PubHlth 540 Fall 2013 Exam II Choice A (Probability) - SOLUTIONS Page 1 of 9
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1 PubHlth 540 Fall 2013 Exam II Choice A (Probability) - SOLUTIONS Page 1 of 9 1. (10 points total) A couple wants to have three children. Assume that the probabilities of a newborn being male or being female are the same and that the gender of one child does not influence the gender of another child. 1a. (5 points) What is the sample space for having three children (gender of the first child, gender of the second child, gender of the third child)? There are 8 equally likely outcomes in the sample space under the assumptions given. {M, M, M} {M, M, F} {M, F, F} {M, F, M} {F, F, F} {F, F, M} {F, M, M} {F, M, F} 1b. (5 points). What is the probability of each elementary outcome (gender of the first child, gender of the second child, gender of the third child) Answer =.125 As the 8 elementary outcomes are all equally likely, each has probability 1/8 =.125
2 PubHlth 540 Fall 2013 Exam II Choice A (Probability) - SOLUTIONS Page 2 of 9 2. (10 points total) The setting is the same as for question #1: A couple wants to have three children. Assume that the probabilities of a newborn being male or being female are the same and that the gender of one child does not influence the gender of another child. This same couple is now wondering how many boys they might get if they have three children. What is the probability model for the number of boys they will get if they have three children. Hint - In specifying your answer, you need to provide two things: the sample space and the probabilities of each outcome in the sample space. Sample Space X = # boys in 3 births Probability [ X = x ] For your reference = Pr[GGG] = Pr[BGG] + Pr[GBG] + Pr[GGB] = Pr[BBG] + Pr[BGB] + Pr[GBB] = Pr[BBB] Exhausts all possibilities Sum = 100% or 1
3 PubHlth 540 Fall 2013 Exam II Choice A (Probability) - SOLUTIONS Page 3 of 9 3. (20 points total) A certain research company analyzed the nutritional content of 77 brands of packaged breakfast cereals. For each cereal, one of the measurements made was the fat content (in grams) per serving. Call this random variable X. The following is the probability model for X = fat content (in grams) per serving. X Probability Consider the following two events: A = { fat content is 4 grams or greater } B = { fat content is 5 grams or less } 3a. (5 points) What outcomes make up the event A? What is Probability [ A ]? Two mutually exclusive outcomes make up the event A: X=4 and X=5 Because X=4 and X=5 are mutually exclusive Pr [A] = Pr[X=4] + Pr[X=5] = =.02 3b. (5 points) What outcomes make up the event B? What is Probability [ B]? Six mutually exclusive outcomes make up the event B: X=0 and X=1 and X=2 and X=3 and X=4 and X=5 Because these are all mutually exclusive Pr [B] = Pr[X=0] + Pr[X=1] + Pr[X=2] + Pr[X=3] + Pr[X=4] + Pr[X=5] = = 1.00
4 PubHlth 540 Fall 2013 Exam II Choice A (Probability) - SOLUTIONS Page 4 of 9 3c. (5 points) What outcomes make up the event A or B?. What is Probability [ A or B ]? There is more than one way to solve this question. Approach #1 Six mutually exclusive outcomes make up the event A or B: X=0 and X=1 and X=2 and X=3 and X=4 and X=5 Pr [A or B] = Pr[X=0] + Pr[X=1] + Pr[X=2] + Pr[X=3] + Pr[X=4] + Pr[X=5] = = 1.00 Approach #2 First note the mutually exclusive outcomes make up the event A and B: These are X=4 and X=5 Pr[A and B] = Pr[X=4] + Pr[X=5] = =.02 Now use the definition of the probability of a union of events and your answers to questions #3a and #3b: Pr [A or B] = Pr[A] + Pr[B] - Pr[A and B] = = d. (5 points) Why is Probability [A or B] Probability[A] + Probability[B]? Because events A and B are not mutually exclusive.
![Approach #1 Six mutually exclusive outcomes make up the event A or B: X=0 and X=1 and X=2 and X=3 and X=4 and X=5 Pr [A or B] = Pr[X=0] + Pr[X=1] + Pr[X=2] + Pr[X=3] + Pr[X=4] + Pr[X=5] =.35 +.39 +.](/docs-images/41/22570986/images/page_4.jpg "18 +.06 +.01 +.01 = 1.00 Approach #2 First note the mutually exclusive outcomes make up the event A and B: These are X=4 and X=5 Pr[A and B] = Pr[X=4] + Pr[X=5] =.01 +.01 =.02 Now use the definition of the probability of a union of events and your answers to questions #3a and #3b: Pr [A or B] = Pr[A] + Pr[B] - Pr[A and B] =.")
5 PubHlth 540 Fall 2013 Exam II Choice A (Probability) - SOLUTIONS Page 5 of 9 4. (10 points total) The following is the probability model for the blood type of a randomly chosen person in the United States. Blood Type Probability O 0.45 A 0.40 B 0.11 AB??? 4a. (5 points) What is the probability that a randomly chosen person has blood type AB? Answer =.04.: Because the four blood types exhaust all possibilities, the probability of blood type AB can be obtained by subtraction from the total of the probabilities which must be 100% or 1. Pr[blood type is AB] = b. (5 points) Suppose Maria has blood type B. She can safely receive blood transfusions from people with blood types O or B. What is the probability that a randomly selected person in the United States can donate blood to Maria?.56 Two mutually exclusive outcomes make up the event of blood type O or B: X= blood type is O and X= blood type is B Pr [blood type is O or B] = Pr[blood type is O] + Pr[blood type is B] = =.56
6 PubHlth 540 Fall 2013 Exam II Choice A (Probability) - SOLUTIONS Page 6 of 9 5. (20 points total) Sickle-cell anemia is a hereditary medical condition affecting red blood cells that is thought to protect against malaria, a debilitating parasitic infection of the liver and blood. That would explain why the sickle-cell trait is found in people who originally came from Africa, where malaria is widespread. A study in Africa tested 543 children for the sickle-cell trait and also for malaria infection. In all, 25% of the children had sickle-cell, and 6.6% of the children had both sickle-cell and malaria. Overall, 34.6% of the children had malaria. Use this information to answer the following. Preliminary: Use the information provided to complete the 2x2 table of probabilities. Information Provided Malaria Sickle Cell 6.6% had both sickle cell and malaria NO sickle cell NO malaria 25% had sickle cell 34.6% had malaria 1 or 100% Probabilities (obtained by subtraction, as appropriate) Malaria NO malaria Sickle Cell NO sickle cell or 100%
7 PubHlth 540 Fall 2013 Exam II Choice A (Probability) - SOLUTIONS Page 7 of 9 5a. (5 points) What is the probability that a given child has neither malaria nor sickle-cell?.47 Pr[neither malaria nor sickle cell] =.47 5b. (5 points) What is the probability that a given child has either malaria or sickle-cell?.53 Approach #1 Pr [EITHER malaria or sickle cell ] = 1 Pr[neither malaria nor sickle cell] = =.53 Approach #2 Pr[EITHER malaria or sickle cell] = Pr[malaria] + Pr[sickle cell] Pr[both] =.346 = =.53 5c. (5 points) Perform the necessary calculations and then answer the following question: Are the events sickle-cell trait and malaria independent? Malaria and sickle cell are independent if: Pr[ malaria sickle cell ] = Pr[malaria] Solution for Pr[malaria sickle cell ] =.066/.25 =.264 Solution for Pr[malaria] =.346 Since the two are not equal, the two events are NOT independent. 5d. (5 points) What does your answer to #5c suggest to you about the relationship between sickle-cell and malaria? Explain your answer in 1-2 sentences. These data suggest that the presence of sickle cell trait confers some protection against malaria, as the probability of malaria is reduced in the presence of sickle cell trait.
![.53 Approach #1 Pr [EITHER malaria or sickle cell ] = 1 Pr[neither malaria nor sickle cell] = 1 -.47 =.53 Approach #2 Pr[EITHER malaria or sickle cell] = Pr[malaria] + Pr[sickle cell] Pr[both] =.](/docs-images/41/22570986/images/page_7.jpg "346 =.25 -.066 =.53 5c. (5 points) Perform the necessary calculations and then answer the following question: Are the events sickle-cell trait and malaria independent?")
8 PubHlth 540 Fall 2013 Exam II Choice A (Probability) - SOLUTIONS Page 8 of 9 6. (30 points total) Enzyme immunoassay tests are used to screen blood specimens for the presence of antibodies to HIV, the virus that causes AIDS. The presence of antibodies indicates the presence of the HIV virus. The test is quite accurate but is not always correct. The following table gives the probabilities of positive and negative test results when the blood tested does and does not actually contain antibodies to HIV. Test Result Positive (+) Negative (-) Antibodies present Antibodies absent Suppose that 1% of a large population carries antibodies to HIV in their blood. 6a. (10 points) Draw a tree diagram for selecting a person from this population (outcomes: antibodies present or absent) and for testing his or her blood (outcomes: test positive or negative)..01 present Probability.9985 Test + (.01)(.9985) = Test - (.01)(.0015) = antibodies Test + (.99)(.006) = absent.994 Test - (.99)(.994) = Total = 1 or 100%
9 PubHlth 540 Fall 2013 Exam II Choice A (Probability) - SOLUTIONS Page 9 of 9 6b. (10 points) What is the probability that the test is positive for a randomly chosen person for this population?.016 The tree shows 4 mutually exclusive outcomes for a person who either has or does not have the antibody and who either tests positive or negative. Thus, the answer is obtained by summing the probability of the mutually exclusive outcomes that satisfy the event of a positive test. Pr[test positive] = Pr[antibody and positive test] + Pr[NO antibody and positive test] = = c. (10 points) What is the probability that a person in this population has the HIV virus given that he or she tests negative? Bayes Rule Pr[antibody test-]= Pr[antibody and test-] Pr[test-] pr[antibody]*pr[test- antibody] = pr[antibody]*pr[test- antibody]+pr[noantibody]*pr[test- Noantibody] (.01)(.0015) = (.01)(.0015)+(.99)(.994) =
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