MATHPOWERTM 12, WESTERN EDITION 8.4.1
Conditional Probability
If A and B are events from an experiment, the conditional
probability of B given A (P(A|B)), is the probability that
Event B will occur given that Event A has already occurred.
The conditional probability is equal to the probability that
B and A will occur divided by the probability that B will occur.
This is given in Bayes’ Formula:
P(B and A)
P(A | B) 
, where P(B)  0
P(B)
8.4.2
Conditional Probability
Determine the conditional probability for each of the following:
a) Given P(B and A) = 0.725 and P(B) = 0.78, find P(A|B).
P(B and A)
P(A | B) 
P(B)
P(A|B) = 0.9295
0.725
P(A | B) 
0.78
a) Given P(blonde and tall) = 0.5 and P(blonde) = 0.73,
find P(A|B).
P(B and A)
P(A | B) 
P(B)
0.5
P(A | B) 
P(A|B) = 0.6849
0.73
8.4.3
Finding Conditional Probability
It is known that 10% of the population has a certain disease.
For a patient without the disease, a blood test for the disease
Shows “not positive” 95% of the time. For a patient with the
Disease, the blood test shows “positive” 99% of the time.
What is the probability that a person whose blood test
is positive for the disease actually has the disease?
0.99
test positive
P(sick and positive)
0.01
test negative
P(sick and negative)
0.05
test positive
sick
0.10
0.90
not
sick
0.95
test negative
= 0.10 x 0.99
= 0.099
= 0.10 x 0.01
= 0.001
P(not sick and positive)
= 0.90 x 0.05
= 0.045
P(not sick and negative)
= 0.90 x 0.95
= 0.855
8.4.4
Finding Conditional Probability [cont’d]
P(B and A)
P(A | B) 
P(B)
P(B and A) = P(sick and positive)
= 0 .099
P(B) = P(positive)
P(positive) = P(sick and positive) or P(not sick and positive)
= 0.099 + 0.045
= 0.144
Therefore, the probability
P(B and A)
of the person testing
P(A | B) 
P(B)
positive and actually
having the disease is
0.099
P(A | B) 
0.6875.
0.144
8.4.5
Finding Conditional Probability
A new medical test for cancer is 95% accurate.
If 0.8% of the population suffer from cancer,
what is the probability that a person selected at
random will test negative and actually have cancer?
0.95
test positive
P(sick and positive)
0.05
test negative
P(sick and negative)
0.05
test positive
cancer
0.008
0.992
not
cancer
0.95
test negative
= 0.008 x 0.95
= 0.0076
= 0.008 x 0.05
= 0.0004
P(not sick and positive)
= 0.992 x 0.05
= 0.0496
P(not sick and negative)
= 0.992 x 0.95
= 0.9424
8.4.6
Finding Conditional Probability [cont’d]
P(B and A)
P(A | B) 
P(B)
P(B and A) = P(cancer and negative)
= 0.0004
P(B) = P(negative)
P(negative) = P(cancer and negative) or P(not cancer and negative)
= 0.0004 + 0.9424
= 0.9428
Therefore, the probability
P(B and A)
of the person testing
P(A | B) 
P(B)
negative and actually
having the disease is
0.0004
P(A | B) 
0.0004.
0.9428
8.4.7