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