Homework 6 Solutions(1a) Let P r(BB) be the probability that two blue balls are drawn. Then using Bayes’ Rulewe know thatP r(m − B|BB) =P r(BB|m−B)P r(m−B)P r(BB)Since the probability of drawing one blue ball when the majority of the balls are blue is2/3 (i.e. P r(B|m − B) = 2/3) and, conditional on the majority being blue, each drawis independent, we can calculate that P r(BB|m − B) = (2/3)2. From the problem weknow that P r(m − B) = 1/2. We then proceed to calculating P r(BB). Knowing thatP r(BB|m − R) = (1/3)2and P r(m − R) = 1/2 (where m-R indicates the majority ofthe balls are red) and using the law of total probability, we get:P r(BB) = P r(BB|m − B)P r(m − B) + P r(BB|m − R)P r(m − R)=(2/3)2· 1/2 + (1/3)2· 1/2 = 5/18Combining we get:P r(m − B|BB) =P r(BB|m−B)P r(m−B)P r(BB)=(2/3)2·(1/2)5/18= 4/5(1b) Let BBR denote the sequence of balls being drawn as Blue, Blue, Red. Once again byapplying Bayes’ Rule we know:P r(m − B|BBR) =P r(BBR|m−B)P r(m−B)P r(BBR)Since P r(B|m − B) = 2/3 and P r(R|m − B) = 1/3 (and draws are independent,conditional on the majority being blue), we know that P r(BBR|m − B) = (2/3)2·1/3. Similarly, since P r(B|m − R) = 1/3 and P r(R|m − R) = 2/3 , this impliesP r(BBR|m − R) = (1/3)2· (2/3). Once again we can use the law of total probabilityto calculate:P r(BBR) = Pr(BBR|m − B)P r(m − B) + P r(BBR|m − R)P r(m − R)= (2/3)2· 1/3 · 1/2 + (1/3)2· 2/3 · 1/2 = 1/9Putting it all together we get:P r(m − B|BBR) =P r(BBR|m − B)P r(m − B)P r(BBR)=(2/3)2· (1/3) · (1/2)1/9= 2/3(2) For this problem let + denote a positive test result and − denotes a negative test result;while D indicates a patient that really has the disease and H indicates a patient thatdoes not have the disease (i.e. is healthy).1(2a) We begin by noting P r(D) = 1/1000 = 0.001 and P r(H) = 999/1000 = 0.999. We areinterested in determining the probability of a person that tests positive actually hasthe disease, P r(D|+). Using Bayes’ Rule we know thatP r(D|+) =P r(+|D)P r(D)P r(+)We know from the problem that Pr(+|D) = 0.99 and P r(D) = 0.001. We can calcu-late:P r(+) = P r(+|D)P r(D) + P r(+|H)Pr(H) = 0.99 · 0.001 + 0.01 · 0.999ThereforeP r(D|+) =P r(+|D)P r(D)P r(+)==0.99 · 0.0010.001 · 0.99 + 0.999 · 0.01= 0.09016(2b) We are now interested in calculating, P r(D|−). Bayes’ Rule yields:P r(D|−) =P r(−|D)P r(D)P r(−)From the problem we know that P r(−|D) = 0.01 and P r(D) = 0.001. We nowcalculateP r(−) = P r(−|D)P r(D) + P r(−|H)Pr(H) = 0.01 · 0.001 + 0.99 · 0.999Giving us:P r(D|−) =P r(−|D)P r(D)P r(−)=0.01 · 0.0010.001 · 0.01 + 0.99 · 0.999= 0.00001011.From the two parts we see that P r(D|+) = 0.09016 and P r(D|−) = 0.00001011.The difference in the two results is because the prior probability that someone hasthe disease is very low and the test is very accurate. So although a positive testresults increases this probability by a large fraction it still remains small. Conversely,a negative test result means that the already low probability of having the diseasebecomes even tinier.(2c) The calculation proceeds exactly as in part a with the exception that Pr(D) now equals0.01 instead of 0.001 and Pr(H)=0.99 instead of 0.999:P r(D|+) =P r(+|D)P r(D)P r(+)=0.99 · 0.010.99 · 0.01 + 0.01 · 0.99= 0.5.2(3a) In the case where the product was actually Bad, an incorrect information cascade wouldtypically occur if the first few individuals received High signals and thus purchased theitem. If enough of these occur in a row, individuals will proceed to purchase the itemeven if they get a Low signal since the weight of the previous evidence is indicatingHigh signals (and correspondingly a Good product) and the payoffs of others cannotbe observed. However in this case, individuals can observe the actual payoffs of othersthrough the wiki. Therefore, adoption cascades probably won’t form, and certainlywill not persist. If the technology is Bad, this will be clear on the wiki through thereporting of the (on average) negative payoffs. New users will not infer bad informationfrom the adoption of previous players, given that their resulting average payoff can beseen.(3b) A cascade can form with the same probability as in the model we discussed in class. Ifthe technology is Good, but the first few people get Low signals, then they will choosenot to buy the product. Therefore, later players get no additional information from thewiki (since there will be no payoffs reported) that would help break the cascade. Laterplayers will only observe the sequence of previous players who chose not to adopt thetechnology and will correspondingly also choose to not adopt the technology (regardlessof their personal signal).(4a) Person 1 will follow his own signal, and Person 2 will follow the majority of his threesignals. Person 2 can infer the signal of person 1, but that doesn’t change his behavior.If the majority (at least 2 out of 3) of his signals goes one way, even with an inferredsignal to the opposite direction from Person 1, the two types of signals at least areequal and Player 2 will choose to follow the majority of his personal signals. This istrue even in the case where Player 2 receives two signals in one direction and one signalin the direction matching Player 1’s direction (giving Player 2 a total two of each typeof signals) since our standard tie breaking rule is that each player follows their ownsignal in cases of ‘’ties”.(4b) Person 2 will follow the majority of his three signals. Person 1 can infer what wasthe majority of Person 2’s three signals, but won’t know if was all identical signals, orone signal one way, and two the other way. However, Person 1 will follow Person 2’sdecision. In cases where Person 1’s signal agrees with Person 2’s action, this is clearlytrue as all the information Person 1 has. To understand why Person 1 should followPerson 2’s action even if Person 1’s signal disagrees with Person 2’s action, consider thefollowing case: Person 2 Accepts the technology and Person 1 receives a Low signal.Person 1 knows that Person 2 received three signals and that since Person 2 Acceptedthe technology then at least two out of the three signals received by Person 2 wereHigh. Therefore, at most two of the four total signals could have been Low (in whichcase Player 1 would be indifferent between the two actions), but if all three of Person2’s signals were High, then