QUESTION 11)

3.2. CONDITIONAL PROBABILITY 107 GUIDED PRACTICE 3.43 Jose visits campus every Thursday evening. However, some days the parking garage is full, often due to college events. There are academic events on 35%% of evenings, sporting events on 20% of G evenings, and no events on 45% of evenings. When there is an academic event, the garage fills up about 25% of the time, and it fills up 70% of evenings with sporting events. On evenings when there are no events, it only fills up about 5% of the time. If Jose comes to campus and finds the garage full, what is the probability that there is a sporting event? Use a tree diagram to solve this problem. 37 EXAMPLE 3.44 Here we solve the same problem presented in Guided Practice 3.43, except this time we use Bayes' Theorem. The outcome of interest is whether there is a sporting event (call this A1), and the condition is that the lot is full (B). Let A2 represent an academic event and Ag represent there being no event on campus. Then the given probabilities can be written as P(A1) = 0.2 P(A2) = 0.35 P(A3) = 0.45 P(B|Al) = 0.7 P(B|A2) = 0.25 P(B|A3) = 0.05 E Bayes' Theorem can be used to compute the probability of a sporting event ( A1) under the condition that the parking lot is full (B): P(B|A1) P(Al ) P(A1 |B) = P(B|A,) P(A1) + P(B|A2) P(A2) + P(B| A3) P(A3) (0.7) (0.2) (0.7) (0.2) + (0.25) (0.35) + (0.05)(0.45) = 0.56 Based on the information that the garage is full, there is a 56% probability that a sporting event is being held on campus that evening. 37The tree diagram, with three primary branches, is shown to Event Garage full the right. Next, we identify two probabilities from the tree dia- Full, 0.25 - - --. 0.35*0.25 = 0.0875 gram. (1) The probability that Academic, 0.35 there is a sporting event and Spaces Available, 0.7 the garage is full: 0.14. (2) The probability the garage is full: Full, O.!...... 0.2 0.7 = 0.14 0.0875 + 0.14 + 0.0225 = 0.25. Sporting, 0.20 Then the solution is the ratio of Spaces Available, 0.3 0.2*0.3 = 0.06 these probabilities: 0:14 = 0.56. Full, 0.05 If the garage is full, there is a None, 0.45 --. 0.45*0.05 = 0.0225 56% probability that there is a Spaces Available, _0: 30 45*0.95 = 0.4275 sporting event.104 CHAPTER 3. PROBABILITY GUIDED PRACTICE 3.41 After an introductory statistics course, 78% of students can successfully construct tree diagrams. G Of those who can construct tree diagrams, 97% passed, while only 57% of those students who could not construct tree diagrams passed. (a) Organize this information into a tree diagram. (b) What is the probability that a randomly selected student passed? (c) Compute the probability a student is able to construct a tree diagram if it is known that she passed. 36 3.2.8 Bayes' Theorem In many instances, we are given a conditional probability of the form P(statement about variable 1 | statement about variable 2) but we would really like to know the inverted conditional probability: P(statement about variable 2 | statement about variable 1) Tree diagrams can be used to find the second conditional probability when given the first. However, sometimes it is not possible to draw the scenario in a tree diagram. In these cases, we can apply a very useful and general formula: Bayes' Theorem. We first take a critical look at an example of inverting conditional probabilities where we still apply a tree diagram. 36(a) The tree diagram is shown to the right. (b) Identify which two joint probabilities represent students who passed, and add them: P(passed) = 0.7566+0.1254 = 0.8820. (c) P(construct tree diagram | passed) = 8-8320 = 0.8578. Able to construct Pass class tree diagrams pass, 0.97 - - - - - 0.78*0.97 = 0.7566 yes, 0.78 fail, 0.03 -- - - 0.78*0.03 = 0.0234 pass, 0.57 0.22*0.57 = 0.1254 no, 0.22 fail, 0.43 - - - - - - - - 0.22*0.43 = 0.09463.2. CONDITIONAL PROBABILITY 105 EXAMPLE 3.42 In Canada, about 0.35% of women over 40 will develop breast cancer in any given year. A common screening test for cancer is the mammogram, but this test is not perfect. In about 11% of patients with breast cancer, the test gives a false negative: it indicates a woman does not have breast cancer when she does have breast cancer. Similarly, the test gives a false positive in 7% of patients who do not have breast cancer: it indicates these patients have breast cancer when they actually do not. If we tested a random woman over 40 for breast cancer using a mammogram and the test came back positive - that is, the test suggested the patient has cancer - what is the probability that the patient actually has breast cancer? Notice that we are given sufficient information to quickly compute the probability of testing positive if a woman has breast cancer (1.00 - 0.11 = 0.89). However, we seek the inverted probability of cancer given a positive test result. (Watch out for the non-intuitive medical language: a positive test result suggests the possible presence of cancer in a mammogram screening.) This inverted probability may be broken into two pieces: P(has BC | mammogram+) = P(has BC and mammogram +) P(mammogram* where "has BC" is an abbreviation for the patient having breast cancer and "mammogram " means the mammogram screening was positive. We can construct a tree diagram for these probabilities: Truth Mammogram positive, 0.89 0.0035*0.89 = 0.00312 cancer, 0.0035 negative, 0.11 0.0035*0.11 = 0.00038 E positive, 0.07 - - . .".".""-". 0.9965*0.07 = 0.06976 no cancer, 0.9965 negative, 0.93 0.9965 0.93 = 0.92675 The probability the patient has breast cancer and the mammogram is positive is P(has BC and mammogram) = P(mammogram | has BC) P(has BC) = 0.89 x 0.0035 = 0.00312 The probability of a positive test result is the sum of the two corresponding scenarios: P(mammogram) = P(mammogram and has BC) + P(mammogram and no BC) = P(has BC)P(mammogram | has BC) + P(no BC) P(mammogram | no BC) = 0.0035 x 0.89 + 0.9965 x 0.07 = 0.07288 Then if the mammogram screening is positive for a patient, the probability the patient has breast cancer is P(has BC | mammogram ) = - P(has BC and mammogram +) P(mammogram) 0.00312 0.07288 ~0.0428 That is, even if a patient has a positive mammogram screening, there is still only a 4% chance that she has breast cancer.106 CHAPTER 3. PROBABILITY Example 3.42 highlights why doctors often run more tests regardless of a first positive test result. When a medical condition is rare, a single positive test isn't generally definitive. Consider again the last equation of Example 3.42. Using the tree diagram, we can see that the numerator (the top of the fraction) is equal to the following product: P(has BC and mammogram) = P(mammogram | has BC)P(has BC) The denominator - the probability the screening was positive - is equal to the sum of probabilities for each positive screening scenario: P(mammogram) = P(mammogram and no BC) + P(mammogram and has BC) In the example, each of the probabilities on the right side was broken down into a product of a conditional probability and marginal probability using the tree diagram. P(mammogram) = P(mammogram and no BC) + P(mammogram and has BC) = P(mammogram | no BC) P(no BC) + P(mammogram* | has BC) P(has BC) We can see an application of Bayes' Theorem by substituting the resulting probability expressions into the numerator and denominator of the original conditional probability. P(has BC | mammogram) P(mammogram | has BC)P(has BC) P(mammogram | no BC) P(no BC) + P(mammogram | has BC) P(has BC) BAYES' THEOREM: INVERTING PROBABILITIES Consider the following conditional probability for variable 1 and variable 2: P(outcome Aj of variable 1 | outcome B of variable 2) Bayes' Theorem states that this conditional probability can be identified as the following frac- tion: P(B|Al) P(A1) P(B|Al) P(Al) + P(B A2) P(A2) + .. . + P(B Ak)P(Ak) where A2, A3, ..., and Ax represent all other possible outcomes of the first variable. Bayes' Theorem is a generalization of what we have done using tree diagrams. The numerator identifies the probability of getting both A, and B. The denominator is the marginal probability of getting B. This bottom component of the fraction appears long and complicated since we have to add up probabilities from all of the different ways to get B. We always completed this step when using tree diagrams. However, we usually did it in a separate step so it didn't seem as complex. To apply Bayes' Theorem correctly, there are two preparatory steps: (1) First identify the marginal probabilities of each possible outcome of the first variable: P(A]), P(A2), ...; P(Ak). (2) Then identify the probability of the outcome B, conditioned on each possible scenario for the first variable: P(B|A1), P(B|A2), ..., P(B|Ak). Once each of these probabilities are identified, they can be applied directly within the formula. Bayes' Theorem tends to be a good option when there are so many scenarios that drawing a tree diagram would be complex.108 CHAPTER 3. PROBABILITY GUIDED PRACTICE 3.45 G Use the information in the previous exercise and example to verify the probability that there is an academic event conditioned on the parking lot being full is 0.35.38 GUIDED PRACTICE 3.46 G In Guided Practice 3.43 and 3.45, you found that if the parking lot is full, the probability there is a sporting event is 0.56 and the probability there is an academic event is 0.35. Using this information, compute P(no event | the lot is full). 39 The last several exercises offered a way to update our belief about whether there is a sporting event, academic event, or no event going on at the school based on the information that the parking lot was full. This strategy of updating beliefs using Bayes' Theorem is actually the foundation of an entire section of statistics called Bayesian statistics. While Bayesian statistics is very important and useful, we will not have time to cover much more of it in this book. 38 Short answer: P(A2| B) = P(B|A2) P(A2) P(B|A1) P(A1) + P(B|A2)P(A2) + P(B|A3) P(A3) (0.25) (0.35) (0.7) (0.2) + (0.25) (0.35) + (0.05) (0.45) = 0.35 39 Each probability is conditioned on the same information that the garage is full, so the complement may be used: 1.00 - 0.56 - 0.35 = 0.09.Q11) (22-pts) The table below describes the smoking habits of a representative group of US adults (> 18-years-old) who are asthma sufferers. Light Heavy Nonsmoker smoker smoker Total Men 386 66 68 520 Women 328 89 71 488 Total 714 155 139 1008 You might need to construct new probability tables or trees to answer the following questions (show your work), if one of the 1008 subjects is randomly selected, a) convert your table to a probability table. b) find the probability that these asthma sufferers are light smokers or heavy smokers, c) find the probability that these asthma sufferers are light smokers or heavy smokers, d) find the probability that the person chosen is a light smoker. e) find the probability that the person chosen is a woman. f) find the probability that the person chosen is a light smoker and it is a woman. g) find the probability that the person chosen is a nonsmoker given that it is a woman. find the probability that the person chosen is a woman given that it is a nonsmoker. i) find the probability that the person chosen is a man given that it is a smoker (you may need to reconstruct the table to account for collapsed groups). j) find the probability that the person chosen is a smoker given that it is a man (you may need to reconstruct the table). k) Use g as an expression in Bayes' Theorem (section 3.2.8 in your book Bayes' Theorem) to answer h 1) Use i as an expression in Bayes' Theorem (section 3.2.8 in your book Bayes' Theorem) to answer j You now gained additional information. Nearly 14 out of 100 U.S. adults aged 18 years or older (14.0%) currently smokes cigarettes, and the percent of US adults aged 18 and over who currently have asthma is 8.4%. Using the information from the 1008 subjects and the new information (you may have to reformulate a new table), m) find the probability that an adult individual (M or F) randomly chosen from the US population has asthma and smokes (are light smokers or heavy smokers), n) find the probability that an adult individual (M or F) randomly chosen from the US population has asthma and DO NOT smoke, o) find the probability that an adult individual (M or F) randomly chosen from the US population has asthma given they smoke (are light smokers or heavy smokers), p) find the probability that an adult individual (M or F) randomly chosen from the US population has asthma given they DO NOT smoke, q) What can you conclude from o and p