CISC 7700X Midterm Exam 1. a 2. c 3. d 4. c 5. c 6. about 100^10. e.g for(a in all-values-of-A) for(b in all-values-of-B) for(c in all-values-of-C) for(d in ...etc. 7. 0.75 all outcomes: 0 wins: 0.125, e.g. \$1 * 0.5 * 0.5 * 0.5 1 wins: 0.375,0.375,0.375 2 wins: 1.125,1.125,1.125 3 wins: 3.375 median is (0.375+1.125)/2 = 0.75 8. 0.649519 eg. exp((log(0.125)+3*log(0.375)+3*log(1.125)+log(3.375))/8) 9. 1. 10. c 11. c 12. d 13. b 14. e, it is how conditional probability is defined. 15. 0.4000 given: P(V) = 0.1, P(-V) = 0.9 P(Y|V) = 0.6, P(-Y|V) = 0.4 P(Y|-V) = 0.1, P(-Y|-V) = 0.9 P(V|Y) = P(Y|V)P(V) / (P(Y|V)P(V) + P(Y|-V)P(-V) ) = (0.6 * 0.1) / ( 0.6 * 0.1 + 0.1 * 0.9) = 0.4000 16. 0.021739 given: P(NA|V) = 0.01, P(-NA|V) = 0.99 P(NA|-V) = 0.05, P(-NA|-V) = 0.95 P(V|NA) = P(NA|V) P(V) / ( P(NA|V) P(V) + P(NA|-V) P(-V) ) = ( 0.01*0.1 ) / ( 0.01*0.1 + 0.05*0.9 ) = 0.021739 17. P(V|NA,Y) = P(NA,Y|V)P(V) / P(NA,Y) We cannot calculate it since we are not given P(NA,Y|V) or P(NA,Y). 18. 0.1176 We need P(V|NA,Y) = P(NA,Y|V)P(V) / P(NA,Y), we assume NA and P are independent. e.g.: P(NA,Y|V) = P(NA|V)P(Y|V). P(V|NA,Y) = P(NA|V)P(Y|V)P(V) / ( P(NA|V)P(Y|V)P(V)+P(NA|-V)P(Y|-V)P(-V)) = (0.01*0.6*0.1) / ( 0.01*0.6*0.1 + 0.05*0.1*0.9) = 0.1176 19. c 20. everyone gets 5-points.