Probability and Logic
There is a close relationship between probability and symbolic logic. Consider a truth table for the following statements:
(A) The team will make the playoffs.
(B) The coach will be rehired.
The truth table is:
|Statement A||Statement B|
The rules of logic are that the statement "A and B" is true on the first line of the table and false everywhere else. However, the statement "A or B" is true on the first three lines of the table.
Now, we assign a probability to each of the four lines. Let us assign a probability w to the first line, x to the second line, etc. Thus, we have
|Probability||Statement A||Statement B|
In words, w represents the probability that the team will make the playoffs and the coach will be rehired. x is the probability that the team will make the playoffs and the coach will not be rehired.
What is the probability that the team will make the playoffs? We have to add w + x. We can write,
In addition, we know the following:
P(B) = w + y
w + x + y + z = 1
P(A and B) = w
P(A or B) = w + x + y = 1-z
Suppose that the playoffs and the coach have nothing to do with one another. The coach could be the coach of our high school basketball team, and the playoffs could be the professional football playoffs. In that case, we say that A and B are independent by assumption.
If we assume indepence, this means that P(A and B) = P(A)P(B). That means that w = P(A)P(B). Now, we have four equations in four unknowns:
w = P(A)P(B)
w + x = P(A)
w + y = P(B)
w + x + y + z = 1
Suppose that P(A) = .7 and P(B) = .4. Then, we have w = .28, x = .42, y = .12, and z = .18
Be careful! Of the four equations above, the first equation is only true if you can assume that the two events are independent.
If we cannot assume independence, then we need additional information to solve for all of the probabilities. For example, suppose that we are talking about the coach of a pro basketball team and the prospects for that team making the playoffs. In that case, we cannot say that the probability that the coach will be rehired is independent of the probability that the team will make the playoffs.
One possibility is that we are given the joint probability of the coach being rehired and the team making the playoffs. That is, suppose that we were told that P(A and B) = .3, in addition to being told that P(A) = .7 and P(B) = .4.
In that case, we are given that w = .3, and we can use the other three equations to obtain x = .4, y = .1, and z = .2
Another possibility is that we are given the conditional probability of the coach being rehired if the team makes the playoffs. We write this as P(A|B), and it is equal to P(A and B)/P(B), which means that it equals w/(w+y).
Suppose that the conditional probability is .8. What that means is that w/(w+y) = .8
If we know that P(A) = .7 and P(B) = .4, then we know that (w+y) = .4, so that a = (.8)(.4) = .32. Then we know that x = .38, y = .02, and z = .28.