Bayes Theorem

BAYES' THEOREM AND CONDITIONAL PROBABILITIES by Klara Grodzinsky Suppose an experiment is conducted in two stages, where the first stage has four possible outcomes and the second stage has two possible outcomes. For example, say there are four different sections of your Math 1711 course (let's label them A, B, C, and D), and for each section we want to see how many students pass or fail. When we have probability problems that involve experiments such as the one given above, we begin by drawing a tree diagram for our experiment. Let's start by drawing a tree for the above example:

Now, let's put some numbers in our problem. Suppose we know that section A contains 40 students, of whom 32 pass; section B has 50 students, of whom 45 pass; section C has 20 students, of whom 1 fails; and section D has 36 students, of whom 27 pass. Our diagram now has the following form:

Let's change everything on our tree to probabilities, and work some problems. There are a total of 146 students enrolled in Math 1711 for this problem. Changing all the numbers to probabilities gives the following diagram:

Using the above tree, let's work some probability problems. 1. What is the probability that a student will pass the class? We need to find all the branches that lead to "Pass", and add up the probabilities of each of these branches. We have four branched leading to pass, so Pr(pass) = (20/73)*(0.8) + (25/73)*(0.9) + (10/73)*(0.95) + (18/73)*(0.75), which is approximately 84.25%. So, overall, about 84% of the students will pass the class. 2. What is the probability that a student will pass the class if s/he is in section D? Here, we examine the above tree diagram. For section D, 75% of the class will pass, so the probability of passing given that a student is in section D is 0.75. In other words, Pr(pass | section D) = 0.75. Notice that we did not need Bayes' theorem here (why not?). 3. What is the probability that if a student passes, s/he is in section D? It is VERY important to notice the difference between problems 2 and 3. In problem 2, we wanted to find the conditional probability Pr(pass | D), which can be read off the tree. For this problem, we wish to find the conditional probability Pr(D | pass), which CANNOT be read off

the tree. Here, we need Bayes' theorem: find the probability of the intersection and divide by the probability of what we are given. In other words, we find the probability that a student is in section D and passes and divide that by the probability of passing the class. Pr(section D | pass) = [Pr(section D and pass)] / [Pr(pass)] = [(18/73)*(0.75)] / [(20/73)*(0.8) + (25/73)*(0.9) + (10/73)*(0.95) + (18/73)*(0.75)] = 0.2195 (approximately). So, there is a 21.95% chance that a student who passes Math 1711 is in section D. 4. What is the probability that if a student fails, s/he is in section A? This problem is similar to problem 3, but now we need the probability of failing. Since we know that the probability of passing and the probability of failing must sum to 1, Pr(fail) = 1 - Pr(pass) = 0.1575. (You could have also found this number by working the problem similarly to problem 1, replacing the "pass" probabilities with the "fail" probabilities.) Now, we want the probability that a student who fails is in section A: Pr(section A | fail) = [Pr(A and fail)] / [Pr(fail)] = [(20/73)*(0.2)] / [0.1575] = 0.3479 (approximately), so there is a 34.79% chance that the student was in section A. One last comment on Bayes' theorem: if you have a difficult time remembering how to apply the formula, just remember that EVERY Bayes' theorem problem can be worked using the formula for conditional probability (as in the last two examples).