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Solution: 2018 Winter Final - 15

Author: Michiel Smid

Question

You flip a fair coin four times; these four flips are independent. Define the events
  • A = "the first two flips result (in this order) in $HT$",
  • B = "the second and third flips result in $TT$".
Which of the following is correct?
(a)
All of the above.
(b)
None of the above.
(c)
The events $A$ and $B$ are not independent.
(d)
The events $A$ and $B$ are independent.

Solution

We’ll take a slow and systematic approach to this question

  • Let S be the set of all possible outcomes
    $ |S| = 2^4 = 16$
  • Let's determine $A$
    The first two flips are fixed: $HT$
    The last 2 flips can be either heads or tails: $ 2^2 = 4 $
    $ |A| = 1 \cdot 4 = 4 $
    $ Pr(A) = \frac{4}{16} = \frac{1}{4} $
  • Let's determine $B$
    The second and third flips are fixed: $TT$
    The first flip can be either heads or tails: $ 2^1 = 2 $
    The last flip can be either heads or tails: $ 2^1 = 2 $
    $ |B| = 2 \cdot 2 = 4 $
    $ Pr(B) = \frac{4}{16} = \frac{1}{4} $
  • Let's determine $A \cap B$
    If the first two flips are $HT$ and the second and third flips are $TT$, then it looks like this: $HTTX$
    X can either be heads or tails: $ 2^1 = 2 $
    $ |A \cap B| = 1 \cdot 2 = 2 $
    $ Pr(A \cap B) = \frac{2}{16} = \frac{1}{8} $

Now, let’s check whether it’s independent

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

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

$ \frac{1}{8} = \frac{1}{16} $

Since the two sides are not equal, the events are not independent