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Probability
14.1 Experimental Probability
14.2 Principles of Counting
14.3 Theoretical Probability
14.1
Experimental Probability
Slide 14-2
DEFINITION:
EXPERIMENTAL PROBABILITY
Suppose an experiment with a number of
possible outcomes is performed repeatedly –
say, n times – and that a specific outcome E
occurs r times. The experimental
probability, denoted by Pe  E  , that E will
occur on any given trial of the experiment is
given by
r
Pe  E   . Experimental outcomes
n Total number or outcomes
Slide 14-3
THE LAW OF LARGE NUMBERS
If an experiment is performed repeatedly,
the experimental probability of a particular
outcome more and more closely
approximates a fixed number as the
number of trials increases.
Slide 14-4
THE LAW OF LARGE NUMBERS
Ratio of the number of heads to the number of
tosses in a coin-tossing experiment:
Slide 14-5
Example - Computing Experimental
Probability from Data
The final examination scores of students in a
precalculus class are as shown. Compute the
experimental probability that a student chosen
at random from the class had a score in the
70s?
Slide 14-6
Example : continued
Since 6 of the 28 students scored in the 70s,
the experimental probability that a student
chosen at random had a score in the 70s is
6
 0.21.
28
Slide 14-7
Example: Determining Experimental
Probabilities from an Experiment
Christine has five pennies. She is curious how
often she should expect to see at most one
head when all five coins are flipped onto the
floor. To find an answer, she repeatedly flips the
five pennies and counts the number of heads
that turn up. After repeating the experiment 50
times, she obtains the following frequency table:
Slide 14-8
Example: continued
On the basis of Christine’s data, what is the
experimental probability that a flip of five coins
results in at most one head?
The data shows that exactly one head appeared
on seven of the trials ad no heads appeared
once. This means that the outcome of at most
one head occurred 1 + 7 = 8 times in the 50
trials, giving the experimental probability of
8
Pe  at most 1 head  
 0.16.
50
Slide 14-9
THE TERMINOLOGY OF PROBABILITY
OUTCOME:
a result of one trial of an experiment
SAMPLE SPACE S:
the set of all outcomes of an experiment
Slide 14-10
THE TERMINOLOGY OF PROBABILITY
MUTUALLY EXCLUSIVE EVENTS A and B:
two events A and B such that the occurrence
of A does not affect its occurrence in B, and
vice-versa. In other words they cannot
happen at the same time. E.g. Turning left
and turning right.
Slide 14-11
THE TERMINOLOGY OF PROBABILITY
NON-MUTUALLY EXCLUSIVE EVENTS A
and B:
If A and B are any two events, they can
happen at the same time. E.g. Turning left
and scratching your head can happen at the
same time.
Slide 14-12
Example: Computing the Experimental
Probability of Non-Mutual Exclusive Events
A penny and a dime are flipped, with the
results as shown in the table.
Determine the experimental probability that
the dime shows a head or both coins land
with the same side up.
Slide 14-13
EXPERIMENTAL PROBABILITY OF
NON-MUTUALLY EXCLUSIVE EVENTS
Let A denote the event “the dime shows a
head.” Let B denote the event “both coins
land with the same side up.”
Pe ( A) 
Pe ( B) 
24
50
31
50
Pe ( A  B) 
14
50
and
Slide 14-14
14.2
Principle of Counting
Slide 14-15
PRINCIPLES OF COUNTING
To determine the theoretical probabilities of
events, we will need to determine the
number of possible outcomes.
Thus, the calculation of theoretical
probability is a matter of counting.
Slide 14-16
DEFINITION:
INDEPENDENT EVENTS
Events A and B are independent events
if the occurrence or nonoccurrence of
event A does not affect the occurrence or
nonoccurrence of event B, and vice versa.
Slide 14-17
ADDITION PRINCIPLE OF COUNTING
If A and B are events, then
n( A or B)  n( A)  n( B)  n( A  B).
and
Slide 14-18
Example 14.10: Counting and the Word
Or
In how many ways can you select a red
card or an ace from an ordinary deck of
playing cards?
Solution 1: There are 26 red cards in the
deck (including two red aces) as well as two
black aces. Thus, the desired answer is 28.
n( A or B)  n( A)  n( B).
Slide 14-19
Example 14.10: continued
In how many ways can you select a red card or an
ace from an ordinary deck of playing cards?
Solution 2: Let R denote the set of red cards and
let A denote the set of aces. Then n(R) = 26,
n(A) = 4, n(R ∩ A) = 2, and
n( R or A)  n( R)  n( A)  n  R
26  4  2
A
and
 28
Slide 14-20
ADDITION PRINCIPLE OF COUNTING
FOR MUTUALLY EXCLUSIVE EVENTS
If A and B are mutually exclusive events,
then
n( A or B)  n( A)  n( B).
Slide 14-21
Example 14.12: Choosing a Chocolate
A box of 40 chocolates contains 14 cremes (A), 16
caramels (B), and 10 chocolate-covered nuts. In
how many ways can you select a crème or a
caramel from the box.
n( A or B)  n( A)  n( B).
14 + 16 = 30.
Thus, the number of ways of choosing a creme or
a caramel is 30.
Slide 14-22
MULTIPLICATION PRINCIPLE OF
COUNTING FOR INDEPENDENT EVENTS
When there are A ways to do one thing,
and B ways to do another, then there are A
x B ways of doing both. If A and B are
independent events, then the number of
ways that A and B can occur in a two-stage
experiment is given by
n( A and B)  n( A)  n( B).
Slide 14-23
Example 14.14: Counting the Number of
Ways to Draw Two Aces
How many ways can two aces be drawn in
succession from a deck of ordinary cards
-- With replacement?
If we replace the drawn card, then each
choice is independent, so
n(two aces)  n(1st ace)  n(2nd ace)
 4
 16

4
Slide 14-24
Example 14.14: continued
How many ways can two aces be drawn in
succession from a deck of ordinary cards
-- Without replacement?
If we don’t replace the first drawn ace, then
the number of ways to draw a second ace is
impacted. So
n(two aces)  n(1st ace)  n(2nd ace|1st was ace)
 4
 12

3
Slide 14-25
14.3
Theoretical Probability
Slide 14-26
DEFINITION:
THEORETICAL PROBABILITY
Let S, the sample space (total outcomes),
and let E, an event (desired outcomes), Let
n(S) and n(E) denote the number of
outcomes in S and E, respectively. The
probability of E, denoted by P(E), is given
by
n( E )
PE 
.
n( S )
Slide 14-27
DETERMINING PROBABILITY
Determine the probability of obtaining a total
score of 3 or 4 on a single throw of two dice.
List the sample space as ordered pairs.
(1, 1) (1, 2) (1, 3) (1, 4) (1, 5) (1, 6)
(2, 1) (2, 2) (2, 3) (2, 4) (2, 5) (2, 6)
(3, 1) (3, 2) (3, 3) (3, 4) (3, 5) (3, 6)
(4, 1) (4, 2) (4, 3) (4, 4) (4, 5) (4, 6)
(5, 1) (5, 2) (5, 3) (5, 4) (5, 5) (5, 6)
(6, 1) (6, 2) (6, 3) (6, 4) (6, 5) (6, 6)
P  sum of 3 or 4 
n(3 or 4)

n( S )
5

36
Slide 14-28
Example
Five black marbles, seven white marbles,
eight red marbles are placed in an bag. If
one is chosen at random, what is the
probability of a red marble?
P(white) = desired outcome / total outcomes
P(white) = 8/20 = 2/5 = 0.4 = 40%
Slide 14-29
PROPERTIES OF PROBABILITY
1.
P( A)  0 if, and only if, A cannot occur.
E.g. Six sided die, the probability of
rolling a seven.
2.
P( A)  1 if, and only if, A always occurs.
E.g. Six sided die, the probability of
rolling 1,2,3,4,5, or 6.
Slide 14-30
PROPERTIES OF PROBABILITY
3. For any event A, 0  P  A  1.
E.g. all probabilities are between 0 and 1
and add up to 1 or 100%
Slide 14-31
DEFINITION:
ODDS
Then the odds in favour of A are
favourable outcomes : unfavourable outcomes
Then the odds against A are
unfavourable outcomes : favourable outcomes
Slide 14-32
Example 14.34: Determining the Odds in
Favor of Rolling a 7 or an 11
In the game of craps, one wins on the first roll of
the pair of dice if a 7 or an 11 is thrown. What are
the odds of winning on the first roll?
Let W be the set of outcomes that result 7 or 11.
Since W = {(1,6), (2,5), (3,4), (4,3), (5,2), (6,1),
(5,6), (6,5)} and these are 36 ways two dice can
come up.
Thus the odds in favor of W are 8:28 or, 2:7.
Slide 14-33
DEFINITION:
EXPECTED VALUE OF GAMBLING
Average expected payoff in the long run.
positive results = win, negative results =
loss, zero = breakeven.
EV = P(win) x gain – P(lose) x loss
Gain = Winnings – Initial Bet
Slide 14-34
DEFINITION:
EXPECTED VALUE OF GAMBLING
You pay $2 to play the game, if you pull
an ace from a deck you win $10, what is
the expected value?
EV = (4/52)(8) – (48/52)(2)
EV = 0.6154 – 1.8462
EV = -1.2308
Should you play this game? If you play it
100 times what will the results be?
Slide 14-35