Solution: 2015 Winter Final - 18

Author: Michiel Smid

Question

Consider a uniformly random bitstring of length 5. Define the events

and

Which of the following is true?

(a)

The events $A$ and $B$ are independent.

(b)

None of the above.

(c)

The events $A$ and $B$ are not independent.

Let S be the set of all bitstrings of length 5.
$ |S| = 2^5 $
A occurs when there are 1, 3, or 5 0s in the bitstring.
$ |A| = \binom{5}{1} + \binom{5}{3} + \binom{5}{5} $
$ |A| = 5 + 10 + 1 $
$ |A| = 16 $
$ Pr(A) = \frac{\binom{5}{1} + \binom{5}{3} + \binom{5}{5}}{2^5} $
$ Pr(A) = \frac{16}{2^5} $
$ Pr(A) = \frac{2^4}{2^5} $
$ Pr(A) = \frac{1}{2} $
B occurs when the first 3 bits are 111.
This leaves the remaining 2 bits free to be any combination of 0s and 1s: $2^2$
$ |B| = 2^2 $
$ Pr(B) = \frac{2^2}{2^5} $
$ Pr(B) = \frac{1}{2^3} $
$ A \cap B $ occurs when the first 3 bits are 111 and there is 1 0
The 0 can be at either position 4 or 5: 2
The other bit is forced to be a 1: 1
$ |A \cap B| = 2 $
$ Pr(A \cap B) = \frac{2}{2^5} $
$ Pr(A \cap B) = \frac{1}{2^4} $

If they are independent, the probability of both events occurring is the product of the probabilities of each event occurring.

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

$ \frac{1}{2^4} = \frac{1}{2} \cdot \frac{1}{2^3} $

$ \frac{1}{2^4} = \frac{1}{2^4} $

$ 1 = 1 $

Since the equation is true, A and B are independent.