# Ideas in Mathematics and Probability: Conditional Probabilities and Bayes’ Theorem (2007) – Article by G. Stolyarov II

**G. Stolyarov II**

**Note from the Author:**

*This article was originally published on Associated Content (subsequently, Yahoo! Voices) in 2007. The article earned over 2,100 page views on Associated Content/Yahoo! Voices, and I seek to preserve it as a valuable resource for readers, subsequent to the imminent closure of Yahoo! Voices. Therefore, this essay is being published directly on The Rational Argumentator for the first time. ****

*~ G. Stolyarov II, July 18, 2014*

*given that*event B has occurred. In mathematical notation, the probability of A given B is expressed as P(A|B).

Bayes’ Theorem enables us to determine the probability of *both* of two dependent events occurring when we know the conditional probability of one of the events occurring, given that the other has occurred. Bayes’ Theorem states the probability of A and B occurring is equal to the product of the probability of B and the conditional probability of A given B or the product of the probability of A and the conditional probability of B given A:

P(A and B) = P(B)* P(A|B) = P(A)*P(B|A).

This theorem works for both independent and dependent events, but for independent events, the result is equivalent to what is given by the multiplication rule: P(A and B) = P(B)*P(A). Why is this the case? When two events are independent, the occurrence of one has no effect on whether the other will occur, so the probability of the event taking place should be equal to the conditional probability of that event given that the other event has taken place. So for independent events A and B: P(A) = P(A|B) and P(B) = P(B|A). If one ever wishes to determine *whether* two events are independent, it is possible to do so by computing their individual probabilities and their conditional probabilities and seeing if the former equal the latter.

The following sample problem can illustrate the kinds of probability questions that Bayes’ Theorem can be used to answer. This particular problem is of my own invention, but the first actuarial exam (Exam P) has been known to have other problems of this sort, which are virtually identical in format.

**Problem:** A company has four kinds of machines: A, B, C, and D. The probabilities that a machine of a given type will fail on a certain day are: 0.02 for A, 0.03 for B, 0.05 for C, and 0.15 for D. 10% of a company’s machines are of type A, 25% are of type B, 30% are of type C, and 35% are of type D. Given that a machine has failed on a certain day, what is the probability of the machine being of type B?

**Solution:** First, let us designate the event of a machine’s failure with the letter F. Thus, from the given information in the problem, P(A) = 0.10, P(B) = 0.25, P(C) = 0.3, and P(D) = 0.35. P(F|A) = 0.02, P(F|B) = 0.03, P(F|C) = 0.05, and P(F|D) = 0.15. We want to find P(B|F). By Bayes’ Theorem, P(B and F) = P(F)* P(B|F). We can transform this to

P(B|F) = P(B and F)/P(F). To solve this, we must determine P(B and F). By another application of Bayes’ Theorem, P(B and F) = P(B)* P(F|B) = 0.25*0.03 = 0.0075. Furthermore,

P(F) = P(A and F) + P(B and F) + P(C and F) + P(D and F)

P(F) = P(A)*P(F|A) + P(B)* P(F|B) + P(C)* P(F|C) + P(D)* P(F|D)

P(F) = 0.10*0.02 + 0.25*0.03 + 0.3*0.05 + 0.35*0.15 = 0.077So P(B|F) = P(B and F)/P(F) = 0.0075/0.077 = **15/154 or about 0.0974025974.** Thus, if a machine has failed on a certain day, the probability that it is of type B is 15/154.