5.4 Conditional Probability

In the previous section, we have seen how to determine the probability of simple events. In this section, we will learn how to calculate the probability of an event knowing that some other event happened before. We call this conditional probability, i.e., what can we say about an event if the sample space changes? Consider the following examples:

  • What is the probability of a person earning more than $150,000 given graduation from Harvard Law School?
  • What is the probability of a person getting arrested given a prior arrest?
  • What is the probability of getting an A in graduate statistics given an undergraduate degree in mathematics?
  • What is the probability of receiving a grant from a funding agency given prior funding from the same agency?

The conditional probability of event \(A\) given that event \(B\) happened is expressed as \(Pr(A|B)\). Given event \(B\) such that \(Pr(B)>0\) and any other event \(A\), we define the conditional probability of \(A\) given \(B\) as: \[Pr(A|B)=\frac{Pr(A\cap B)}{Pr(B)}\] Rearrangement of the terms from the conditional probability definition leads to the multiplication rule for dependent events: \[Pr(A\cap B) = Pr(A|B) \cdot Pr(B)\] Let us consider a couple of examples to illustrate this concept. Consider the following table relating the quality of the service (from an electrician) received to the years of experience.

Good service Bad Service Total
Over 10 years 16 4 50
Below 10 years 10 20 30
Total 26 24 50

Define the event “good service” as \(G\) and “more than 10 years of experience” as \(E\). What is \(Pr(G)\), \(Pr(G|E)\)?, \(Pr(E)\)?

Next, consider rolling a die and the interest in the probability of a 1 given that an odd number was obtained. Let event \(A\) be “observe a 1” and event \(B\) be “observe an odd number”. The probability of interest is \(Pr(A|B)\). The event \(A\cap B\) requires the observance of both a 1 and an odd number. In this instance, \(A\subset B\) so \(A\cap B=A\) and \(Pr(A \cap B)=Pr(A)=1/6\). Also, \(Pr(B)=1/2\) and, using the definition, \[Pr(A|B) = \frac{Pr(A \cap B)}{Pr(B)} =\frac{1/6}{1/2} = \frac{1}{3}\] In the last example, a box with red and black balls is considered. The box contains r red balls labeled \(1,2,3,\dots,r\) and b black balls labeled \(1,2,3,\dots,b\). If a ball from the box is known to be red, what is the probability it is the red ball labeled 1, i.e., \(Pr(B|A)\)? Let event \(A\) be “observe a red ball” and event \(B\) be “observe a 1”. The probability of \(A\) is \(Pr(A)=r/(r+b)\) and the probability of a red ball with the number 1 on it is \(Pr(A \cap B)=1/(r+b)\). Then the probability that the ball is red and labeled 1 given that it is red is given by \[Pr(B|A)=\frac{Pr(A\cap B)}{Pr(A)} =\frac{1/(r+b)}{r/(r+b)} = \frac{1}{r}\] This differs from the probability of \(B\) (a 1 on the ball) which is given by \(Pr(B)=2/(r+b)\).