Solution: 2015 Fall Final - 13

• Let S be the set of all possible bitstrings of length 5.
  The size of S is the number of bitstrings of length 5: $ |S| = 2^{5} = 32 $
• In the instance of A, the first 3 bits can be fixed as 101 or 110: 2 ways.
  The last 2 bits can be any combination of 0s and 1s: $2^{2}$.
  $ |A| = 2 \times 2^{2} = 8 $
  $ Pr(A) = \frac{8}{32} = \frac{1}{4} $
• In the instance of B, the first 2 bits can be any combination of 0s and 1s: $2^{2}$.
  The last 3 bits are fixed as 111: 1 way.
  $ |B| = 2^{2} = 4 $
  $ Pr(B) = \frac{4}{32} = \frac{1}{8} $
• Let $A \cap B$ be the event that the first 3 bits are 101 or 110 AND the last 3 bits are 111.
  When the last 3 bits are 111, the first 3 bits cannot be 110
  This leaves us with 10111 as the only bitstring that satisfies both conditions.
  $ |A \cap B| = 1 $
  $ Pr(A \cap B) = \frac{1}{32} $

Now, we check whether they are independent or not

$ Pr(A \cap B) = Pr(A) \cdot Pr(B) $

$ \frac{1}{32} = \frac{1}{4} \cdot \frac{1}{8} $

$ \frac{1}{32} = \frac{1}{32} $

The events A and B are independent.