Solution: 2017 Winter Final - 13

• Let S be all possible outcomes of the coin flips: $ |S| = 2^5 = 32 $
• Let's determine B
  There is a fifth coin flip is heads: 1
  The other 4 coin flips can be any combination of heads and tails: $ 2^4 = 16 $
  $ |B| = 16 $
  $ Pr(B) = \frac{16}{32} = \frac{1}{2} $
• Let's determine $ A \cap B $
  Since the fifth coin flip results in heads, we choose 1 tails for one of the first 4 coin flips: 4
  $ | B \cap A | = 4 $
  $ Pr(A \cap B) = \frac{4}{32} = \frac{1}{8} $

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

$ Pr(A|B) = \frac{ \frac{1}{8} }{ \frac{1}{2} } $

$ Pr(A|B) = \frac{1}{8} \cdot \frac{2}{1} $

$ Pr(A|B) = \frac{1}{4} $