Probability Examples
I. Blood types: All human blood can be “ABO-typed” as one of O, A, B, or AB, but the distribution of the types varies a bit among groups of people. Here is the distribution of blood types for a randomly chosen person in the U.S.
|
Blood Type
|
O
|
A
|
B
|
AB
|
|
U.S. probability
|
0.45
|
0.40
|
0.11
|
0.04
|
Consider a married couple.
1. What is the probability that the wife has type A blood and the husband type B blood?
p(A) ·p(B)=0.4x0.1=0.04
The probability is 0.04.
2. What is the probability that one of the couple has type A blood and the other has type B blood?
0.4x0.11+0.11x0.4=0.088
The probability is 0.088
3. What is the probability that a wife and husband share the same blood type?
p(OO)+p(AA)+p(BB)+p(AB AB)=(0.45)(0.45)+(0.4)(0.4)+(0.11)(0.11)+(0.04)(0.04)
=0.3762
II: Heart Disease is one of the leading causes of death in the U.S – approximately 25% of deaths in the U.S. can be attributed to Heart Disease. Medical researchers have determined that there are several risk factors, including High Blood Pressure, high cholesterol and diabetes. Approximately 49% of the population has at least one of these risk factors. ( Note: some individuals with these risk factors never do have a heart attack and some individuals who die of heart disease have no risk factors.). Information collected from the Center for Disease Control tells us the following:
P(dying Heart Disease) = P(H) = 0.25
P(having at least one risk factor for Heart Disease) = P(R)= 0.49
P(having at least one risk factor and dying of Heart disease) = P( H and R) = 0.16
(Use a Venn Diagram to model this data.)
(a.) What is the probability that an individual randomly selected from the population will NOT die of heart disease?
p(H)=0.25
p(H’)=1-0.25=0.75
(b.) What is the probability that an individual will die of heart disease and he/she has no risk factors?
p(H and not R)=0.25-0.16=0.09
(c.) What is the probability that an individual will die of Heart disease, given that the individual has at least one risk factor?
p(not H and R)=0.49-0.16=0.33
(d.) What is the probability that an individual had at least one risk factor, given that the individual died of Heart Disease?
p(R/H)=016/0.25=0.64=64%
III. Testing for HIV: Enzyme immunoassay (EIA) tests are used to screen blood specimens for the presence of antibodies to HIV, the virus that causes AIDS. Antibodies indicate the presence of the virus. The test is quite accurate but not always correct. Here are approximate probabilities of positive and negative EIA outcomes when the blood tested does and does not actually contain antibodies to HIV:
Test Result:
|
|
+
|
-
|
|
Antibodies Present
|
0.9985
|
0.0015
|
|
Antibodies absent
|
0.0060
|
0.9940
|
Suppose that 1% of a large population carries antibodies to HIV in their blood.
(a.) Draw a tree diagram for selecting person from the population and testing his or her blood.
(b.) What is the probability that the EIA test is positive for a randomly chosen person from this population?
P(T)=(0.9985×0.01)+(0.0060×0.99)=0.015925=1.59%
(c.) What is the probability that a person has the antibody, given that the EIA test is positive?
p(A/T)0.0985x0.01/0.015925=0.6268
Note: This illustrates a fact that is important when considering proposals for widespread testing for HIV, illegal drugs, or agents of biological warfare: if the condition being tested is uncommon in the population, many positives will be false positives.