We rely on probability to draw conclusions about our population of interest
Event: a potential outcome
Probability is a measure of how likely it is that an event occurs
The sum of the probabilities of all events in a sample space is 1
\(0 \leq Pr(A) \leq 1\)
The complement of event \(A\) is denoted \(A^c\) and it indicates that event \(A\) does not occur. \(Pr(A) = 1-Pr(A^c)\)
The occurrence of two events \(A\) and \(B\) at the same time is denoted by \(A \cap B\) and \(Pr(A\cap B)\) is known as the joint probability of \(A\) and \(B\)
The event that either \(A\) or \(B\) or both occur is denoted by \(A \cup B\). \(Pr(A\cup B) = Pr(A) + Pr(B) - Pr(A \cap B)\)
Mutually exclusive events: \(Pr(A\cap B)=0\)
Conditional probability: \(Pr(A|B)=\frac{Pr(A\cap B)}{Pr(B)}\)
If knowing about the occurrence of event \(A\) does not give information about the probability of event \(B\), then events \(A\) and \(B\) are independent.
Mathematically: \(Pr(A\cap B) = Pr(A)Pr(B)\)
and \(Pr(A|B)=Pr(A)\)
Bayes Theorem allows us to invert conditional probabilities. We can use it to update a belief based on prior knowledge.
\(Pr(B|A) = \frac{Pr(A|B)Pr(B)}{Pr(A)}\)
You have 3 coins. 2 of them are fair and one is rigged so that it always lands on heads. You choose a coin at random and flip it; it lands on heads. Is the coin you chose likely to be fair or rigged?
\(Pr(F)=2/3\), \(Pr(F^c)=1/3\)
\(Pr(H|F)=1/2\), \(Pr(H|F^c)=1\)
\(Pr(F|H)=\frac{Pr(H|F)Pr(F)}{Pr(H)}\)
We can break down the denominator: \(Pr(H)=Pr(H|F)Pr(F)+Pr(H|F^c)Pr(F^c)\)
So then we have: \(Pr(F|H)=\frac{(1/2)(2/3)}{(1/2)(2/3)+1(1/3)}\)
\(Pr(F|H)=1/2\)