1. Probability Distributions and Expected Value
Chapter 5.1 – Probability Distributions and Predictions
Mathematics of Data Management (Nelson)
MDM 4U

2. Probability Distributions of a Discrete Random Variable
a discrete random variable X is a variable that can take on only a finite set of values
for example, rolling a die can only produce numbers in the set {1,2,3,4,5,6}
rolling 2 dice can produce only numbers in the set {2,3,4,5,6,7,8,9,10,11,12}
choosing a card from a standard deck (ignoring suit) can produce only the cards in the set {A,2,3,4,5,6,7,8,9,10,J,Q,K}

3. Probability Distribution
a probability distribution of a random variable x, is a function which provides the probability of each possible value of x
this function may be represented as a table of values, a graph or a mathematical expression
for example, rolling a die:

4. Probability Distribution for 2 Dice

5. What would a probability distribution graph for three dice look like?
We will try it! Using three dice, figure out how many outcomes there are
Then find out how many possible ways there are to create each of the possible outcomes
Fill in a table like the one below
Now you can make the graph
Outcome 3 4 5 6 7 8 9 …
# ways 1

6. Probability Distribution for 3 Dice
Outcome 3 4 5 6 7 8 9 10
# cases 1 15 21 28

7. So what does an experimental distribution look like?
A simulated dice throw was done a million times using a computer program and generated the following data
What is/are the most common outcome(s)?
Does this make sense?

8. Back to 2 Dice What is the expected value of throwing 2 dice?
How could this be calculated?
So the expected value of a discrete variable X is the sum of the values of X multiplied by their probabilities

9. Example 1a: tossing 3 coins
0 heads
1 head
2 heads
3 heads
P(X) ⅛ ⅜
What is the likelihood of at least 2 heads?
It must be the total probability of tossing 2 heads and tossing 3 heads
P(X = 2) + P(X = 3) = ⅜ + ⅛ = ½
so the probability is 0.5

10. Example 1b: tossing 3 coins
0 heads
1 head
2 heads
3 heads
P(X) ⅛ ⅜
What is the expected number of heads?
It must be the sums of the values of x multiplied by the probabilities of x
0P(X = 0) + 1P(X = 1) + 2P(X = 2) + 3P(X = 3)
= 0(⅛) + 1(⅜) + 2(⅜) + 3(⅛) = 1½
So the expected number of heads is 1.5

11. Example 2a: Selecting a Committee of three people from a group of 4 men and 3 women
What is the probability of having at least one woman on the team?
There are C(7,3) or 35 possible teams
C(4,3) = 4 have no women
C(4,2) x C(3,1) = 6 x 3 = 18 have one woman
C(4,1) x C(3,2) = 4 x 3 = 12 have 2 women
C(3,3) = 1 have 3 women

12. Example 2a cont’d: selecting a committee
0 women
1 woman
2 women
3 women
P(X)
4/35
18/35
12/35
1/35
What is the likelihood of at least one woman?
It must be the total probability of all the cases with at least one woman
P(X = 1) + P(X = 2) + P(X = 3)
= 18/ /35 + 1/35 = 31/35

13. Example 2b: selecting a committee
0 women
1 woman
2 women
3 women
P(X)
4/35
18/35
12/35
1/35
What is the expected number of women?
0P(X = 0) + 1P(X = 1) + 2P(X = 2) + 3P(X = 3)
= 0(4/35) + 1(18/35) + 2(12/35) + 3(1/35)
= 1.3 (approximately)

14. MSIP / Homework
p. 277 #1, 2, 3, 4, 5, 9, 12, 13