# Doubt about a simple conditional probability problem

I am having some trouble with this probability problem. In a presidential election, exit polls from Washington State provided the following results:

                            John        Bush
Not Employee(65%)           32%         68%
Employee(35%)               61%         39%


If a randomly selected respondent voted for John, what is the probability that this person is an employee?

I thought of it as a conditional probability, where E denotes Employee, and J denotes John's votes. So P(E|J). And I found to be P(E|J)=P(EnJ)/P(J), where P(J)= The sum of the numbers in John's column; 32%+61%=93%=0.93

So P(E|J)=0.61/0.93=0.656.

Is this a correct solution? And what did I understand wrong?

What you are given is this: E = employee, N = non-employee, and P(J/N) = .32, P(B/N) = .68, P(J/E) = .61, P(B/E) = .39, P(E) = .35, and P(N) = .65. So you are to calculate: P(E/J). We use Bayes' rule to find the answer. Namely:

P(E/J)*P(J) = P(J/E)*P(E) ===> P(E/J) = P(J/E)*P(E)/P(J) =

= P(J/E)*P(E)/(P(J/E)*P(E) + P(J/N)P(N)) = (.61)(.35)/(.61*.35 + .32*.65) = .506

You have the formula correct, but are misreading the table. Notice that the numbers in each row adds to 100%, but the numbers in the columns do not. Also, notice the percentages in the labels for Not Employees and Employees, respectively.

What you need to do is create a table that expresses the probability of each combination.

John Bush

0.208 0.442 | 0.65 (Not Employee)

0.2135 0.1365| 0.35 (Employee)

0.4215 0.5785| 1.00

I think that is correct. Now you can do your calculation.