Transcript Section 6.2: Definition of Probability

Section 6.2: Definition of
Probability
• Probability of an event E denoted P(E) is the ratio of the
number of outcomes favorable to E to the total number
of outcomes in the sample space:
• This method is appropriate only when the outcome is
equally likely
number of outcomes favorable to E
P( E ) 
number of outcomes in the sample space
Example: Calling the Toss
• On some football teams, the honor of calling the
toss at the beginning of a football game is
determined by random selection. Suppose that
this week a member of the offensive team will
call the toss. There are 5 interior linemen on the
11-player offensive team. If we define the event
L as the event that lineman is selected to call the
toss, 5 of the 11 possible outcomes are included
in L. The probability that a lineman will be
selected is then P(L) = 5/11
Example: Math Contest
• Four students (Adam, Betty, Carlos, and Debra)
submitted correct solutions to a math contest
with two prizes. The contest rules specify that if
more than two correct responses are submitted,
the winners will be selected at random from
those submitting correct responses. In this case,
the set of possible outcomes for the chance
experiment that consists of selecting the two
winners from the four correct responses is:
{(A,B), (A,C), (A,D), (B,C), (B,D), (C,D)}
• Because the winners are selected at
random, the six outcomes are equally
likely and the probability of each individual
outcome is 1/6.
• Let E be the event that both selected
winners are the same gender. Then:
E = {(A,C), (B,D)}
• Because E contains two outcomes,
P(E)=2/6 = .333.
• If F denotes the event that at least one of
the selected winners is female, the F
consists of all outcomes except (A,C) and
P(F) = 5/6 = .833
Probability – Empirical Approach
• Law of large numbers – As the number of
repetitions of a chance experiment
increases, the chance that the relative
frequency of occurrence for an event will
differ from the true probability of the event
by more than any small number
approaches 0.
Relative Frequency Approach to
Probability
• The probability of an event E, denoted by P(E) is
defined to be the value approached by the
relative frequency of occurrence of E in a very
long series of trials of a chance experiment.
Thus, if the number of trials is quite large,
number of times E occurs
P( E ) 
number of trials
Methods for Determining
Probability
1. The classical approach: Appropriate for
experiments that can be described with equally
likely outcomes.
2. The subjective approach: Probabilities
represent an individual’s judgment based on
facts combined with personal evaluation of
other information.
3. The Relative Frequency Approach: An
estimate is based on an accumulation of
experimental results. This estimate, usually
derived empirically, presumes a replicable
chance experiment.
Section 6.3: Basic Properties
of Probability
Basic Properties of Probability
•
•
•
•
For any event E, 0P(E) 1.
If S is the sample space for an experiment,
P(S)=1.
If two events E and F are disjoint, then
P(E or F) = P(E) + P(F).
For any event E,
P(E) + P(not E) = 1 so,
P(not E) = 1 – P(E) and P(E) = 1 – P(not E).
Property One:
• Suppose we are flipping a bottle cap to
find the success of caps landing face up.
• Suppose that after N times, we have x
successes. What are the possible values
of x?
• The least amount it can be is 0. The most
it could be is N. The relative frequency
between them is 0 to 1.
Property Two:
• Because the probability of any event is the
proportion of time an outcomes in the
event will occur in the long run and
because the sample space consists of all
possible outcomes for a chance
experiment, in the long run an outcome in
S must occur 100% of the time. Thus
P(S)=1
Example: Cash or Credit?
• Customers at a certain department store pay for
purchases with either cash or one of four types
of credit card. Store records, kept for a long
period of time, indicate that 30% of all purchases
involve cash, 25% are made with the store’s own
credit card, 18% with MasterCard (MC), 15%
with Visa (V), and the remaining 12% with
American Express (AE). The following table
displays the probabilities of the simple events for
the chance experiment in which the mode of
payment for a randomly selected transaction is
observed:
Simple event O1(Cash) O2(Store) O3(MC) O4(V) O5(AE)
Probability
.30
.25
.18
.15
.12
Let’s create event E, the event a randomly selected
purchase is made with a nationally distributed credit
card. This event consists of the outcomes MC, V, and
AE. Therefore,
P(E)=P(03 ∪ O4 ∪ O5)=P(03)+P(O4)+P(O5)=.18+.15+.12=.45
45% of all purchases are made using one of the three
national cards.
• In addition,
P(not E) = 1 – P(E) = 1 - .45 = .55
You could also say that not E consists of
outcomes 1 and 2 with is .30 + .25 = .55
Calculating Probabilities When
Outcomes are Equally Likely
Consider an experiment that can result in any
one of N possible outcomes. Denote the
corresponding simple events by O1, O2,… On.
If these simple events are equally likely to
occur, then
1
1
1
1. P(O1 )  ,P(O2 )  , ,P(ON ) 
N
N
N
2. For any event E,
number of outcomes in E
P(E) 
N
Example: How likely is it you will be the mayor in Mafia?
Consider the experiment consisting of randomly
picking a card from an ordinary deck of playing
cards (52 card deck).
Let A stand for the event that the card chosen is a
King.
The sample space is given by S =
{A, K,,2, A , K  , , 2, A,, 2, A,…, 2}
and consists of 52 equally likely outcomes.
The event is given by
A={K, K, K, K}
and consists of 4 outcomes, so
Addition Rule for Disjoint Events
• Let E and F be two disjoint events. One of the basic
properties of probability is
P(E or F) = P(E ∪ F) = P(E) + P(F)
This property of probability is known as the addition rule for disjoint
events.
P( E1 or E2 or or Ek )  P( E1  E2    Ek )  P( E1 )  P( E2 )    P( Ek )
Example
Consider the experiment consisting of rolling two
fair dice and observing the sum of the up faces.
Let E stand for the event that the sum is 7 and
F stand for the event that the sum is 11.
6
2
P(E) 
& P(F) 
36
36
Since E and F are disjoint events
P(E F)  P(E)  P(F) 
6 2
8


36 36 36