Solution: 2014 Winter Final - 15

• Let S be the set of all outcomes when rolling two dice.
  The size of S is the number of outcomes when rolling two dice
  $ |S| = 6 \times 6 = 36 $
  
• Let's find the size of the set of outcomes for event $A$.
  The set of A is $ { (3,1), (3,2), (3,3), (3,4), (3,5), (3,6), (1,3), (2,3), (4,3), (5,3), (6,3) } $
  $ |A| = 11 $
  $ Pr(A)= \frac{|A|}{|S|} = \frac{11}{36} $
• Let's find the size of the set of outcomes for event $B$.
  The set of B is $ { (1,4), (2,3), (3,2), (4,1) } $
  $ |B| = 4 $
  $ Pr(B)= \frac{|B|}{|S|} = \frac{4}{36} = \frac{1}{9} $
• Let's find the size of the set of outcomes for event $ A \cap B $.
  The set of $ A \cap B $ is $ { (3,2), (2,3) } $
  $ |A \cap B| = 2 $
  $ Pr(A \cap B) = \frac{|A \cap B|}{|S|} = \frac{2}{36} = \frac{1}{18} $

Let’s check whether the solution is correct

$ Pr(A|B) = \frac{Pr(A \cap B)}{Pr(B)} $

$ Pr(A|B) = \frac{ \frac{1}{18}}{ \frac{1}{9}} $

$ Pr(A|B) = \frac{1}{18} \times \frac{9}{1} $

$ Pr(A|B) = \frac{1}{2} $