Warmup
Our Sample Space for rolling one die is 𝑆𝑆 = {1,2,3,4,5,6}
This means, our Sample Space for rolling 10 die is
𝑆𝑆10 = {lists of 10 items, where order matters}
The probability of getting exactly six 1’s is:
10
6
⋅ 54
610
Different ways you can
get six 1s
Different the 4 other die
rolls can turn out. (No 1s)
Total size of 𝑆𝑆10
Random Variable – Definition and Example
𝑋𝑋 (Capital X) is the random variable for the number of heads shown
The probabilities of each of these should add up to one.
3 2 22 1 1 1 0
𝒙𝒙 0 1 2 3
𝑃𝑃(𝑋𝑋 = 𝑥𝑥) 1/8 3/8 3/8 1/8
Random Variable – Another Example
Try it on your own first. Solution on the next slide
Random Variable – Another Example
𝒙𝒙 2 3 4 5 6 7 8 9 10 11 12
𝑃𝑃(𝑋𝑋 = 𝑥𝑥) 1/36 2/36 3/36 4/36 5/36 6/36 5/36 4/36 3/36 2/36 1/36
Hint: Look at the diagonals to
see how many times you can
get each sum
Exercises
Try these on your own first. Solutions on the next slides
Exercises
𝑋𝑋 ≤ 1 is the same as 𝑋𝑋 = 0 or 𝑋𝑋 = 1
𝑃𝑃 𝑋𝑋 ≤ 1 = 𝑃𝑃 𝑋𝑋 = 1 + 𝑃𝑃 𝑋𝑋 = 2
1
8
+
3
8
=
4
8
=
1
2
𝒙𝒙 0 1 2 3
𝑃𝑃(𝑋𝑋 = 𝑥𝑥) 1/8 3/8 3/8 1/8
Exercises
Let 𝑀𝑀 be the event that 𝑆𝑆 is a multiple of 3.
𝑃𝑃 𝑀𝑀 = 𝑃𝑃 𝑆𝑆 = 3 ∪ 𝑆𝑆 = 6 ∪ 𝑆𝑆 − 9 ∪ 𝑆𝑆 = 12
= 𝑃𝑃 𝑆𝑆 = 3 + 𝑃𝑃 𝑆𝑆 = 6 + 𝑃𝑃 𝑆𝑆 − 9 + 𝑃𝑃 𝑆𝑆 = 12
2
36
+
5
36
+
4
36
+
1
36
=
12
36
=
1
3
𝒙𝒙 2 3 4 5 6 7 8 9 10 11 12
𝑃𝑃(𝑋𝑋 = 𝑥𝑥) 1/36 2/36 3/36 4/36 5/36 6/36 5/36 4/36 3/36 2/36 1/36
3 divides 𝑆𝑆
Recall
Note that independent
events are NOT
necessarily disjoint
(and vice versa).
Example
Try it on your own first. Solution on the next slide
Example
𝑃𝑃 𝑋𝑋 = 𝑎𝑎 ∩ 𝑌𝑌 = 𝑏𝑏 =
1
52
=
1
13
×
1
4
Therefore, these events are independent.
𝑃𝑃 𝑋𝑋 = 𝑎𝑎 =
4
52
=
1
13
For example, probability of getting a
Queen
𝑃𝑃 𝑌𝑌 = 𝑏𝑏 =
13
52
=
1
4
For example, probability of
getting a Diamond
Examples
Try it on your own first. Solution on the next slide
Examples
For this one, try looking at a counterexample.
Suppose 𝑆𝑆 = 12 and 𝑇𝑇 = 1.
This 𝑆𝑆 means, you rolled two dice that sum up to 12
This 𝑇𝑇 means you got 1 three that showed up.
𝑃𝑃 𝑆𝑆 = 12 = 1
36
and 𝑃𝑃 𝑇𝑇 = 1 = 11
36
The probability of getting a 12, and having a 3 we know is zero.
But
1
36
× 11
36
≠ 0. So we know that these events are not independent
Exercise
From the textbook (33.1)
Let (𝑆𝑆,𝑃𝑃) be a sample space with 𝑆𝑆 = {𝑎𝑎, 𝑏𝑏, 𝑐𝑐,𝑑𝑑} and
𝑃𝑃 𝑎𝑎 = 0.1,𝑃𝑃 𝑏𝑏 = 0.2,𝑃𝑃 𝑐𝑐 = 0.3,𝑃𝑃 𝑑𝑑 = 0.4
Define random variables 𝑋𝑋,𝑌𝑌 on this sample space according to the below
table:
Complete (a) through (g) in text
𝒔𝒔 𝑿𝑿(𝒔𝒔) 𝒀𝒀(𝒔𝒔)
𝑎𝑎 1 -1
𝑏𝑏 3 3
𝑐𝑐 5 6
𝑑𝑑 8 10
Motivation and Videos
Watch these videos on Expected Value and Variance:
• https://www.youtube.com/watch?v=kZTKuMBJP7Y
• This one is long, but fun, and provides good motivation for this section
• https://www.youtube.com/watch?v=OvTEhNL96v0
• https://www.youtube.com/watch?v=sheoa3TrcCI
Definitions
Examples
Try these on your own first. Solution on the next slides
Examples
𝒙𝒙 1 2 3 4 5 6
𝑃𝑃(𝑋𝑋 = 𝑥𝑥) 1
6
1
6
1
6
1
6
1
6
1
6
𝐸𝐸 𝑋𝑋 = 1
1
6
+ 2
1
6
+ 3
1
6
+ 4
1
6
+ 5
1
6
+ 6
1
6
= �
𝑖𝑖=1
6
𝑥𝑥𝑖𝑖𝑃𝑃(𝑥𝑥𝑖𝑖)
=
21
6
= 3.5
Examples
𝐸𝐸 𝑋𝑋 = 1
1
2
+ 2
1
4
+ 3
1
8
+ 4
1
8
=
15
8
= 1.875
𝒙𝒙 1 2 3 4
𝑃𝑃(𝑋𝑋 = 𝑥𝑥) 1
2
1
4
1
8
1
8
Exercise
Try it on your own first. Solution on the next slide
Exercise
= �
𝑖𝑖=2
12
𝑥𝑥𝑖𝑖𝑃𝑃(𝑥𝑥𝑖𝑖)
…do a lot of arithmetic…
= 7
𝒙𝒙 2 3 4 5 6 7 8 9 10 11 12
𝑃𝑃(𝑋𝑋 = 𝑥𝑥) 1/36 2/36 3/36 4/36 5/36 6/36 5/36 4/36 3/36 2/36 1/36
But, there is a quicker
way to get this answer!
Algebra of Random Variables
Basically, what this tells us, is that when we’re dealing
with Expected Value, we can break things down into
individual events and put them back together and
we’ll get the same answer.
To answer this previous question,
it required a lot of work!
What linearity allows us to do, is take the Expected Value
of 1 die (which we already calculated), and multiply that by
2 to get the expected value of rolling 2 dice.
Linearity of Expectation
Fancy math for what the previous slide says. The previous
slide is intuition. This slide is formal.
From 2 to n
Try it on your own first. Solution on the next slide
From 2 to n
1
𝑛𝑛
1 + 2
𝑛𝑛
2 + 3
𝑛𝑛
3 + ⋯+ 𝑛𝑛
𝑛𝑛
𝑛𝑛 = 𝑛𝑛2
𝑛𝑛−1
This equality can be proven with a combinatorial proof. Check for yourself
before moving on.
Product of Random Variables
Examples
Try these on your own. Solutions on the next slide
Examples
A straightforward application gives the solution
1
4
A straightforward application gives the solution
7
2
2
= 49
4
Variance
Try this on your own. Solution on the next slide
Variance
Recall: 𝐸𝐸 𝑋𝑋 = 7
2
𝒙𝒙 1 2 3 4 5 6
𝑃𝑃(𝑋𝑋 = 𝑥𝑥) 1
6
1
6
1
6
1
6
1
6
1
6
𝑥𝑥 − 𝐸𝐸(𝑋𝑋)
1 −
7
2
2 −
7
2
3 −
7
2
4 −
7
2
5 −
7
2
6 −
7
2
𝑥𝑥 − 𝐸𝐸 𝑋𝑋
2 25
4
9
4
1
4
1
4
9
4
25
4
𝐸𝐸 𝑥𝑥 − 𝐸𝐸 𝑋𝑋
2 1
6
⋅
25
4
1
6
⋅
9
4
1
6
⋅
1
4
1
6
⋅
1
4
1
6
⋅
9
4
1
6
⋅
25
4
𝑉𝑉𝑎𝑎𝑉𝑉 𝑋𝑋 =
35
12
Read this table from left to right, and then work your way
down the columns after you understand where everything
is coming from
In general, calculating the Variance
this way is a lot of arithmetic, even
for a relatively small example like
this one
Another [long] Example
Another [long] Example
Another [long] Example