Previously on Conditional Probability…

• Independence vs Dependence

• Conditional probability Notation

• Introduction to law of total probability and Bayes’ theorem

Independence vs Dependence

Let \(A\) and \(B\) be events.

10.10-Minute Activity (1/3)

• Question - Draw 2 cards from the top of a standard deck. Let event \(A\) be that the first card is an ace, event \(B\) is that the second card is an ace. What is \(P(A \cap B)\) and \(P(A \cup B)\)? You can use the law of total probability to find \(P(B)\).

• Timer starts

10.10-Minute Activity (2/3)

• Step 1: \[P(A) = \frac{4}{52}, P(A^C) = \frac{48}{52}\]

• Step 2: The \(P(B)\) term is computed using the law of total probability. \[ \begin{align} P(B) & = P(A \cap B) + P(A^C \cap B) \\ & = P(A)(B|A) + P(A^C)P(B|A^C) \\ & = \frac{4}{52}\frac{3}{51} + \frac{48}{52}\frac{4}{51} \\ & = \frac{4}{52}\left( \frac{3}{51} + \frac{48}{51} \right) \\ & = \frac{4}{52} \end{align} \]

10.10-Minute Activity (3/3)

• Step 3: \[P(B|A) = \frac{3}{51}\] \[P(A \cap B) = P(A)P(B|A) = \frac{4}{52}\frac{3}{51}\]

• Step 4: \[ \begin{align} P(A \cup B) & = P(A) + P(B) - P(A \cap B) \\ & = \frac{4}{52} + \frac{4}{52} - \frac{4}{52}\frac{3}{51} \\ \end{align} \]

Bayes’ Theorem (1/2)

Let \(A\) and \(B\) be events.

\[P(A|B) = \frac{P(B|A)P(A)}{P(B)}\]

where \(P(B) \ne 0\).

Using the law of total probability, we can write Bayes’ theorem as \[P(A|B) = \frac{P(B|A)P(A)}{P(B)} = \frac{P(B|A)P(A)}{\sum_{i=1}^{n} P(B \cap A_i)} = \frac{P(B|A)P(A)}{\sum_{i=1}^{n} P(B|A_i)P(A_i)}.\]

Bayes’ Theorem (2/2)

Image Source: “Bayes’ rule with a simple and practical example” by Tirthajyoti Sarkar

10.10-Minute Activity (1/2)

• Question - Let \(A\) be the event of having COVID and \(B\) be the event of testing positive. Suppose about 1% of Oregon’s population would test positive for COVID if tested today. Suppose that the false positive rate is about 1%. We randomly select a person, and that person tests positive. What is the chance that the person has COVID?

• Timer starts

10.10-Minute Activity (2/2)

• Answer: Suppose that all people who have COVID will test positive, so likelihood is \(P(B|A)=1\), which is the true positive rate - this information is missing from the problem. Assuming this the prior probability of having COVID (also known as prevalence rate) is \(P(B) = 0.01\) - this is what’s given but the language used in the problem might be weird. The given false positive rate is \(P(B|A^C) = 0.01\). We want the posterior probability \(P(A|B)\). \[ \begin{align} P(A|B) = & \frac{P(A \cap B)}{P(B)} \\ = & \frac{P(A)P(B|A)}{P(A)P(B|A) + P(A^C)P(B|A^C)} \\ = & \frac{(0.01)(1)}{(0.01)(1) + (0.99)(0.01)} \\ P(A|B) \approx & \frac{1}{2} \end{align} \]

Summary

Today, we discussed the following:

• Law of total probability

• Bayes’ theorem

• COVID testing example

Next, we will discuss:

• More on Bayes’ theorem

• More on the COVID testing example