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.