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Solution: 2018 Winter Final - 13

Author: Michiel Smid

Question

You are given a uniformly random bitstring of length five. Define the events

A = "the bitstring contains at most four 1's",
B = "the bitstring contians an odd number of 1's".

What is $\Pr(A|B)$?

(a)

13/16

(b)

15/16

(c)

12/16

(d)

14/16

Solution

Let S be the set of all bitstrings
$ |S| = 2^5 $
Let's determine B
We could choose 1 of the 5 positions to be a 1: $ \binom{5}{1} $
We could choose 3 of the 5 positions to be a 1: $ \binom{5}{3} $
We could choose 5 of the 5 positions to be a 1: $ \binom{5}{5} $
$ |B| = \binom{5}{1} + \binom{5}{3} + \binom{5}{5} $
$ |B| = 5 + 10 + 1 $
$ |B| = 16 $
$ Pr(B) = \frac{16}{32} = \frac{1}{2} $
Let's determine $A \cap B $. The only valid cases are where the bitstring has 1 1's or 3 1's:
We could choose 1 of the 5 positions to be a 1: $ \binom{5}{1} $
We could choose 3 of the 5 positions to be a 1: $ \binom{5}{3} $
$ |A \cap B| = \binom{5}{1} + \binom{5}{3} $
$ |A \cap B| = 5 + 10 $
$ |A \cap B| = 15 $
$ Pr(A \cap B) = \frac{15}{32} $

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

$ Pr(A|B) = \frac{ \frac{15}{32}}{ \frac{1}{2}} $

$ Pr(A|B) = \frac{15}{32} \cdot 2 $

$ Pr(A|B) = \frac{15}{16} $