1. Conditional Probability
Chapter 8 Probability
8.4
Conditional
Probability
MATHPOWERTM 12, WESTERN EDITION
8.4.1

2. 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:
8.4.2

3. Conditional Probability
Determine the conditional probability for each of the following:
a) Given P(B and A) = and P(B) = 0.78, find P(A|B).
P(A|B) =
Given P(blonde and tall) = 0.5 and P(blonde) = 0.73,
find P(A|B).
P(A|B) =
8.4.3

4. 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)
= 0.10 x 0.99
= 0.099
sick
0.01
test negative
P(sick and negative)
0.10
0.90
= 0.10 x 0.01
= 0.001
0.05
test positive
P(not sick and positive)
not
sick
= 0.90 x 0.05
= 0.045
test negative
P(not sick and negative)
0.95
= 0.90 x 0.95
= 0.855
8.4.4

5. Finding Conditional Probability [cont’d]
P(B and A) = P(sick and positive) =
P(B) = P(positive)
P(positive) = P(sick and positive) or P(not sick and positive) =
= 0.144
Therefore, the probability
of the person testing
positive and actually
having the disease is
8.4.5

6. 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)
= x 0.95 =
cancer
0.05
test negative
P(sick and negative)
0.008
0.992
= x 0.05 =
0.05
test positive
P(not sick and positive)
not
cancer
= x 0.05 =
test negative
P(not sick and negative)
0.95
= x 0.95 =
8.4.6

7. Finding Conditional Probability [cont’d]
P(B and A) = P(cancer and negative) =
P(B) = P(negative)
P(negative) = P(cancer and negative) or P(not cancer and negative) = =
Therefore, the probability
of the person testing
negative and actually
having the disease is
8.4.7

8. Assignment Suggested Questions: Pages 391 and 392
1-10, 11 a, 12 a, 13 b
8.4.8