AP Statistics Flashcards: The Geometric Distribution

What this deck covers

This deck focuses on The Geometric Distribution, giving you a quick way to review the definitions, rules, and examples that matter most for AP Statistics.

How to use these flashcards

Work through these flashcards in short sessions. Try to answer each prompt before flipping the card, then revisit any cards you miss until the explanation feels automatic.

Flashcard 1: What does the geometric distribution assume about trial independence?

Answer: Trials are independent. Each trial outcome doesn't affect subsequent trial probabilities.

Flashcard 2: What type of variable is used in a geometric distribution?

Answer: Discrete random variable. Counts whole number trials, not continuous values.

Flashcard 3: What is the probability of success in a geometric distribution?

Answer: The constant probability $p$. Success probability remains fixed across all trials.

Flashcard 4: What is the skewness of a geometric distribution?

Answer: Positively skewed. Most probability mass at $X=1$, decreasing for higher values.

Flashcard 5: How does the probability of success $p$ affect the mean of a geometric distribution?

Answer: As $p$ increases, the mean decreases. Higher $p$ means success comes sooner on average.

Flashcard 6: State the formula for the variance of a geometric distribution.

Answer: $Var(X) = \frac{1-p}{p^2}$. Standard variance formula derived from geometric distribution theory.

Flashcard 7: Calculate $P(X \leq 5)$ for a geometric distribution with $p = 0.2$.

Answer: $P(X \leq 5) = 1 - (0.8)^5$. CDF calculation for success within 5 trials.

Flashcard 8: What is the mean (expected value) formula for a geometric distribution?

Answer: $E(X) = \frac{1}{p}$. Higher success probability means fewer expected trials needed.

Flashcard 9: Find $P(X \leq 4)$ for a geometric distribution with $p = 0.3$.

Answer: $P(X \leq 4) = 1 - (0.7)^4$. Uses CDF formula with success probability $p=0.3$.

Flashcard 10: What is the cumulative distribution function of a geometric distribution?

Answer: $P(X \leq k) = 1 - (1-p)^k$. CDF gives probability of success by trial $k$.

Flashcard 11: Find $P(X > 3)$ for a geometric distribution with $p = 0.2$.

Answer: $P(X > 3) = (1-0.2)^3$. Probability of needing more than 3 trials equals $(0.8)^3$.

Flashcard 12: For $p = 0.25$, what is the mean $E(X)$ of the distribution?

Answer: $E(X) = \frac{1}{0.25} = 4$. Expected value calculation using mean formula.

Flashcard 13: Determine $Var(X)$ for a geometric distribution with $p = 0.4$.

Answer: $Var(X) = \frac{1-0.4}{0.4^2}$. Uses standard variance formula $\frac{1-p}{p^2}$ with $p=0.4$.

Flashcard 14: Find the probability of the first success on the 4th trial with $p = 0.3$.

Answer: $P(X = 4) = (0.7)^3 \times 0.3$. Three failures $(1-0.3)^3$ then success $0.3$ on trial 4.

Flashcard 15: Which parameter defines a geometric distribution?

Answer: Probability of success $p$. Single parameter $p$ completely specifies the distribution.

Flashcard 16: Find the probability of the first success on or before the 3rd trial with $p = 0.5$.

Answer: $P(X \leq 3) = 1 - (0.5)^3$. CDF formula gives probability of success by trial 3.

Flashcard 17: Calculate $P(X \leq 5)$ for a geometric distribution with $p = 0.2$.

Answer: $P(X \leq 5) = 1 - (0.8)^5$. CDF calculation for success within 5 trials.

Flashcard 18: Identify the random variable $X$ in a geometric distribution.

Answer: The number of trials until the first success. Counts trials needed to achieve one success in independent trials.

Flashcard 19: What happens to $Var(X)$ as $p$ increases in a geometric distribution?

Answer: Variance decreases. Higher success probability reduces variability in waiting time.

Flashcard 20: What condition must be met for a geometric distribution to be applicable?

Answer: Trials are independent with constant probability $p$ of success. Each trial must have same success chance and be unaffected by others.

Flashcard 21: Express the cumulative probability $P(X > k)$ for a geometric distribution.

Answer: $P(X > k) = (1-p)^k$. Survival function giving probability of exceeding $k$ trials.

Flashcard 22: Find the probability of the first success on or before the 3rd trial with $p = 0.5$.

Answer: $P(X \leq 3) = 1 - (0.5)^3$. CDF formula gives probability of success by trial 3.

Flashcard 23: If trial success probability is 0.15, what is $E(X)$?

Answer: $E(X) = \frac{1}{0.15}$. Mean equals $\frac{1}{p}$ for geometric distribution.

Flashcard 24: What is the standard deviation of a geometric distribution with $p = 0.4$?

Answer: $\frac{\sqrt{1-0.4}}{0.4}$. Standard deviation equals $\sqrt{\frac{1-p}{p^2}}$.

Flashcard 25: State the relationship between geometric and negative binomial distributions.

Answer: Geometric is a special case of negative binomial with $r=1$. Negative binomial generalizes to $r$ successes; geometric has $r=1$.

Flashcard 26: Identify the key property of a geometric distribution related to memory.

Answer: Memoryless property. Previous failures don't influence future trial outcomes.

Flashcard 27: What is the cumulative distribution function of a geometric distribution?

Answer: $P(X \leq k) = 1 - (1-p)^k$. CDF gives probability of success by trial $k$.

Flashcard 28: How does the probability of success $p$ affect the mean of a geometric distribution?

Answer: As $p$ increases, the mean decreases. Higher $p$ means success comes sooner on average.

Flashcard 29: What is the formula for the probability mass function of a geometric distribution?

Answer: $P(X = k) = (1-p)^{k-1}p$. Standard PMF where $k$ is trial number and $p$ is success probability.

Flashcard 30: Find $P(X = 3)$ if $p = 0.4$ in a geometric distribution.

Answer: $P(X = 3) = (0.6)^2 \times 0.4$. Two failures $(0.6)^2$ followed by success $0.4$.

Flashcard 31: If $X \sim \text{Geom}(0.25)$, find $P(X = 2)$.

Answer: $P(X = 2) = (1-0.25)^1 \times 0.25$. One failure $(0.75)^1$ then success $0.25$ on trial 2.

Flashcard 32: Determine $Var(X)$ for $p = 0.2$ in a geometric distribution.

Answer: $Var(X) = \frac{1-0.2}{0.2^2}$. Standard variance formula applied with given probability.

Flashcard 33: What is the mode of a geometric distribution?

Answer: Mode = 1. First trial always has highest probability of success.

Flashcard 34: Identify the key property of a geometric distribution related to memory.

Answer: Memoryless property. Previous failures don't influence future trial outcomes.

Flashcard 35: Is the geometric distribution memoryless? Yes or No.

Answer: Yes. Past failures don't affect future success probabilities.

Flashcard 36: What is the mode of a geometric distribution?

Answer: Mode = 1. First trial always has highest probability of success.

Flashcard 37: Find the probability of the first success on the 4th trial with $p = 0.3$.

Answer: $P(X = 4) = (0.7)^3 \times 0.3$. Three failures $(1-0.3)^3$ then success $0.3$ on trial 4.

Flashcard 38: State the relationship between geometric and negative binomial distributions.

Answer: Geometric is a special case of negative binomial with $r=1$. Negative binomial generalizes to $r$ successes; geometric has $r=1$.

Flashcard 39: Find $P(X = 3)$ if $p = 0.4$ in a geometric distribution.

Answer: $P(X = 3) = (0.6)^2 \times 0.4$. Two failures $(0.6)^2$ followed by success $0.4$.

Flashcard 40: If $p = 0.3$, find $P(X > 5)$ for a geometric distribution.

Answer: $P(X > 5) = (0.7)^5$. Complement of CDF gives probability of exceeding 5 trials.

Flashcard 41: Find $P(X > 3)$ for a geometric distribution with $p = 0.2$.

Answer: $P(X > 3) = (1-0.2)^3$. Probability of needing more than 3 trials equals $(0.8)^3$.

Flashcard 42: What is the formula for the probability mass function of a geometric distribution?

Answer: $P(X = k) = (1-p)^{k-1}p$. Standard PMF where $k$ is trial number and $p$ is success probability.

Flashcard 43: Describe the shape of the probability mass function of a geometric distribution.

Answer: Decreasing exponential shape. Highest probability at $X=1$, decreasing exponentially thereafter.

Flashcard 44: What does the geometric distribution assume about trial independence?

Answer: Trials are independent. Each trial outcome doesn't affect subsequent trial probabilities.

Flashcard 45: If trial success probability is 0.15, what is $E(X)$?

Answer: $E(X) = \frac{1}{0.15}$. Mean equals $\frac{1}{p}$ for geometric distribution.

Flashcard 46: Determine $Var(X)$ for a geometric distribution with $p = 0.4$.

Answer: $Var(X) = \frac{1-0.4}{0.4^2}$. Uses standard variance formula $\frac{1-p}{p^2}$ with $p=0.4$.

Flashcard 47: If $X \sim \text{Geom}(0.25)$, find $P(X = 2)$.

Answer: $P(X = 2) = (1-0.25)^1 \times 0.25$. One failure $(0.75)^1$ then success $0.25$ on trial 2.

Flashcard 48: What type of trials are used in a geometric distribution?

Answer: Bernoulli trials. Independent trials with constant success probability.

Flashcard 49: What type of trials are used in a geometric distribution?

Answer: Bernoulli trials. Independent trials with constant success probability.

Flashcard 50: Identify the random variable $X$ in a geometric distribution.

Answer: The number of trials until the first success. Counts trials needed to achieve one success in independent trials.

Flashcard 51: What does the geometric distribution model in terms of failures?

Answer: Number of failures before the first success. Alternative interpretation: failures before achieving success.

Flashcard 52: Calculate $E(X)$ for a geometric distribution with $p = 0.5$.

Answer: $E(X) = \frac{1}{0.5} = 2$. Expected trials equals reciprocal of success probability.

Flashcard 53: Calculate $E(X)$ for a geometric distribution with $p = 0.5$.

Answer: $E(X) = \frac{1}{0.5} = 2$. Expected trials equals reciprocal of success probability.

Flashcard 54: Which parameter defines a geometric distribution?

Answer: Probability of success $p$. Single parameter $p$ completely specifies the distribution.

Flashcard 55: For $p = 0.25$, what is the mean $E(X)$ of the distribution?

Answer: $E(X) = \frac{1}{0.25} = 4$. Expected value calculation using mean formula.

Flashcard 56: Determine $Var(X)$ for $p = 0.2$ in a geometric distribution.

Answer: $Var(X) = \frac{1-0.2}{0.2^2}$. Standard variance formula applied with given probability.

Flashcard 57: Find $P(X \leq 4)$ for a geometric distribution with $p = 0.3$.

Answer: $P(X \leq 4) = 1 - (0.7)^4$. Uses CDF formula with success probability $p=0.3$.

Flashcard 58: What does the geometric distribution model in terms of failures?

Answer: Number of failures before the first success. Alternative interpretation: failures before achieving success.

Flashcard 59: Calculate the probability of the first success on the 1st trial with $p = 0.6$.

Answer: $P(X = 1) = 0.6$. Success on first trial occurs with probability $p$.

Flashcard 60: What is the main application of the geometric distribution?

Answer: Modeling the number of trials until the first success. Primary use case in probability and statistics applications.

Flashcard 61: Calculate the probability of the first success on the 1st trial with $p = 0.6$.

Answer: $P(X = 1) = 0.6$. Success on first trial occurs with probability $p$.

Flashcard 62: Describe the shape of the probability mass function of a geometric distribution.

Answer: Decreasing exponential shape. Highest probability at $X=1$, decreasing exponentially thereafter.

Flashcard 63: Express the cumulative probability $P(X > k)$ for a geometric distribution.

Answer: $P(X > k) = (1-p)^k$. Survival function giving probability of exceeding $k$ trials.

Flashcard 64: Is the geometric distribution memoryless? Yes or No.

Answer: Yes. Past failures don't affect future success probabilities.

Flashcard 65: What is the main application of the geometric distribution?

Answer: Modeling the number of trials until the first success. Primary use case in probability and statistics applications.

Flashcard 66: What is the mean (expected value) formula for a geometric distribution?

Answer: $E(X) = \frac{1}{p}$. Higher success probability means fewer expected trials needed.

Flashcard 67: State the formula for the variance of a geometric distribution.

Answer: $Var(X) = \frac{1-p}{p^2}$. Standard variance formula derived from geometric distribution theory.

Flashcard 68: Which distribution models the number of trials until first success?

Answer: The geometric distribution. Specifically models waiting time until first success occurs.

Flashcard 69: What happens to $Var(X)$ as $p$ increases in a geometric distribution?

Answer: Variance decreases. Higher success probability reduces variability in waiting time.

Flashcard 70: What is the skewness of a geometric distribution?

Answer: Positively skewed. Most probability mass at $X=1$, decreasing for higher values.

Flashcard 71: What condition must be met for a geometric distribution to be applicable?

Answer: Trials are independent with constant probability $p$ of success. Each trial must have same success chance and be unaffected by others.

Flashcard 72: What is the probability of success in a geometric distribution?

Answer: The constant probability $p$. Success probability remains fixed across all trials.

Flashcard 73: What type of variable is used in a geometric distribution?

Answer: Discrete random variable. Counts whole number trials, not continuous values.

Flashcard 74: Which distribution models the number of trials until first success?

Answer: The geometric distribution. Specifically models waiting time until first success occurs.