AMH - g_ud.3_conditional

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?

An Explanation Based On Set Theory

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

Solution: By the definition of conditional probability, $P(A|B) = \frac{P(A \cap B)}{P(B)} = \frac{0.1}{0.5} = 0.2$

2. If $P(A) = 0.3,$ $P(B) = 0.5,$ and $P(A \cup B) = 0.7,$ what is $P(A|B)?$

Unanswered

Solution: By the definition of conditional probability, $P(A|B) = \frac{P(A \cap B)}{P(B)}$ By inclusion-exclusion, $\begin{align} P(A \cap B) & = P(A) + P(B) - P(A \cup B) \\ & = 0.3 + 0.5 - 0.7 \\ & = 0.1 \end{align}$ So, $\frac{P(A \cap B)}{P(B)} = \frac{0.1}{0.5} = 0.2$

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

Solution: Let $D$ be the event that a resident owns a dog and let $C$ be the event that a resident owns a cat. Then $P(D) = 0.3$ and since residents that own dogs have a $50\%$ chance of owning a cat, $P(C|D) = 0.5.$ The probability that a resident owns both a cat and a dog is $P(C \cap D) = P(D)P(C|D) = 0.3 \cdot 0.5 = 0.15$

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

Solution: Let $S$ be the event that a viewer likes Star Wars and $I$ be the event that a viewer likes Indianna Jones. Then $P(S \cap I) = 0.8,$ and $P(I|S) = 0.9.$ By definition of conditional probability, $P(I|S) = \frac{P(I \cap S)}{P(S)}$ Using the values given, $0.9 = \frac{0.8}{P(S)}$ which implies $P(S) = 0.889.$