CISC 7700X Final Exam 1. b 2. 1 # ( (1 + 0.35) + (1 + 0.35) + (1 -0.70)) / 3 = 1, or 0% net-return. 3. 0.8177 # (1.35 * 1.35 * 0.30)^(1/3) 4. c 5. b 6. b, or e if ~10 # sqrt(2*7^2) 7. b # we found a random widget, there is a 50% chance that the serial number 1234 is within the interquartile range of all serial numbers. 8. d 9. e # invalid; p(x,y)=p(x|y)p(y)=p(y|x)p(x) 10. d 11. b 12. a 13. c 14. 0.3571 # P(speeding) = 0.1, P(-speeding) = 0.9, P(bmw|speeding) = 0.5, P(bmw|-speeding) = 0.1 # P(speeding|bmw) = P(bmw|speeding)P(speeding) / ( P(bmw|speeding)P(speeding) + P(bmw|-speeding)P(-speeding)) # = (0.5 * 0.1) / ((0.5 * 0.1) + (0.1 * 0.9)) = 0.3571 15. 0.4255 # P(speeding) = 0.1, P(-speeding) = 0.9, P(2door|speeding) = 1/3, P(2door|-speeding) = 1/20 # P(speeding|bmw) = P(2door|speeding)P(speeding) / ( P(2door|speeding)P(speeding) + P(2door|-speeding)P(-speeding)) # = ( (1/3) * 0.1 ) / ( ((1/3) * 0.1) + ((1/20) * 0.9) = 0.4255) 16. not enough information. # P(speeding|two-door-coupe,bmw) = P(2door,bmw|speeding)P(speeding) / ( P(2door,bmw|speeding)P(speeding) + P(2door,bmw|-speeding)P(-speeding)) # we don't have P(2door,bmw|speeding) nor P(2door,bmw|-speeding) 17. 0.7874 # P(2door,bmw|speeding) = P(2door|speeding)*P(bmw|speeding) # P(speeding|2door,bmw) = P(2door,bmw|speeding)P(speeding) / ( P(2door,bmw|speeding)P(speeding) + P(2door,bmw|-speeding)P(-speeding)) # P(speeding|2door,bmw) = P(2door|speeding)*P(bmw|speeding)*P(speeding) / # ( P(2door|speeding)*P(bmw|speeding)*P(speeding) + P(2door|-speeding)*P(bmw|-speeding)*P(-speeding)) # = ( (1/3) * 0.5 * 0.1 ) / ( ( (1/3) * 0.5 * 0.1 ) + ( (1/20) * 0.1 * 0.9 )) = 0.7874 18. d 19. c 20. 0.64952; # exp( 3*( 0.5*log(1+0.5)+0.5*log(1-0.5)) )