CISC 7700X Midterm Exam

1. c
2. d
3. a
4. c
5. b
6. a
7. b
8. c
9. d
10. M*M size ( X with N rows and M columns, size of X^T * X )
11. N*N size (size of X * X^T)
12. X * X^T
13. It depends on problem we are solving, there's no best measure that works for all scenarios.
14. d
15. b
16. b
17. 0.90323
  P(L) = 0.7, P(-L) = 0.3, P(I|L)=0.4, P(I|-L)=0.1
  P(L|I) = P(I|L)P(L) / ( P(I|L)P(L) + P(I|-L)P(-L) )
         = 0.4 * 0.7 / ( 0.4 * 0.7 + 0.1 * 0.3 ) = 0.90323

18. 0.82353
  P(L) = 0.7, P(-L) = 0.3, P(S|L)=0.2, P(S|-L)=0.1
  P(L|S) = P(S|L)P(L) / ( P(S|L)P(L) + P(S|-L)P(-L) )
         = 0.2 * 0.7 / ( 0.2 * 0.7 + 0.1 * 0.3 ) = 0.82353

19. Not enough info.
  The bayes rule would be: 
   P(L|S,I) = P(S,I|L)P(L) / P(S,I)
   we don't know P(S,I|L), all we have is P(S|L) and P(I|L)

20. 0.94915
  P(L) = 0.7, P(-L) = 0.3, P(I|L)=0.4, P(I|-L)=0.1,  P(S|L)=0.2, P(S|-L)=0.1

  assume S and I are independent, e.g.: P(S,I|L) = P(S|L)*P(I|L) 
  
  P(L|S,I) = P(S|L)*P(I|L)*P(L) / ( P(S|L)*P(I|L)*P(L) + P(S|-L)*P(I|-L)*P(-L) )
           = 0.2 * 0.4 * 0.7 / ( 0.2 * 0.4 * 0.7 + 0.1* 0.1* 0.3  ) = 0.94915
           
  simlarly, we can do it in sequence,

     pretend first we learned that IronMan is in the movie, from q17, our P(L) turns to 0.90323,
     we then plug that into q18 (learned that Spiderman is in movie) and solve, e.g.: 
         = 0.2 * 0.90323 / ( 0.2 * 0.90323 + 0.1 * (1-0.90323) ) = 0.94915
         
    OR
    
     pretend first we learned that Spiderman is in the movie, from q18, our P(L) turns to 0.82353,
     we then plug that into q17 (learned that Ironman is in movie) and solve, e.g.: 
         = 0.4 * 0.82353 / ( 0.4 * 0.82353 + 0.1 * (1-0.82353) ) = 0.94915