Conditional Probability Chapter 8 Probability 8.4 Conditional Probability 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: 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(A|B) = 0.9295 Given P(blonde and tall) = 0.5 and P(blonde) = 0.73, find P(A|B). P(A|B) = 0.6849 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) = 0.10 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

Finding Conditional Probability [cont’d] 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 of the person testing positive and actually having the disease is 0.6875. 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) = 0.008 x 0.95 = 0.0076 cancer 0.05 test negative P(sick and negative) 0.008 0.992 = 0.008 x 0.05 = 0.0004 0.05 test positive P(not sick and positive) not cancer = 0.992 x 0.05 = 0.0496 test negative P(not sick and negative) 0.95 = 0.992 x 0.95 = 0.9424 8.4.6

Finding Conditional Probability [cont’d] 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 of the person testing negative and actually having the disease is 0.0004. 8.4.7

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