CISC 7700X Final Exam 1. b 2. 1 # ( (1 + 0.25) + (1 + 0.25) + (1 -0.50)) / 3 = 1, or 0% net-return. 3. 0.9210 # (1.25 * 1.25 * 0.50)^(1/3) 4. b 5. c 6. b 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. c 11. b 12. a 13. c 14. 0.2941 # P(speeding) = 0.2, P(-speeding) = 0.8, P(bmw|speeding) = (1/3), P(bmw|-speeding) = (1/5) # P(speeding|bmw) = P(bmw|speeding)P(speeding) / ( P(bmw|speeding)P(speeding) + P(bmw|-speeding)P(-speeding)) # = ((1/3) * 0.2) / (((1/3) * 0.2) + ((1/5) * 0.8)) = 0.2941 15. 0.4545 # P(speeding) = 0.2, P(-speeding) = 0.9, P(2door|speeding) = 1/3, P(2door|-speeding) = 1/10 # P(speeding|2door) = P(2door|speeding)P(speeding) / ( P(2door|speeding)P(speeding) + P(2door|-speeding)P(-speeding)) # = ( (1/3) * 0.2 ) / ( ((1/3) * 0.2) + ((1/10) * 0.8)) = 0.4545 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. 0.5814 # 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) * (1/3) * 0.2 ) / ( ( (1/3) * (1/3) * 0.2 ) + ( (1/10) * (1/5) * 0.8 )) = 0.5814 18. d 19. c 20. 0.64952 # exp( 3*( 0.5*log(1+0.5)+0.5*log(1-0.5)) ) =0.64952