Bayes rule states that
P(A|B)=P(B|A)P(A)/(P(B|A)P(A)+P(B|A')P(A'))
Given one set of conditional probabilities, it enables one to calculate conditional probabilities with the reverse conditioning. However, we shall focus on the fact that Bayes rule enables one to calculate other probabilities when three probabilities (in this case: P(B|A), P(B|A'), and P(A); P(A')=1-P(A)) are given.
A Venn diagram for two events divides the sample space into four disjoint subsets: AB, A'B, AB', A'B'. The probabilities of these four events can be concisely represented with a square:
A A'
_____________
| | |
B | x | y | x+y=P(B)
|______|______|
| | |
B' | z | w | z+w=P(B')
|______|______|___
x+z= y+w= |
P(A) P(A')| 1
In accordance with the row and column labels, this square means that
P(AB)=x, P(A'B)=y, P(AB')=z, and P(A'B')=w. P(A)=x+z and P(B)=x+y as
indicated above. P(A|B) = P(AB)/P(B) = x/(x+y); and the other conditional
probabilities can be represented in a similar manner.
There are four unknowns (x, y, z, and w) in the above square, in terms of which all the probabilities we are interested in can be calculated. One constraint is that x+y+z+w=1 (P(S)=1); hence it is reasonable that three further equations would enable us to solve for x, y, z, and w, hence all probabilities.
Examples:
A A'
_____________
| | |
B | x | y | x+y=P(B)
|______|______|
| | |
B' | z | w | z+w=P(B')
|______|______|___
|
.5 .5 | 1
Then P(B|A)=.2 becomes x ÷ .5 = .2 which provides x=.1; x+z=.5
then provides z=.4, and we are now looking at:
A A'
_____________
| | |
B | .1 | y | x+y=P(B)
|______|______|
| | |
B' | .4 | w | z+w=P(B')
|______|______|___
|
.5 .5 | 1
Finally, P(B|A')=.6 becomes y ÷ .5 =.6 which provides y=.3, and
w=.2 follows. summing x+y and z+w lets us complete the square:
A A'
_____________
| | |
B | .1 | .3 | .4
|______|______|
| | |
B' | .4 | .2 | .6
|______|______|___
|
.5 .5 | 1
From this we can readily calculate P(A|B)=.1/.4=.25, or any other probability
we wish.
A A'
_____________
| | |
B | x | y | .6
|______|______|
| | |
B' | z | w | .4
|______|______|___
|
.4 .6 | 1
P(B|A)=.5 then becomes x ÷ .4 = .5, hence x = .2.
A A'
_____________
| | |
B | .2 | y | .6
|______|______|
| | |
B' | z | w | .4
|______|______|___
|
.4 .6 | 1
The rest of the square can be completed by .2+z=.4, .2+y=.6, and z(now known)+
w=.4.
A A'
_____________
| | |
B | .2 | .4 | .6
|______|______|
| | |
B' | .2 | .2 | .4
|______|______|___
|
.4 .6 | 1
A A'
_____________
| | |
B | x | y | x+y=P(B)
|______|______|
| | |
B' | z | w | z+w=P(B')
|______|______|___
|
2/5 3/5 | 1
Now P(B|A)=1/4 becomes x/(2/5)=1/4, or x=1/10. (Indeed x/(x+y)=1/3 is
somewhat intimidating). z=3/10 since x+z=2/5.
A A'
_____________
| | |
B | 1/10 | y | x+y=P(B)
|______|______|
| | |
B' | 3/10 | w | z+w=P(B')
|______|______|___
|
2/5 3/5 | 1
Now P(A|B)=1/3 is (1/10)/((1/10)+y)=1/3 which provides (1/10)+y=3/10 or
y=2/10 and the square can be completed.
A A'
_____________
| | |
B | 1/10 | 2/10 | 3/10
|______|______|
| | |
B' | 3/10 | 4/10 | 7/10
|______|______|___
|
2/5 3/5 | 1