Solution: 2016 Fall Final - 22

For questions like these, we can answer by checking for a value $i$ if the following is true: $ Pr(X=i \cap Y=i) = Pr(X=i) \cdot Pr(Y=i) $

• Let's determine $ Pr(X=1) $
  The coin that's heads has a probability of $ \frac{1}{2} $
  The other 6 coins have a probability of $ \frac{1}{2} $
  $ Pr(X=1) = { left( \frac{1}{2} \right)}^1 {\left( \frac{1}{2} \right) }^6 $
  $ Pr(X=1) = \frac{1}{2^7} $
• Let's determine $ Pr(Y=1) $
  The coin that's tails has a probability of $ \frac{1}{2} $
  The other 6 coins have a probability of $ \frac{1}{2} $
  $ Pr(Y=1) = { left( \frac{1}{2} \right) }^1 { left( \frac{1}{2} \right) }^6 $
  $ Pr(Y=1) = \frac{1}{2^7} $
• Let's determine $ Pr(X=1 \cap Y=1) $
  If we get exactly 1 heads, then the remaining 6 coins must be tails
  but that means we can't have exactly 1 tails as well
  Since having both happen at the same time is impossible
  $ Pr(X=1 \cap Y=1) = 0 $

$ Pr(X=i \cap Y=i) = Pr(X=i) \cdot Pr(Y=i) $

$ 0 = \frac{1}{2^7} \cdot \frac{1}{2^7} $

Since the equation is false, X and Y are not independent.