Back

Solution: 2017 Fall Final - 13

Author: Michiel Smid

Question

Let $X = \{1,2,3,\dots,10\}$. Let $Y$ be a uniformly random subset of $X$. Define the events

A = "$Y$ contains at least 4 elements",
B = "all elements of $Y$ are even".

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

(a)

4/16

(b)

6/16

(c)

3/16

(d)

5/16

Solution

Let’s find the values

Let S be the set of all subsets of X
$ |S| = 2^{10} $
Let's determine $ Pr(B) $
The even numbers in X are $ { 2, 4, 6, 8, 10 } $
There are 5 elements to make subsets out of, so let's get this bread then
$ |B| = 2^25 $
$ Pr(B) = \frac{|B|}{|S|} $
$ Pr(B) = \frac{2^5}{2^{10}} $
$ Pr(B) = \frac{1}{2^5} $
Let's determine $ Pr( A \cap B) $
The even numbers in X are $ { 2, 4, 6, 8, 10 } $
The number of subsets of X that contain at least 4 even numbers is $ \binom{5}{4} + \binom{5}{5} = 6 $
$ |A \cap B| = 6 $
$ Pr(A \cap B) = \frac{|A \cap B|}{|S|} $
$ Pr(A \cap B) = \frac{6}{2^{10}} $
$ Pr(A \cap B) = \frac{3}{2^9} $

Now, let’s find $ Pr(A|B) $

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

$ Pr(A|B) = \frac{ \frac{3}{2^9}}{ \frac{1}{2^5}} $

$ Pr(A|B) = \frac{3}{2^4} $

$ Pr(A|B) = \frac{3}{16} $