NOTES 3.9 – THEORETICAL
PROBABILITY – DAY 1
Each time an experiment such as one toss of a coin,
one roll of a dice, one spin on a spinner etc.
is performed, the result is called an ___________.
OUTCOME
Theoretical Probability is a measure of what you
________
EXPECT to occur.
A _______________
SAMPLE SPACE for an experiment is the set of
possible outcomes for that experiment.
Example 1:
Create a sample space for the following situation: Mr.
And Mrs. Sanderson are expecting triplets. Assume
there is an equally likely chance that the Sandersons
will have a boy or girl.
BBB
GBB
BGB
BBG
Total Number of Outcomes:
GGB
GBG
BGG
8
GGG
Theoretical Probability =
Number of FAVORABLE outcomes in the sample space
Number of TOTAL outcomes in the sample space
P( E )  FAVORABLE
TOTAL
Notation: The probability of a certain event occurring
P( E )
is notated by _____.
Where P stands for _________
probability and E is the ______
event
occurring.
Example 2 - Using the sample space above, if the
couple has 3 children, what is the probability of
having 2 boys and 1 girl?
P( of 2 boys) =
3
8
P( of 1 boys) =
3
8
P( of 3 girls) =
1
8
P( of 2 girls) =
3
8
Example 3 - Out of 100 families with 3 children
how many would you expect to have all girls?
1
P( of 3 girls) = 8
1 n

8 100
8n  100
8n 100

8
8
n  12.5
or about 13 families
INDEPENDENT EVENTS:
The above situation is an example of an
______________
INDEPENDENT event because the outcome of one
event does not affect the probability of the other
events occurring.
Example 4 - Radcliff is playing a game where he
spins the spinner below and tosses
1
2
and coin right after. Create a sample
space for all possible outcomes, and
4
3
then answer the questions below.
Spinner
Coin
1 1 2 2 3 3 4 4
H T H T H T H T
4 1

P(Tails)? 8 2
2 1

P(1)? 8 4
P(1 and Tails)?
1
8
0
0
P(5 and Tails)? 8
2 1

P(Even and Heads)?_______
8 4
6 3

P(Odd or Heads)? 8 4
MathematicallyAND means we can MULTIPLY each
individual probability.
OR means we can ADD the probabilities. (But
don’t count an event twice!)
P(Even and Heads) = P(E) x P(H) =
1
2

1
2
=
1
4
P(Odd or Heads) = P(O) + P(H) - P(O and H) =
1
2
HW :

1
2
Section 3.9
-
1
4
=
3
4
pages 193-194
#’s 6-29
NOTES 3.9 – THEORETICAL
PROBABILITY – DAY 2
WITH OR WITHOUT REPLACEMENT
Some probability events require the act of
____________
REPLACING an item back before choosing another
item. These events are called events
_____________________.
WITH REPLACEMENT
Other probability events require the act of
_________________
NOT REPLACING an item before choosing another
item. These events are called events ___________
WITHOUT
_______________.
REPLACEMENT
Example 1 - Suppose a bag contains 12 marbles: 6 red (R), 4
Green (G), and 2 yellow (Y). Two marbles are randomly
drawn. Use a grid to find the following probabilities:
First marble returned
First marble not returned
(dependent event)
(independent event)
P(R, then R)
P(R, then G)
P(R, then Y)
P(G, then R)
P(G, then G)
P(G, then Y)
6 6 1 1 1
   
12 12 2 2 4
6 4 1 1 1
   
12 12 2 3 6
6 2 1 1 1
   
12 12 2 6 12
4 6 1 1 1
   
12 12 3 2 6
4 4 1 1 1
   
12 12 3 3 9
4 2 1 1 1
   
12 12 3 6 18
6 5
 
12 11
6 4
 
12 11
6 2
 
12 11
4 6
 
12 11
4 3
 
12 11
4 2
 
12 11
1 5 5
 
2 11 22
1 4
2
 
2 11 11
1 2 1
 
2 11 11
1 6
2
 
3 11 11
1 3
1
 
3 11 11
1 2 2
 
3 11 33
First marble returned
First marble not returned
(dependent event)
(independent event)
P(Y, then R)
P(Y, then G)
P(Y, then Y)
2 6 1 1 1
   
12 12 6 2 12
2 4 1 1 1
   
12 12 6 3 18
2 2 1 1 1
   
12 12 6 6 36
2 6 1 6 1
   
12 11 6 11 11
2 4 1 4 2
   
12 11 6 11 33
2 1 1 1
1
   
12 11 6 11 66
CARDS
A standard deck of playing cards consist of ___
52 cards.
There are __
26 of each.
2 colors; RED and BLACK. ____
There are __
4 suits; HEARTS, DIAMONDS, CLUBS, and
SPADES.
____
13 of each.
Each suit consist of the cards 2, 3, 4, 5, 6, 7, 8, 9, 10, Jack,
Queen, King, and Ace.
__
4 of each.
Example 2 - Suppose you are going to pull two cards from a
standard deck of 52, one right after the other WITHOUT
replacing the first card. Find the following probabilities:
26
26

=
51
1.) P(A red and then a black) = 52
13
51
13
13

=
51
2.) P(Spade and then a Heart) = 52
13
204
4
4

=
3.) P(Jack and then an ACE) = 52
51
26
25

=
51
4.) P(2 Reds) = 52
4
3

=
51
5.) P(2 Kings) = 52
HW : Section 3.9
4
663
25
102
1
221
pages 193-194
#’s 30-45