AP Statistics: Probability rules and distributions

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Questions and Answers

Consider a scenario where a novel stochastic process exhibits asymptotic behavior such that the long-run relative frequency of an event $E$ converges to a value dictated by a transcendental equation involving the Riemann zeta function. Which of the following statements best describes the limitations of applying the Law of Large Numbers in this specific context?

  • The Law of Large Numbers guarantees convergence to a frequentist probability, but provides no insight into the sample size required to reach a specific level of accuracy when transcendental functions are involved in defining probabilities.
  • The Law of Large Numbers is inapplicable because the transcendental nature of the limit violates the axioms of probability theory.
  • While the Law of Large Numbers still holds, obtaining an accurate estimate within a reasonable computational time becomes exceptionally challenging due to the slow rate of convergence dictated by the properties of the Riemann zeta function. (correct)
  • The Law of Large Numbers is only valid for discrete random variables, and the Riemann zeta function implies a continuous probability distribution.

Given two events, A and B, within a sample space $\Omega$, the assertion that $P(A \cup B) = P(A) + P(B) - P(A \cap B)$ always provides an accurate computation of the probability of either A or B occurring, assuming all probabilities are well-defined and measurable according to Kolmogorov's axioms.

True (A)

In the context of Bayesian epistemology, where prior probabilities are updated based on observed evidence, what is the philosophical significance of assigning a non-zero prior probability to every conceivable hypothesis, even those considered highly implausible?

Prevents absolute certainty and allows for continuous refinement of beliefs.

In quantum probability theory, which deviates from classical probability by allowing for non-commutative events, the violation of ______'s inequality serves as a key indicator of quantum entanglement and non-classical correlations. If the inequality is not violated, the system can usually be described by classical physical systems.

<p>Bell</p> Signup and view all the answers

Match the following probability concepts with their applications in advanced statistical modeling:

<p>Bayesian Inference = Updating prior beliefs with evidence through Bayes' Theorem for parameter estimation and model comparison. Markov Chain Monte Carlo (MCMC) = Employing Markov chains to sample from complex probability distributions, crucial in Bayesian computation and simulation. Dirichlet Process = A stochastic process used as a prior in Bayesian nonparametrics, allowing for flexible modeling of unknown distributions. Copula Functions = Modeling the dependence structure between random variables independently from their marginal distributions.</p> Signup and view all the answers

Consider a complex system where the probability of event A occurring depends on a continuous-time Markov process $X(t)$ governing environmental conditions, such that $P(A|X(t)) = e^{-\lambda X(t)}$, where $\lambda$ is a positive constant. Given that $X(t)$ follows an Ornstein-Uhlenbeck process with mean reversion level $\mu$ and volatility $\sigma$, what is the most appropriate method for estimating the long-term average probability of event A?

<p>Using a Monte Carlo simulation to approximate the time average of $P(A|X(t))$ over a long time horizon, effectively calculating $\frac{1}{T} \int_0^T e^{-\lambda X(t)} dt$. (C)</p> Signup and view all the answers

If two events, A and B, are independent according to one probability measure P, they must necessarily also be independent under any other probability measure Q defined on the same sample space, provided that both P and Q assign non-zero probabilities to both A and B.

<p>False (B)</p> Signup and view all the answers

In the context of algorithmic fairness, explain how disparate impact can arise even when a classification algorithm is explicitly designed to be 'color-blind' (i.e., not directly using sensitive attributes), and suggest a method to mitigate this issue.

<p>Proxy variables correlate with protected attributes. Use a causal inference framework to build a model.</p> Signup and view all the answers

The ______ paradox demonstrates that aggregating conditional probabilities across different subgroups can yield results that contradict the marginal probabilities, highlighting the importance of considering confounding variables when interpreting statistical data.

<p>Simpson</p> Signup and view all the answers

Consider a scenario where a clinical trial is designed to evaluate the efficacy of a new drug. The probability of a patient recovering with the drug is denoted as $P(R|D)$, and without the drug, the probability is $P(R|¬D)$. However, there exists an unobserved confounder $C$ (e.g., genetic predisposition) that affects both the likelihood of receiving the drug and the probability of recovery. Which of the following statistical techniques is most appropriate for estimating the causal effect of the drug on recovery, accounting for the unobserved confounder?

<p>Instrumental variables (IV) regression using a valid instrument that affects drug assignment but is independent of recovery except through its effect on drug assignment. (D)</p> Signup and view all the answers

Flashcards

Random Process

A process where outcomes are uncertain and determined by chance.

Long Run Relative Frequency

The viewpoint that probability is the long-run relative frequency of an outcome.

Law of Large Numbers

Simulated probabilities get closer to the true probability with more trials.

Sample Space

The list of all possible non-overlapping outcomes of a random process.

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Complement of an Event

The event does not occur.

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Conditional Probability

The probability of event A occurring given event B has already occurred.

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Addition Rule

P(A or B) = P(A) + P(B) - P(A and B).

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Multiplication Rule

P(A and B) = P(A) * P(B|A)

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Independent Events

Knowing that event A has occurred doesn't impact the probability that event B will occur. P(A) = P(A|B).

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Mutually Exclusive Events

Events that cannot happen at the same time; their joint probability is zero.

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Study Notes

Introduction to Probability

  • This unit focuses on probability, random variables, and probability distributions in AP Statistics.
  • The unit is broken down into three parts: basic probability rules, discrete random variables and their probability distributions, and binomial/geometric probability distributions.

Random Processes and Probability

  • A random process generates results that are random or unknown, determined by chance.
  • Randomness implies uncertainty and unpredictability of individual outcomes.
  • An outcome is the result of a random process, and an event is a collection of outcomes.
  • Probability quantifies the uncertainty in a random process.

Long Run Relative Frequency

  • One viewpoint of probability is the long-run relative frequency of an outcome occurring.
  • The long-run relative frequency is found after a large number of repetitions of a random process.
  • It is calculated by dividing the number of times an outcome occurs by the total number of repetitions.
  • Simulations can find these long run relative frequencies, also known as probabilities.

Law of Large Numbers

  • The law of large numbers states that simulated probabilities get closer to the true probability with more trials.
  • Accuracy increases with the number of trials performed.
  • The long-run relative frequency definition of probability can never be perfect as infinite simulations are impossible.

Simulations with Numbers

  • Simulations with numbers involve using random numbers to represent customers or outcomes of a random process.
  • Numbers may represent customers filling water cups with something using a random number table or generator.
  • Instead of looking at customers, this simulation is fast using numbers and generates good probabilities.
  • After repetitions, count the number of trials with less than four people filling the cup with water to estimate a probability.
  • Simulations are useful to find or begin to find the true probability of an event.

Basic Probability Concepts and Rules

  • The sample space of a chance process is a list of all non-overlapping outcomes.
  • Observable events are labeled with Latin letters.
  • The probability of an event is the total number of outcomes in favor of the event divided by the total number of outcomes in the sample space.
  • The probability of an event is always between zero and one, inclusive.
  • The complement of an event is that the event does not happen.
  • To find the probability of its complement, subtract the probability that the event will happen from one.
  • The probability of event A and event B occurring is the joint probability.
  • If the two events cannot happen at the same time, they are known as disjoint or mutually exclusive.
  • If two events are mutually exclusive, the probability of A and B is zero.
  • The probability of event a occurring OR the probability of event B occurring is known as the union of event a and event b.
  • Probability of A or B looks for the probability that event A happens, or event B happens, or both A and B happen.
  • The complement of the probability of A or B is the probability that neither event happens.

Conditional Probability

  • Conditional probability finds the probability of event A occurring given event B has already or will occur.
  • Expressed as P(A|B), where A is the event of interest and B is the given condition.
  • Formula: Take the probability of both events happening at the same time (A and B) and divide by the probability of the condition (B).

Addition Rule

  • The addition rule helps find the probability of A or B.
  • Formula: P(A or B) = P(A) + P(B) - P(A and B)
  • The word "or" means to add, but the overlap must be subtracted if the two events are not mutually exclusive.
  • If events are mutually exclusive, the probability of A and B equals zero; hence, P(A or B) = P(A) + P(B).

Multiplication Rule

  • The multiplication rule finds the probability of A and B.
  • Formula: P(A and B) = P(A) * P(B|A)
  • "And" in probability means to multiply, considering how the second event is impacted by the first event.

Independence between Events

  • Two events A and B are independent if knowing that event a has occurred doesn't impact the probability that event B will occur.
  • Two events are independent if P(A) = P(A|B).
  • If independent, then P(A and B) = P(A) * P(B).
  • If not independent, then P(A and B) = P(A) * P(B|A).
  • Recommendation: Write out everything that can happen explicitly to easily figure out probabilities.

Two-Way Table Example Problems

  • Two-way tables are common in AP statistics exams.
  • A survey of 500 high school students about their high school and favorite slushy flavor shown in results table
  • Example Question 1: Probability that selected student is from high school A?
    • 140 students from high school A out of 500 total
    • Probability = 140/500 = 0.28 or 28%
  • Example Question 2: Probability that selected student prefers a red slushy?
    • 275 kids prefer red slushies out of 500 total
    • Probability = 275/500 = 55%
  • Example Question 3: Probability that selected student loves a red slushy and is from high school A?
    • Joint probability: find where the row and column join together
    • 91 students are both in high school A and love red slushies
    • Probability = 91/500 = 18.2%
    • Since 18.2% of kids both go to high school a and love a red slushie, these events are not mutually exclusive
  • Example Question 4: Probability that selected student is from high school A or likes a red slushy?
    • Or means add but beware of overlap
    • Red slushy: 275/500, high school A: 140/500
    • 91 kids are double counted.
    • Probability = (275/500) + (140/500) - (91/500) = 324/500 = 64.8%
  • Example Question 5: Probability that the student prefers a red slushy, given they are from high school A?
    • Working with conditional probability given the keyword "given"
    • The formula for conditional probability is P(A and B) / P(B)
    • Numerator is the probability of them both A and B, high school A and red slushy, 91/500
    • Denominator, probability there from high school A is 140/500
    • (91/500) / (140/500) those 500s will cancel
    • 91/140 or 65%
    • The idea is that given, the students from high school A, with a conditional probability understand, that means that we can only look at that row for high school a.
    • Changing our dominator to 140, for only those kids from high school A from those kids from high school a 91 love a red slushy for 91 out of 140.

Independence Between Two Events

  • To check if events A and B are independent, compare the probability of A with the probability of A given B.
  • If P(A) = P(A|B), then B doesn't affect A, indicating independence.
  • If P(A) ≠ P(A|B), the events are not independent.
  • Example: Check if "going to high school A" and "liking a red slushie" are independent.
  • The probability of liking a red slushie is 275/500 = 55%.
  • The probability of liking a red slushie given the student is from high school A is 91/140 = 65%.
  • Since 55% ≠ 65%, the events are not independent; being from high school A increases the likelihood of preferring a red slushie.

Probability of Exactly One Event

  • If Andrea purchases car one, the probability is 42% (0.42).
  • If she purchases car two, the probability is 26% (0.26).
  • The decision to purchase each car is made independently.
  • Sample space includes:
    • Purchasing car one and not car two.
    • Purchasing car one and car two.
    • Purchasing car two and not car one.
    • Purchasing neither car one nor car two.
  • Calculate probabilities for each scenario:
    • P(Car 1 and not Car 2) = 0.42 * 0.74 = 0.3108.
    • P(Car 1 and Car 2) = 0.42 * 0.26 = 0.1092.
    • P(Car 2 and not Car 1) = 0.26 * 0.58 = 0.1508.
    • P(Neither Car 1 nor Car 2) = 0.58 * 0.74 = 0.4292.
  • To find the probability of purchasing exactly one car, add probabilities where only one car is purchased:
    • P(Exactly one car) = P(Car 1 and not Car 2) + P(Car 2 and not Car 1) = 0.3108 + 0.1508 = 0.4616 (approximately 0.46).

Conditional Probability and Tree Diagrams

  • The probability of purchasing car one is 42% (0.42).
  • If Andrea purchases car one, the probability of purchasing car two is 15% (0.15).
  • If Andrea does not purchase car one, the probability of purchasing car two is 60% (0.60).
  • Car purchases are dependent, unlike the previous example.
  • Use a tree diagram to illustrate:
    • First branch: Car 1 (0.42) or not Car 1 (0.58).
    • Second branch: Car 2 or not Car 2, with conditional probabilities based on the first branch.
  • Conditional probabilities:
    • P(Car 2 | Car 1) = 0.15, P(not Car 2 | Car 1) = 0.85.
    • P(Car 2 | not Car 1) = 0.60, P(not Car 2 | not Car 1) = 0.40.
  • Calculate final probabilities by multiplying along branches:
    • P(Car 1 and not Car 2) = 0.42 * 0.85 = 0.357.
    • P(not Car 1 and Car 2) = 0.58 * 0.60 = 0.348.
  • Probability of purchasing exactly one car:
    • P(Exactly one car) = P(Car 1 and not Car 2) + P(not Car 1 and Car 2) = 0.357 + 0.348 = 0.705.

Generic Probability with Mutually Exclusive Events

  • Given events A and B with:
    • P(A) = 0.35.
    • P(B) = 0.20.
  • Find P(A or B) when A and B are mutually exclusive.
  • Mutually exclusive means A and B cannot occur at the same time; P(A and B) = 0.
  • P(A or B) = P(A) + P(B) - P(A and B) = 0.35 + 0.20 - 0 = 0.55.

Generic Probability with Independent Events

  • Given events A and B with:
    • P(A) = 0.35
    • P(B) = 0.20
  • Find P(A or B) when A and B are independent.
  • If A and B are independent, P(A and B) = P(A) * P(B).
  • P(A and B) = 0.35 * 0.20 = 0.07.
  • P(A or B) = P(A) + P(B) - P(A and B) = 0.35 + 0.20 - 0.07 = 0.48.

Generic Problems: Solving for Unknown Probability

  • Given:
    • P(A) = 0.35
    • P(A or B) = 0.80
  • Find P(B) when A and B are mutually exclusive.
  • Since A and B are mutually exclusive, P(A and B) = 0.
  • P(A or B) = P(A) + P(B) - P(A and B).
  • 0.80 = 0.35 + P(B) - 0.
  • P(B) = 0.80 - 0.35 = 0.45.

Generic Problems: Independent Events

  • Given events A and B with:
    • P(A) = 0.35
    • P(A or B) = 0.80
  • Find P(B) when A and B are independent.
  • For independent events, P(A and B) = P(A) * P(B).
  • P(A or B) = P(A) + P(B) - P(A and B)
  • 0.80 = 0.35 + P(B) - (0.35 * P(B))
  • Combine like terms: 0.80 = 0.35 + P(B) * (1 - 0.35)
  • 0.80 = 0.35 + 0.65 * P(B)
    1. 45 = 0.65 * P(B)
  • P(B) = 0.45 / 0.65 ≈ 0.692.

Probability with Disease Testing

  • 2% of men have a disease.
  • Test accuracy:
    • If a man has the disease, the test is positive 95% of the time.
    • If a man does not have the disease, the test is positive 12% of the time (false positive).
  • Find the probability of a positive result.
  • Calculate combined probabilities:
    • Has disease and tests positive: 0.02 * 0.95 = 0.019.
    • Does not have disease and tests positive: 0.98 * 0.12 = 0.1176.
  • P(Positive result) = 0.019 + 0.1176 = 0.1366.

Conditional Probability: Positive Test Result and Having the Disease

  • Given a man tests positive, find the probability he has the disease.
  • Use conditional probability formula: P(Disease | Positive) = P(Disease and Positive) / P(Positive)
  • P(Disease and Positive) = 0.02 * 0.95 = 0.019 (numerator).
  • P(Positive) = 0.1366 (from previous calculation, the denominator).
  • P(Disease | Positive) = 0.019 / 0.1366 ≈ 0.1391.
  • Interpretation: A man with a positive test result has approximately a 14% chance of actually having the disease.

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