This is used to find conditional probabilities
based on a weighted average.
What is a weighted average?
Example
For a store, 40% of the customers are male.
The males spend $274 on average while the females spend $186 on average. What
is the overall average spent?
If the gender split were 50/50, we could add
$274 and $186 and divide the result by 2 to get $230. Note that we could
rewrite (274 + 186)/2 as (0.5)(274) + (0.5)(186).
However, the gender split is 40/60. Thus, the answer is (0.4)(274)
+ (0.6)(186) = 221.20. Note that $221.20 is closer to $186 than $274 since the
majority of customers are female.
Multiplication Rule
We begin with the formula for conditional
probability:
Crossmultiplying, we get:
This equation is used to find a weighted
average.
Example
10% of people live in the target market
9% of those living in the target market
respond to an offer
5% of those living outside the target market
respond to an offer
What percentage of people overall respond to
the offer?
To find the weighted average, we construct a
tree.
P(TM) = 0.1 P(R  TM) = 0.09
P(TM and R) = (0.1)(0.09) = 0.009
P(Not TM) = 0.9 P(R
 Not TM) = 0.05
P(Not TM and R) = (0.9)(0.05) = 0.045
P(R) = 0.009 + 0.045 = 0.054 = 5.4%
Some notes:
·
Since 10% live in
the target market, that means that 90% live outside the target market. These
two probabilities are the first terms on the left side of the multiplication
rule for their respective branches.
·
The second term
on the left side is the conditional probability of responding to the offer
given that the person either lives in the target market or lives outside it.
·
When we multiply
the two terms on the left side, we get the joint probability of the two events
on the left side. When we multiply P(TM) and P(R 
TM), we get P(TM and R).
·
The final percentage
of those who respond at 5.4% is between P(R  TM) = 9%
and P(R  not TM) = 5%. Note that the 5.4% is closer to 5% than 9% since the
majority live outside the target market (90% versus 10%).
What percentage of those who respond to the
offer live in the target market?
We want P(TM  R). In
its infancy, Bayes theorem was known as inverse conditional probability. This
is because a conditional probability, in this case P(R
 TM), is used to construct a weighted average. We then use the weighted
average to find the inverse conditional probability. For this problem, we can
use the probabilities from the tree:
Note that 16.67% is equivalent to 1 in 6. This
is greater than the 10% who live in the target market. This is due to the fact
that the response rate is higher in the target market.
Of those who do not respond, what percentage
live in the target market?
In theory, we could construct another tree.
However, we have enough information to construct a crosstab:

TM 
Not TM 
Total 
R 
0.009 
0.045 
0.054 
Not R 
0.091 
0.855 
0.946 
Total 
0.1 
0.9 
1.0 
We want P(TM  not
R).
Note that the 9.62% is less than the 10% who
live in the target market. Again, this is due to the fact that the response
rate is higher in the target market.
Example
A province is divided into three regions A, B
and C. Onehalf live in A, onethird in B and onesixth in C.
In the last election, the percentage in each
district who voted was 50.2% in A, 57.3% in B and 60.9% in C.
Of those who voted, the percentage in each
district who voted Liberal was 47.3% in A, 32.9% in B and 25.4% in C.
What percentage overall voted?
In this problem, the tree has 3 branches since
there are 3 regions.
P(A) = 1 / 2 P(V  A)_= 0.502
P(A and V) = 0.502/2 = 0.251
P(B) = 1 /3 P(V  B) = 0.573
P(B and V) = 0.573/3 = 0.191
P(C) = 1 /6 P(V  B) = 0.609
P(B and V) = 0.609/6 = 0.1015
P(V) = 0.251 + 0.191 + 0.1015 = 0.5435
= 54.35%
Note that 54.35% is between the smallest
voting percentage of 50.2% in region A and the largest voting percentage of
60.9% in region C.
Of those who voted, what percentage voted
Liberal?
To solve this problem, the joint probabilities
from the previous problem form the first terms of the left side of the
multiplication rule.
P(A and V) = 0.251 P(L  A and V) = 0.473
P(A and V and L) = (0.251)(0.473) = 0.118723
P(B and V) = 0.191 P(L  B and V) = 0.329
P(B and V and L) = (0.191)(0.329) = 0.062839
P(C and V) = 0.1015 P(L  C and V) = 0.254
P(C and V and L) = (0.1015)(0.254) = 0.025781
P(V and L) = 0.118723 + 0.062839 + 0.025781 =
0.207343 = 20.7343%
Note that in constructing this tree, the joint
probabilities have two common events, namely, that the person voted and that
the person voted Liberal. The only event that varies is the region. Thus, when
we add the probabilities, we get P(V and L).
However, 20.7343% is not the answer to the
problem. This is the percentage of the entire population that voted Liberal
which includes those who did not vote. We want P(L 
V).
Note that 38.15% is between the lowest
percentage of 25.4% in region C and the highest percentage of 47.3% in region
A.
Create a tree of voting patterns to determine
the percentage of voters who did not vote Liberal.
P(V) = 0.5435 P(V and L) = 0.207343
P(V and not L) = 0.5435 – 0.207343 = 0.336157
P(not V) = 0.4565
Note that P(not V) is
simply 1 – P(V) = 1 – 0.5435 = 0.4565. However, for this branch, this is the
end of the line since it is nonsensical to speak of a person voting Liberal if
the person did not vote. However, of those who did vote, we can divide them
into those who voted Liberal and those who did not. We want P(not
L  V).
Note that 38.15% and 61.85% sum to 100%.