Example 22.
Roll a fair D6 two times. Let \(A = \{\text{sum } \geq 10\}\) and \(B = \{\text{first roll is } 6\}\text{.}\) \(A\) feels more likely if we already know \(B\) has occurred.
Section Thursday, Jan 15
This is an outline of the topics we covered in class. These notes are not a substitute for your own note-taking. I highly recommend that you take your own notes during class. If you ever miss a class for any reason, reach out to another student in class to get a copy of their notes.
Subsection Sec 1.3: Conditional Probability
Question: How does evidence (e.g., knowledge of one event occurring) change our knowledge of probabilities for other events?
Definition 23.
The conditional probability of \(A\) given \(B\) is: ,
\begin{gather*}
\Pr(A \mid B) = \frac{\Pr(A\cap B)}{\Pr(B)}
\end{gather*}
Two overlapping circles representing events \(A\) and \(B\) sit inside a rectangle representing the sample space \(\Omega\text{.}\) The circle labeled \(B\) is shaded. The portion of that circle which is overlapped by the \(A\) circle is also filled in with slanted lines.
Example 25.
Continuing from the previous example, \(|\Omega| = 36\text{.}\)
\begin{align*}
A \amp = \{(4, 6), (5, 5), (5, 6), (6, 4), (6, 5), (6, 6)\} \\
B \amp = \{(6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)\} \\
A \cap B \amp = \{(6, 4), (6, 5), (6, 6)\}
\end{align*}
So \(\Pr(A) = \frac{6}{36}, \Pr(B) = \frac{6}{36}, \text{and } \Pr(A\cap B) = \frac{3}{36}\text{.}\) Then:
\begin{gather*}
\Pr(A \mid B) = \frac{\Pr(A\cap B)}{\Pr(B)} = \frac{3/36}{6/36} = \frac{3}{6} = \frac{1}{2}.
\end{gather*}
Notice that \(\Pr(A\mid B)\) is significantly larger than \(\Pr(A)\text{.}\)
Subsection Diagnostic Testing
Setup: A patient takes a diagnostic test. Let \(P\) be the event that they test positive. Let \(D\) be the event that they have the disease.
Definition 26.
The sensitivity of a diagnostic test is \(\Pr(P \mid D)\text{.}\) The specificity of a diagnostic test is \(\Pr(P^c \mid D^c)\text{.}\)
But, what the patient really wants to know is \(\Pr(D \mid P)\text{.}\)
Example 27.
A disease has a prevalence of 1%. A test has sensitivity of 90% and specificity of 91%. For a patient who gets a positive test result, what is the probability that they have the disease?
\begin{align*}
\text{A) } 9/10 \amp \amp \text{B) } 8/10 \amp \amp \text{C) } 1/10 \amp \amp \text{D) } 1/100
\end{align*}
Theorem 28. Bayes’ Theorem (v1).
For events \(A, B\) with nonzero probability:
\begin{gather*}
\Pr(B \mid A) = \frac{\Pr(A \mid B)\Pr(B)}{\Pr(A)}
\end{gather*}
For example, if a patient sees a positive diagnostic test result, they might try to calculate:
\begin{gather*}
\Pr(D \mid P) = \frac{\Pr(P \mid D)\Pr(D)}{\Pr(P)}
\end{gather*}
\(\Pr(P \mid D)\) is the sensitivity. \(\Pr(D)\) could be the prevalence. We don’t have direct access to \(\Pr(P)\text{.}\)
Observation 2:
\begin{align*}
\Pr(P \mid D) \amp \frac{\Pr(P\cap D)}{\Pr(D)} \amp \amp \Rightarrow \amp \Pr(P \cap D) \amp = \Pr(P\mid D)\Pr(D) \\
\Pr(P \mid D^c) \amp \frac{\Pr(P\cap D^c)}{\Pr(D^c)} \amp \amp \Rightarrow \amp \Pr(P \cap D^c) \amp = \Pr(P\mid D^c)\Pr(D^c)
\end{align*}
So:
\begin{gather*}
\Pr(P) = \Pr(P\mid D)\Pr(D) + \Pr(P\mid D^c)\Pr(D^c)
\end{gather*}
Theorem 30. Bayes’ Theorem (v2).
\begin{gather*}
\Pr(B \mid A) = \frac{\Pr(A \mid B)\Pr(B)}{\Pr(A \mid B)\Pr(B) + \Pr(A \mid B^c)\Pr(B^c)}
\end{gather*}
Example 31.
Continuing from the previous example:
\begin{align*}
\Pr(D\mid P) \amp = \frac{\Pr(P\mid D)\Pr(D)}{\Pr(P\mid D)\Pr(D) + \Pr(P\mid D^c)\Pr(D^c)} \\
\amp = \frac{(0.9)(0.01)}{(0.9)(0.01) + (1 - 0.91)(1 - 0.01)} \\
\amp \approx 0.092
\end{align*}
What if the patient got a negative test result instead? In that case, what is the probaiblity they do not have the disease?
\begin{align*}
\Pr(D^c\mid P^c) \amp = \frac{\Pr(P^c\mid D^c)\Pr(D^c)}{\Pr(P^c\mid D^c)\Pr(D^c) + \Pr(P^c\mid D)\Pr(D)} \\
\amp = \frac{(0.91)(1 - 0.01)}{(0.91)(1 - 0.01) + (1 - 0.9)(0.01)} \\
\amp \approx 0.999
\end{align*}
Subsection Sec 1.4: Independent Events
Question: \(\Pr(A \mid B)\) is supposed to capture how information about \(B\) affects the probability of \(A\text{.}\) What if it doesn’t?
Definition 32.
Observation: If \(A, B\) have nonzero probability and are independent, then:
\begin{align*}
\Pr(A\mid B) \amp = \Pr(A) \\
\frac{\Pr(A\cap B)}{\Pr(B)} \amp = \Pr(A) \\
\Pr(A\cap B) \amp = \Pr(A)\Pr(B)
\end{align*}
We can take this last equation as a definition of independence.
Example 33.
Continuing Example 25, recall \(A = \{\text{sum} \geq 10\}\) and \(B = \{\text{1st roll is } 6\}\text{.}\) We found that \(\Pr(A \mid B) \neq \Pr(A)\text{,}\) so \(A\) and \(B\) are not independent.
Now consider the event \(C = \{\text{sum} = 7\}\text{.}\) We have:
\begin{align*}
C \amp = \{(1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1)\} \\
\Pr(C) \amp = \frac{6}{36} = \frac{1}{6} \\
B\cap C \amp = \{(6, 1)\} \\
\Pr(B\cap C) \amp = \frac{1}{36} \\
\text{therefore: } \Pr(C\mid B) \amp = \frac{\Pr(C\cap B)}{\Pr(B)} = \frac{1/36}{1/6} = \frac{1}{6} = \Pr(C)
\end{align*}
Therefore events \(B, C\) are independent.
