MATHPOWERTM 12, WESTERN EDITION 8.4.1

8.4.2

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:

P(A | B)

P(B and A)P(B)

, where P(B) 0

8.4.3

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(A | B)

P(B and A)P(B)

P(A | B)

0.725

0.78

P(A|B) = 0.9295

a) Given P(blonde and tall) = 0.5 and P(blonde) = 0.73, find P(A|B).

P(A | B)

P(B and A)P(B)

P(A | B)

0.5

0.73P(A|B) = 0.6849

8.4.4

Finding Conditional ProbabilityIt 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?

0.10

0.90

sick

notsick

test positive

test positive

test negative

test negative

P(sick and positive)

P(sick and negative)

P(not sick and positive)

P(not sick and negative)

0.99

0.01

0.95

0.05

= 0.10 x 0.99= 0.099

= 0.10 x 0.01= 0.001

= 0.90 x 0.05= 0.045

= 0.90 x 0.95= 0.855

8.4.5

Finding Conditional Probability [cont’d]

P(A | B)

P(B and A)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

P(A | B)

P(B and A)P(B)

P(A | B)

0.099

0.144

Therefore, the probabilityof the person testingpositive and actuallyhaving the disease is0.6875.

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?

8.4.6

Finding Conditional Probability

0.008

0.992

cancer

notcancer

test positive

test positive

test negative

test negative

P(sick and positive)

P(sick and negative)

P(not sick and positive)

P(not sick and negative)

0.95

0.05

0.95

0.05

= 0.008 x 0.95= 0.0076

= 0.008 x 0.05= 0.0004

= 0.992 x 0.05= 0.0496

= 0.992 x 0.95= 0.9424

8.4.7

Finding Conditional Probability [cont’d]

P(A | B)

P(B and A)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

P(A | B)

P(B and A)P(B)

P(A | B)

0.0004

0.9428

Therefore, the probabilityof the person testingnegative and actuallyhaving the disease is0.0004.

Suggested Questions:Pages 656-657#5-22 all

8.4.8