C2 w12 (#28)

* bug fixed.

docs/index.rst

@@ -19,6 +19,7 @@ Contents
    :maxdepth: 1
 
    probability/what_is_probability
+   probability/conditional_probability_and_bayes_theorem
    probability/random_variable
    probability/discrete_random_variables
    probability/continuous_random_variables

docs/probability/conditional_probability_and_bayes_theorem.md

@@ -0,0 +1,77 @@
+---
+file_format: mystnb
+kernelspec:
+  name: python3
+---
+
+```{title} What is conditional probability and bayes theorem?
+```
+
+# Bayes Theorem
+
+## Definition
+Bayes theorem is also known as the formula for the **probability of causes**. 
+
+Theorem states that the conditional probability of an event, based on the occurrence of another event, is equal to the
+likelihood of the second event given the first event multiplied by the probability of the first event.
+
+### Conditional Probability
+
+Two events A and B from the `same sample space S`. Calculate the probability of event A knowing that event B has occurred.
+B is the “conditioning event”. $P(A|B)$
+
+Conditional Probability is $P(A \mid B)=\frac{P(A \cap B)}{P(B)}, \quad P(B)>0$
+
+### Multiplication Rule
+This leads to the multiplication rule  $P(A \cap B) = P(B) P(A \mid B) = P(A) P(B \mid A)$
+
+**Bayes Theorem** $P(A \mid B) = \frac{P(B \mid A)P(A)} {P(B)}$
+
+## Law of Total Probability
+
+$B=(B \cap A) \cup\left(B \cap A^{c}\right)$
+
+$P(B)=P(B \cap A)+P\left(B \cap A^{c}\right)=P(B \mid A) P(A)+P\left(B \mid A^{c}\right) P\left(A^{c}\right)$
+
+## Independence and Mutually Exclusive Events
+
+Two events are `independent` if knowing the outcome of one event does not change the probability of the other.
+
+- Flip a two-sided coin repeatedly. Knowing the outcome of one flip does not change the probability of the next.
+
+Two events, A and B, are independent if $P(A|B) = P(A)$, or equivalently $P(B|A) = P(B)$.
+
+`Recall:` $P(A \mid B)=\frac{P(A \cap B)}{P(B)}$
+
+then, if A and B are independent, we get the multiplication
+rule for independent events:
+
+$P(A \cap B)=P(A) P(B)$
+
+### Example
+Suppose your company has developed a new test for a disease. Let event A be the event that a randomly selected 
+individual has the disease and, from other data, you know that 1 in 1000 people has the disease.
+Thus, P(A) = .001. Let B  be the event that a positive test result is received for the randomly selected individual.
+Your company collects data on their new test and finds the following:
+
+- $P(B|A)$ = .99
+- $P(B^c |A)$ = .01
+- $P(B|A^c )$ = .02
+
+Calculate the probability that the person has the disease, given a positive test result. That is,
+
+find $P(A|B)$.
+
+$$
+\begin{aligned}
+P(A \mid B) &=\frac{P(A \cap B)}{P(B)}  \\
+&=\frac{P(B \mid A) P(A)}{P(B)} \leftarrow \text { Bayes theorem } \\
+&=\frac{ P(B\mid A) P(A)}{P(B \mid A) P(A)+P\left(B \mid A^{c}\right) P\left(A^{c}\right)} \leftarrow \text { Law of Total Probability } \\
+&=\frac{(.99)(.001)}{(.99)(.001)+(.02)(.999)} \\
+&=.0472
+\end{aligned}
+\\
+P(A) =.001 \leftarrow \text { prior prob of } A \\
+P(A \mid B) =.0472 \leftarrow \text { posterior prob of } A
+$$
+