Conditional probability is a well known, but unintuitive, concept. This post gives a set theoretic description of conditional probability that may be helpful for developing intuition. The definition of conditional probability is the following formula: \[P(A|B)=\frac{P(A \cap B)}{P(B)}\] The formula reads "The probability of \(A\) given \(B\) is the probability of \(A\) and \(B\) divided by the probability of \(B\)." But, why should this be the definition?
We are going to represent the entire probability, \(\Omega\) space by a white square:
The sample space is all points in the square. Consider all points to be equally likely.
An event is a set in the space. This red circle will represent the event \(A:\)
We will use a blue circle to represent \(B\). The probability of an event depends on its size. Since \(B\) is larger than \(A\), it is more likely that a random point will be in \(B\) than in \(A\).
The purple shows where \(A\) and \(B\) overlap. If a point is in the purple area, then both \(A\) and \(B\) occur. This is also called the intersection of \(A\) and \(B\), and is written \(A \cap B\).
Now we want to find \(P(A|B)\). The meaning of "given \(B\)" is that we know the point lands in \(B\). So, we can ignore all of the points outside of \(B\). If we ignore the points outside of \(B\), then we are ignoring the points of \(A\) that are outside of \(A\) and \(B\).
What is the probability that a point in the space is in \(A\)? Since the point has to be in \(B\), a point in \(A\) is in both \(A\) and \(B\). In other words, the point is in \(A \cap B\). The amount of the total space that \(A \cap B\) takes up is \((A \cap B)\)/\(B\). Thus, \[P(A|B) = \frac{P(A \cap B)}{P(B)}.\]
Check your understanding:
1. If \(P(A) = 0.3,\) \(P(B) = 0.5,\) and \(P(A \cap B) = 0.1,\) what is \(P(A|B)?\)
Unanswered
2. If \(P(A) = 0.3,\) \(P(B) = 0.5,\) and \(P(A \cup B) = 0.7,\) what is \(P(A|B)?\)
Unanswered
3. In a particular neighborhood, a randomly chosen resident has a \(30\%\) chance of owning a dog. Residents that own dogs have a \(50\%\) chance of owning a cat. What is the probability that a randomly chosen resident has both a dog and cat?
Unanswered
4. \(80\%\) of viewers report liking both Star Wars and Indiana Jones. Of those who report liking Star Wars, \(90\%\) report liking Indiana Jones. What is the probability that a randomly chosen viewer likes Star Wars?
Unanswered