Recitation 22 Solution. Now we want Ex( C), the expected score a lecturer would give Employing linearity again, we have Ex(x iC)=Ex(product of dice IC) +Ex(general impression C) (because the dice are independent) 50+ 60 (c) What is the expected score on a 6.042 exam? Solution. Let X equal the true exam score. The Total Expectation Theorem implies Ex(X)=Ex(X | C)Pr(C)+Ex(X I C)Pr(c) 4 2 +49 +84 4 40+(-+14)+12=6� � � � � � | � ¯ � � � � � � � | � � � � Recitation 22 5 Solution. Now we want Ex X ¯ | C , the expected score a lecturer would give. Employing linearity again, we have: Ex X C ¯ = Ex product of dice ¯ | C + Ex general impression | C � �2 7 = (because the dice are independent) 2 4 3 3 + 40 + 50 + 60 10 · 10 · 10 · 49 1 = + 49 = 61 4 4 (c) What is the expected score on a 6.042 exam? Solution. Let X equal the true exam score. The Total Expectation Theorem implies: Ex (X ¯ ¯ ) = Ex (X C) Pr (C) + Ex X | C Pr C 4 49 2 1 = 70 · + + 49 + 84 7 4 · 7 · 7 7 1 = 40 + + 14 + 12 = 69 2 2