2:

An airline has a historical on-time departure rate of 80%. Let X be the number of flights checked until the first delayed flight is found.

(a) On average, how many flights would you expect to check to find the first delayed flight?

(b) What is the probability that the first delayed flight is one of the first 3 flights checked?

Solution:

Part (a):

Step 1: Define the Variable and State the Distribution

Let X = the number of flights checked until the first delayed flight is found. A delayed flight is a "success" in this context. The probability of a flight being on time is 0.80, so the probability of a flight being delayed (success) is p = 1 - 0.80 = 0.20. X follows a geometric distribution with p = 0.20.

Step 2: Show the Calculation (for Mean)

We are asked for the expected value (mean) of X.

$E(X)=1/p$

$E(X)=1/0.20=5$

Step 3: State the Conclusion in Context

On average, we would expect to check 5 flights to find the first delayed flight.

Part (b):

Step 1: State the Goal

We want to find the probability that the first delayed flight is one of the first 3, which is $P(X\le 3)$. This is the same as $P(X=1)+P(X=2)+P(X=3)$.

Step 2: Show the Calculation

Using the cumulative probability formula:

$P(X\le 3)=1-(1-p{)}^3$

$P(X\le 3)=1-(0.80{)}^3$

$P(X\le 3)=1-0.512=0.488$

(Calculator check: geometcdf(p:0.20, x value:3) = 0.488)

Step 3: State the Conclusion in Context

There is a 48.8% probability that the first delayed flight will be found within the first 3 flights checked.

Common Mistakes to Avoid

• Confusing Geometric and Binomial: This is the most common error. Remember: Geometric = trials until first success. Binomial = count of successes in a fixed number of trials. If the problem gives you a fixed $n$, it's likely binomial. If it asks "how long until...", it's geometric.

• Incorrect Exponent in the Formula: The formula $P(X=k)=(1-p{)}^{(}k-1)\ast p$ has an exponent of $k-1$, representing the number of failures before the first success. A common mistake is to use $k$ as the exponent.

• Off-by-One Errors with Inequalities: Be very careful with "at least," "more than," and "at most."

  • $P(X\ge 5)$ (at least 5) = $1-P(X\le 4)$.

  • $P(X>5)$ (more than 5) = $1-P(X\le 5)$.

  • Failing to subtract 1 from $k$ when using $geometcdf$ for "at least" calculations is a frequent error.

• Forgetting to Define 'Success': Sometimes a "success" is a negative event, like finding a defective item or a delayed flight (as in Problem 2). Always clearly define what a success means for your problem and correctly identify its probability, $p$.

• Using $np$ for the Mean: The formula for the mean of a binomial distribution is $\mu =np$. For a geometric distribution, the mean is simply $\mu =1/p$. Do not mix them up.