# 10. Bayes' Theorem

Let E1 and E2 be two mutually exclusive events forming a partition of the sample space S and let E be any event of the sample space such that P(E) ≠ 0.

### Example 1

The sample space S is described as "the integers 1 to 15" and is partitioned into:

E1 = "the integers 1 to 8" and

E2 = "the integers 9 to 15".

If E is the event "even number" then we have the following:

[Recall from Conditional Probability that the notation P(E1 | E) means "the probability of the event E1 given that E has already occurred".]

## Statement of Bayes' Theorem

The probabilities for the situation described above is given by Bayes' Theorem, which can be calculated in two ways:

### Method 1

P(E_1 | E)=(P(E_1nnnE))/(P(E_1nnnE)+P(E_2nnnE))

=(P(E_1)xxP(E|E_1))/(P(E_1)xxP(E|E_1)+P(E_2)xxP(E|E_2)

So for our example above, checking both items of this equation:

P(E_1|E)=(P(E_1nnnE))/(P(E_1nnnE)+P(E_2nnnE))

=(4/15)/(4/15+3/15)

=4/7

### Method 2

We get the same result using the second form:

P(E_1|E)=(P(E_1)xxP(E|E_1))/(P(E_1)xxP(E|E_1)+P(E_2)xxP(E|E_2))

=(8/15xx4/8)/(8/15xx4/8+7/15xx3/7)

=(4/15)/(7/15)

=4/7

## Extending Bayes' Theorem for Mutually Exclusive Events

Bayes' Theorem can be extended as follows:

If E1, E2, ... , Ek are mutually exclusive events forming partitions of the sample space S and if E is any event of S such that P(E) ≠ 0, then

P(E_i|E)=(P(E_i nnnE))/(P(E_1nnnE)+P(E_2nnnE)+...+P(E_knnnE))

### Example 2

Of all the smokers in a particular district, 40% prefer brand A and 60% prefer brand B. Of those smokers who prefer brand A, 30% are females, and of those who prefer brand B, 40% are female. What is the probability that a randomly selected smoker prefers brand A, given that the person selected is a female?

S = "smokers in the district",

E1 = "prefer brand A",

E2 = "prefer brand B" and

E is the event "female".

P(E_1|E)=(P(E_1nnnE))/(P(E_1nnnE)+P(E_2nnnE))

=(12/100)/(12/100+24/100)

=1/3

Or equivalently:

P(E_1|E) = (P(E_1)xxP(E|E_1))/(P(E_1)xxP(E|E_1)+P(E_2)xxP(E|E_2))

=(0.4xx0.3)/(0.4xx0.3+0.6xx0.4)

=1/3

Easy to understand math videos:
MathTutorDVD.com

### Example 3

There are 3 urns A, B and C each containing a total of 10 marbles of which 2, 4 and 8 respectively are red. A pack of cards is cut and a marble is taken from one of the urns depending on the suit shown - a black suit indicating urn A, a diamond urn B, and a heart urn C. What is the probability a red marble is drawn?

If somebody secretly cut the cards and drew out a marble and then announced to us a red marble had in fact been drawn, could we compute the probability of the cut being, say, a heart (or more generally, can we compute the probability of a specified prior event given that the subsequent event did occur)?

P(R)=P(A)xxP(R|A)+ P(B)xxP(R|B)+ P(C)xxP(R|C)

=26/52xx2/10+ 13/52xx4/10+ 13/52xx8/10

=2/5

P(H|R) =(P(H)xxP(R|H))/(P(R)) =(13/52xx8/10)/(4/10) =1/2

P(D|R) =(P(D)xxP(R|D))/(P(R)) =(13/52xx4/10)/(4/10) =1/4

P(Bl|R) =(P(Bl)xxP(R|Bl))/(P(R)) =(26/52xx2/10)/(4/10) =1/4

Conclusion: The probabilities add to 1, so we can compute the probability of a specified event given the subsequent event did occur.

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