MTH-361A | Spring 2026 | University of Portland
Understanding Associations:
Basis for Statistical Inference:
Improves Decision-Making:
Clarifies Intuition About Data
Sampling with replacement is a sampling method where each selected unit is returned to the population before the next selection. Each selection is independent of previous selection.
Characteristics:
Sampling without replacement is a sampling method where each selected unit is not returned to the population before the next draw. Each selection depends on previous selections, as the population size decreases.
Characteristics:
| Feature | Sampling with Replacement | Sampling without Replacement |
|---|---|---|
| Independence | Yes | No |
| Duplication | Yes | No |
| Changing Sample Space | No | Yes |
| Changing Probabilities | No | Yes |
You draw cards from a standard \(52\)-card deck.
With Replacement:
\(\star\) Outcomes do not affect future draws.
Example:
You draw cards from a standard 52-card deck.
Without Replacement:
\(\star\) Outcomes change future probabilities.
Example:
Multiplication Rule:
\[ \begin{aligned} P(\text{Ace}_1 \text{ and Ace}_2) & = P(\text{Ace}_1) P(\text{Ace}_2 \text{ given } \text{Ace}_1) \\ & = \left( \frac{4}{52} \right) \left( \frac{3}{51} \right) \\ & = \frac{1}{221} \\ P(\text{Ace}_1 \text{ and Ace}_2) & \approx 0.0045 \end{aligned} \]
Why Conditional Probability?
What is Conditional Probability?
Definition:
Given events \(A\) and \(B\) in the sample space \(S\), \[P(A | B) = \frac{P(A \cap B)}{P(B)}\] where:
\(\star\) Conditional probability helps us update probabilities based on new information.
Independence: Given events \(A\) and \(B\) with non-zero probability in the sample space \(S\), the following are equivalent:
Dependence: Given events \(A\) and \(B\), then by definition:
Not Always Symmetric:
\[P(A|B) \ne P(B|A)\]
Set-up:
Suppose there are two jars \(J_1\) and \(J_2\) continaing colored balls.
Assumptions:
Jar contents as frequencies:
| \(J_1\) | \(J_2\) | |
|---|---|---|
| \(G\) | \(14\) | \(14\) |
| \(B\) | \(11\) | \(6\) |
| Sum | \(25\) | \(20\) |
Based on the given information:
Jar probabilities:
Jar contents probabilities:
| \(J_1\) | \(J_2\) | |
|---|---|---|
| \(G\) | \(P(G|J_1) = \frac{14}{25}\) | \(P(G|J_2) = \frac{14}{20}\) |
| \(B\) | \(P(B|J_1) = \frac{11}{25}\) | \(P(B|J_2) = \frac{6}{20}\) |
| \(J_1\) | \(J_2\) | |
|---|---|---|
| \(G\) | \(0.56\) | \(0.70\) |
| \(B\) | \(0.44\) | \(0.30\) |
Scenario:
Probability tree:
\(\star\) A probability tree is a way to visualize all possible outcomes in the sample space. We look at all the branches of this tree that corresponds to our desired outcome.
What is the probability that the ball drawn is \(G\)?
\[ \begin{aligned} P(G) & = P(J_1 \cap G) + P(J_2 \cap G) \\ & = P(J_1) P(G|J_1) + P(J_2) P(G|J_2) \\ & = \left(\frac{1}{2}\right) \left(\frac{14}{25}\right) + \left(\frac{1}{2}\right) \left(\frac{14}{20}\right) \\ & = \frac{63}{100} \\ P(G) & = 0.63 \longleftarrow \text{equivalent to } P(G) = 1 - P(B) \end{aligned} \]
With the same set-up and assumptions, we now view the same scenario in reverse.
Scenario:
Probability tree in reverse:
\(\star\) The reverse probability tree is way to visualize all possible outcomes given a sample (i.e. an observation or a data point).
\[ \begin{aligned} P(J_1|G) & = \frac{P(J_1 \cap G)}{P(G)} \\ & = \frac{P(G|J_1)P(J_1)}{P(G)} \longleftarrow \text{bayes' Rule} \\ & = \frac{P(G|J_1)P(J_1)}{P(J_1) P(G|J_1) + P(J_2) P(G|J_2)} \\ & = \frac{\left(\frac{1}{2}\right) \left(\frac{14}{25}\right)}{\left(\frac{1}{2}\right) \left(\frac{14}{25}\right) + \left(\frac{1}{2}\right) \left(\frac{14}{20}\right)} \\ P(J_1|G) & = \frac{4}{9} \approx 0.444 \end{aligned} \]
\(\star\) This is consistent with the idea that conditional probability is not always symmetric, meaning \(P(G|J_1) \ne P(J_1|G)\), because \(\displaystyle P(G|J_1) = \frac{14}{25} = 0.56\) is not equal to \(\displaystyle P(J_1|G) \approx 0.444\).
Let \(A\) and \(B\) be events in the sample space \(S\). Then,
\[ \begin{aligned} P(B) & = P(B \cap A) + P(B \cap A^c) \\ \text{or} & \\ P(B) & = P(B|A)P(A) + P(B|A^c)P(A^c) \end{aligned} \]
\(\star\) This law decomposes the probability of \(A\) given \(B\) into two pieces, one where \(B\) happens and one where \(B\) does not happen.
Let \(A\) and \(B\) be events in the sample space \(S\).
\[ \begin{aligned} P(A|B) & = \frac{P(B|A)P(A)}{P(B)} \\ \text{or} & \\ P(A|B) & = \frac{P(B|A)P(A)}{P(B|A)P(A) + P(B|A^c)P(A^c)} \end{aligned} \]
\(\star\) This rule is a simple statement about conditional probabilities that allows the computation of \(P(A|B)\) from \(P(A)\). Due to this rule, events \(A\) and \(B\) are not always symmetric (i.e., \(P(A|B) \ne P(B|A)\)).
With the same set-up and assumptions, we now draw two balls instead of one.
Scenario:
Probability tree:
\(\star\) The probability tree now have three levels instead of two because we drew two balls instead of one. Computing the probabilities now depends on whether we sampled with or without replacement.
Conditional Probabilities:
| Combination | \(J_1\) | \(J_2\) |
|---|---|---|
| \(GG\) | \(P(GG|J_1) = \left(\frac{14}{25}\right) \left(\frac{14}{25}\right)\) | \(P(GG|J_2) = \left(\frac{14}{20}\right) \left(\frac{14}{20}\right)\) |
| \(GB\) | \(P(GB|J_1) = \left(\frac{14}{25}\right) \left(\frac{11}{25}\right)\) | \(P(GB|J_2) = \left(\frac{14}{20}\right) \left(\frac{6}{20}\right)\) |
| \(BG\) | \(P(BG|J_1) = \left(\frac{11}{25}\right) \left(\frac{14}{25}\right)\) | \(P(BG|J_2) = \left(\frac{6}{20}\right) \left(\frac{14}{20}\right)\) |
| \(BB\) | \(P(BB|J_1) = \left(\frac{11}{25}\right) \left(\frac{11}{25}\right)\) | \(P(BB|J_2) = \left(\frac{6}{20}\right) \left(\frac{6}{20}\right)\) |
\(\star\) The notation \(GG\) represents an event where the 1st draw is a \(G\) and the 2nd draw is \(G\). The notations \(GB\), \(BG\), and \(GG\) follows the same notation representation of events.
What is the probability that the two balls drawn contain exactly two \(G\)?
\[ \begin{aligned} P(GG) & = P(J_1)P(GG|J_1) + P(J_2)P(GG|J_2) \\ & = \left(\frac{1}{2}\right) \left(\frac{14}{25}\right)^2 + \left(\frac{1}{2}\right) \left(\frac{14}{20}\right)^2 \\ P(GG) & = \frac{2009}{5000} \approx 0.402 \end{aligned} \]
Conditional Probabilities:
| Combination | \(J_1\) | \(J_2\) |
|---|---|---|
| \(GG\) | \(P(GG|J_1) = \left(\frac{14}{25}\right) \left(\frac{13}{24}\right)\) | \(P(GG|J_2) = \left(\frac{14}{20}\right) \left(\frac{13}{19}\right)\) |
| \(GB\) | \(P(GB|J_1) = \left(\frac{14}{25}\right) \left(\frac{11}{24}\right)\) | \(P(GB|J_2) = \left(\frac{14}{20}\right) \left(\frac{6}{19}\right)\) |
| \(BG\) | \(P(BG|J_1) = \left(\frac{11}{25}\right) \left(\frac{14}{24}\right)\) | \(P(BG|J_2) = \left(\frac{6}{20}\right) \left(\frac{14}{19}\right)\) |
| \(BB\) | \(P(BB|J_1) = \left(\frac{11}{25}\right) \left(\frac{10}{24}\right)\) | \(P(BB|J_2) = \left(\frac{6}{20}\right) \left(\frac{5}{19}\right)\) |
\(\star\) The multiplication rule of dependent events was used on determining these probabilities. Notice that the successive factors for each probability has changed.
What is the probability that the two balls drawn contain exactly two \(G\)?
\[ \begin{aligned} P(GG) & = P(J_1)P(GG|J_1) + P(J_2)P(GG|J_2) \\ & = \left(\frac{1}{2}\right) \left(\frac{14}{25}\right) \left(\frac{13}{24}\right) + \left(\frac{1}{2}\right) \left(\frac{14}{20}\right) \left(\frac{13}{19}\right) \\ P(GG) & = \frac{4459}{11400} \approx 0.391 \\ \end{aligned} \]
With the same set-up and assumptions, we now view the same scenario in reverse.
Scenario:
Probability tree in reverse:
\(\star\) You still need to consider whether the balls are sampled with or without replacement and Bayes’ Rule has to be applied to compute the probabilities.
Assume that we sample without replacement.
\[ \begin{aligned} P(J_1|GG) & = \frac{P(J_1 \cap GG)}{P(GG)} \\ & = \frac{P(GG|J_1)P(J_1)}{P(GG)} \longleftarrow \text{bayes' Rule} \\ & = \frac{P(GG|J_1)P(J_1)}{P(J_1) P(GG|J_1) + P(J_2) P(GG|J_2)} \\ & = \frac{\left(\frac{1}{2}\right) \left(\frac{14}{25}\right) \left(\frac{13}{24}\right)}{\left(\frac{1}{2}\right) \left(\frac{14}{25}\right) \left(\frac{13}{24}\right) + \left(\frac{1}{2}\right) \left(\frac{14}{20}\right) \left(\frac{13}{19}\right)} \\ P(J_1|GG) & = \frac{19}{49} \approx 0.388 \end{aligned} \]
\(\star\) The complement probability \(P(J_2|GG) = 1- P(J_1|GG) = 1 - 0.388 = 0.612\) indicates that the balls are more likely to have come from \(J_2\).
Scenario:
A person takes a test for a rare disease. How do we know the probability they actually have the disease given a positive test result?
Information given:
Bayes’ Rule:
\[ \begin{aligned} P(D|+) & = \frac{P(+|D)P(D)}{P(+)} \\ & = \frac{P(+|D)P(D)}{P(+|D)P(D) + P(+|D^c)P(D^c)} \\ & = \frac{(0.99)(0.01)}{(0.99)(0.01) + (0.05)(1-0.01)} \\ & = \frac{0.0099}{0.0594} \\ P(D|+) & \approx 0.167 \end{aligned} \]
Interpretation:
\(\star\) This highlights the importance of considering prior probabilities (prevalence).
Bayes’ Theorem (or Bayes’ Rule), which provides a formula for updating the probability of a hypothesis based on new evidence, is often described through various colloquialisms in statistics.
Hypothesis statements and data:
Consider hypothesis statements as events and data as an observed evidence.
According to Bayes’ Rule,
\[ \begin{aligned} P(H|E) & = \frac{P(E|H)P(H)}{P(E)} \\ \text{or} & \\ \text{posterior} & = \frac{\text{likelihood} \times \text{prior}}{\text{marginal}}. \end{aligned} \]
\(\star\) In short, learning from data means continuously revising what you think is true as new evidence comes.
Terms explained:
Frequentist Perspective:
\(\dagger\) This course only focuses on the statistical methods based on the frequentist perspective of probability.
In inference:
\(\star\) Most statistical methods introduced in high school and early college are primarily based on the frequentist perspective of probability.
Bayesian Perspective:
\(\dagger\) There is actually different forms of Bayes’ Theorem but let’s just table that until our next life.
In inference:
\(\star\) By combining frequentist and Bayesian approaches, you can perform more flexible and powerful statistical analyses.