CISC 7700X Final Exam

1. b
2. d
3. c
4. a
5. c
6. b
7. c
8. b
9. b
10. c
11. d
12. a
13. 0.066038
  % P(buy) = 0.01, P(-buy) = 0.99
  % P(1car|buy) = 0.7, P(-1car|buy) = 0.3
  % P(-1car|-buy) = 0.9, P(1car|-buy) = 0.1
  % P(buy|1car) = P(1car|buy)P(buy) / (P(1car|buy)P(buy)+P(1car|-buy)P(-buy))
  %        = (0.7*0.01) / (0.7*0.01 + 0.1*0.99) = 0.066038

14. 0.1081
  % P(buy) = 0.01, P(-buy) = 0.99
  % P(family|buy) = 0.6, P(-family|buy) = 0.4
  % P(family|-buy) = 0.05, P(-family|-buy) = 0.95
  %  P(buy|family) = P(family|buy)P(buy) / (P(family|buy)P(buy) + P(family|-buy)P(-buy))
  %                = (0.6*0.01) / ( 0.6*0.01 + 0.05*0.99) = 0.1081

15. Not enough data.
  % we need P(buy|family,1car) = P(family,1car|buy)P(buy) / P(family,1car) 
  % we are not given P(family,1car|buy) or P(family,1car)

16. 0.4590
  % naive assumption: P(family,1car|buy) = P(family|buy)P(1car|buy)
  % P(family,1car|buy) = P(family|buy)P(1car|buy)P(buy) / (P(family|buy)P(1car|buy)P(buy)+P(family|-buy)P(1car|-buy)P(-buy))
  %                    = (0.6*0.7*0.01) / (0.6*0.7*0.01 + 0.05*0.1*0.99) = 0.4590

17. c
18. d
19. a
20. 0.6495
  % exp( 3*( 0.5*log(1+0.5)+0.5*log(1-0.5)) )