Blood Type Probability O 0.42 A 0.43 B 0.11 AB 0.04


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1 PubH 6414 Worksheet 5a: Probability Principles 1 of 6 Example 1: Distribution of Blood Types in the US Blood Type Probability O 0.42 A 0.43 B 0.11 AB 0.04 Use the distribution of blood types in the US to find the probability that a randomly selected person in the US has: a. blood type A or B: P (A or B) = = 0.54 b. blood type O or A P (O or A) = = 0.85 c. a blood type other than A P(not A) = P(A c ) = 1 P(A) = = 0.57 d. a blood type other than AB P(not AB) = P(AB c ) = 1 P(AB) = = 0.96 e. blood type O or AB P(O or AB) = = 0.46 f. blood type AB P(AB) = 0.04 Which of these combinations are complementary events? (more than one answer is possible) 1. a and b 2. b and c 3. a and e 4. c and d 5. d and f 6. e and f
2 PubH 6414 Worksheet 5a: Probability Principles 2 of 6 Example 2: Distribution of Blood Type in the US by Gender Blood Type Males Females Total O A B AB Gender and Blood type are nonmutually exclusive events because they can occur simultaneously. Find the following marginal probabilities: 1. P(Type A) = P(Type AB) = P(Female) = P(Type O) = In this 4 X 2 table there are 6 marginal probabilities 6. The sum of the 4 row marginal probabilities = The sum of the 2 column marginal probabilities = 1.0 Find the following joint probabilities 1. P(Male and Type O) = P(Female and Type B) = P(Male and Type AB) = P(Female and Type A) = In this table there are 8 joint probabilities. 6. The sum of the 8 joint probabilities = 1.0
3 PubH 6414 Worksheet 5a: Probability Principles 3 of 6 Example 3: Distribution of Blood Type in the US by Gender Blood Type Males Females Total O A B AB Calculate the following conditional probabilities: 1. P(Type A Male) = 0.215/0.50 = P(Type B Female) = 0.055/0.50 = P(Type O Male) = 0.21/0.50 = P(Type AB Female) = 0.02/.050 = The sum of all blood type probabilities conditional on being female = The sum of all blood type probabilities conditional on being male = 1.0 Confirm that blood type and gender are independent events using the multiplication rule for independent events: 1. P(Male and O) = P(M)* (Type O) = 0.5*0.42 = P(Female and A) = P(F)*P(Type A) = 0.5*0.43 = P(Female and B) = P(F)*P(Type B) = 0.5*0.11 = P(Male and AB) = P(M)*P(Type AB) = 0.5*0.04=0.02
4 PubH 6414 Worksheet 5a: Probability Principles 4 of 6 Blood Type Males Females Total O A B AB Demonstrate that the probability of blood type is independent of gender by calculating the following conditional probabilities and comparing to the marginal probability: 1. P(Type A Female) = P(A and Female) / P(Female) = / 0.5 = 0.43 = P(A) 2. P(Type B Male) = P(B and Male) / P(M) = / 0.5 = 0.11 = P(B) 3. P(Type AB Female) = P(Female and AB) / P(F) = 0.02 / 0.5 = 0.04 = P(AB) 4. P(Type O Male) = P(Male and O) / P(Male) = 0.21 / 0.5 = 0.42 When marginal probabilities and conditional probabilities are equal, the two events are independent
5 PubH 6414 Worksheet 5a: Probability Principles 5 of 6 Example 4: Distribution of Blood Types by Country for Australia and Finland Blood Type Australia Finland Total O A B AB Total probabilities in this table assume equal proportions of observations are from each country. Calculate the following and demonstrate that blood type and country of origin are not independent events for these two countries: 1. P(Type A Finland) = 0.22 / 0.50 = 0.44 P(A) 2. P(Type A Australia) = 0.19 / 0.5 = 0.38 P(A) 3. P(Type O and Australia) = P(O) *P(Aust) = 0.40*0.5 = P(Type AB and Finland) = 0.04 P(AB) *P (Finland) = 0.055*0.5 = P(Type AB Australia) = / 0.5 = 0.03 P(Type AB) = P(Type AB Finland) = 0.04 / 0.5 = 0.08 P(Type AB) = The two conditional probabilities of Type AB are not equal and do not equal the marginal probability of Type AB. Events are not independent if joint probabilities do not equal the product of the marginal probabilities conditional probabilities do not equal marginal probabilities two conditional probabilities of the same event are not equal
6 PubH 6414 Worksheet 5a: Probability Principles 6 of 6 Example 4 Table of results for survey of daycare status and ear infection by 6 months for 260 infants Ear Infection by 6 Months Daycare Yes No Total Yes No Total Marginal, joint and conditional probabilities can also be calculated from a table of cell counts. The denominator for marginal and joint probabilities is the total number of observations. The denominator for conditional probabilities is the number of observations in the condition group which is the group behind the. Calculate the following from the table counts 1. Marginal probability that an infant had an ear infection by 6 months? P(ear infection) = 103/260 = Marginal probability that an infant was in daycare? P(daycare) = 120/260 = Joint probability of being in daycare and having an ear infection by 6 months? P(ear infection and daycare) = 58/260 = Joint probability of being in daycare and not having an ear infection by 6 months? P(daycare and no ear infection) = 62/260 = Conditional probability of having an ear infection by 6 months given that the infant is in daycare? P(ear infection daycare) = 58/120 = What is the conditional probability of having an ear infection given that the infant was not in daycare? P(ear infection no daycare) = 45/140 = Are daycare status (Yes / No) and ear infection status (Yes / No) independent events? Provide reasons. No the joint probabilities are not products of the marginal probabilities and the two conditional probabilities of having an ear infection given daycare or no daycare are not equal
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