Homework 1 Solutions
7CCSMRVA - Random Variables and Stochastic Processes
October 5, 2018
Problem 1. If A = {2 x 5} and B = {3 x 6}, find A [ B, AB and (A [ B)(AB).
Please note that (AB)c can also be denoted as (AB).
Solution 1.
A [ B = {2 x 5} [ {3 x 6} = {2 x 6} (1)
AB = {2 x 5} \ {3 x 6} = {3 x 5} (2)
AB = {{x < 3} [ {x > 5}} (from (2)) (3)
(AU B)(AB) = {2 x 6} \ {{x < 3} [ {x > 5}} (4)
= {2 x < 3} [ {5 < x 6} (5)
Problem 2. We select at random m objects from a set S of n objects and we denote by
Am the set of the selected objects. Show that the probability that a particular element ⇣0 of
S is in Am equals to p = m/n.
Solution 2. There are two solutions to this problem.
First solution:
Using the concept of combinations, we have
# of m-element subsets containing ⇣0
p= (6)
# of m-element subsets
✓ ◆
n 1
(n 1)!
m 1 (n m+1)!(m 1)! m
= ✓ ◆ = n!
= (7)
n m!(n m)!
n
m
Second solution:
m
Probability that ⇣0 belongs to a specific Am is clearly P (⇣0 |Am ) = n
. Then, we sum all
over sets Am
X X m m X m m
p= P (⇣0 |Am )P (Am ) = P (Am ) = P (Am ) = ·1= (8)
all sets Am all sets Am
n n all sets A n n
m
1
7CCSMRVA - Random Variables and Stochastic Processes
October 5, 2018
Problem 1. If A = {2 x 5} and B = {3 x 6}, find A [ B, AB and (A [ B)(AB).
Please note that (AB)c can also be denoted as (AB).
Solution 1.
A [ B = {2 x 5} [ {3 x 6} = {2 x 6} (1)
AB = {2 x 5} \ {3 x 6} = {3 x 5} (2)
AB = {{x < 3} [ {x > 5}} (from (2)) (3)
(AU B)(AB) = {2 x 6} \ {{x < 3} [ {x > 5}} (4)
= {2 x < 3} [ {5 < x 6} (5)
Problem 2. We select at random m objects from a set S of n objects and we denote by
Am the set of the selected objects. Show that the probability that a particular element ⇣0 of
S is in Am equals to p = m/n.
Solution 2. There are two solutions to this problem.
First solution:
Using the concept of combinations, we have
# of m-element subsets containing ⇣0
p= (6)
# of m-element subsets
✓ ◆
n 1
(n 1)!
m 1 (n m+1)!(m 1)! m
= ✓ ◆ = n!
= (7)
n m!(n m)!
n
m
Second solution:
m
Probability that ⇣0 belongs to a specific Am is clearly P (⇣0 |Am ) = n
. Then, we sum all
over sets Am
X X m m X m m
p= P (⇣0 |Am )P (Am ) = P (Am ) = P (Am ) = ·1= (8)
all sets Am all sets Am
n n all sets A n n
m
1