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