# 10. Bayes' Theorem

Let *E*_{1} and *E*_{2} 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*.*

### On this page...

### Example 1

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

E_{1}= "the integers `1` to `8`" and

E_{2}= "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*(*E*_{1} | *E*) means "the probability of the event *E*_{1} 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 *E*_{1}, *E*_{2}, ... , *E _{k}* 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?

Answer

S= "smokers in the district",

E_{1}= "prefer brand A",

E_{2}= "prefer brand B" and

Eis 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)?

Answer

`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.

Get the Daily Math Tweet!

IntMath on Twitter

## Further explanation of Bayes' Theorem

Here's 2 video explanations of Bayes' Theorem:

### Search IntMath, blog and Forum

### Online Algebra Solver

This algebra solver can solve a wide range of math problems.

Go to: Online algebra solver

### Math Lessons on DVD

Math videos by MathTutorDVD.com

Easy to understand math lessons on DVD. See samples before you commit.

More info: Math videos

### The IntMath Newsletter

Sign up for the free **IntMath Newsletter**. Get math study tips, information, news and updates each fortnight. Join thousands of satisfied students, teachers and parents!