# Section 3.3

```Chapter 3
Section 3.3
Basic Rules of Probability
Numbers and Probability
We have previously said that the probability of and event is
the ration between the number of equally likely outcomes
in the event E over the number of equally likely outcomes
in the sample space S.
n( E )
P( E ) 
n( S )
Facts About the Probability of an Event E
1. The probability of an event E is a fraction, decimal or percentage. In other
words it is a number.
a. The probability number has to be between 0 and 1 if it is a fraction or
decimal (i.e. 0P(E)1).
b. The probability number has to be between 0% and 100% if it is a
percentage (i.e. 0%P(E)100%).
2. The probability of E is zero (i.e. P(E) = 0 or P(E) = 0%) means it is impossible
for the event to happen.
3. The probability of E is one (i.e. P(E) = 1 or P(E) = 100%) means the event is
certain to happen.
Representing Experiments in Venn Diagrams
In experiments that have more than one event associated with them we can fill in
the corresponding regions in the Venn Diagram with any one of the following:
a. The number of outcomes in the region.
b. The percentage of outcomes in the region.
S F
A
c. The fraction of outcomes in that region.
HHT
THH
HTH
HHH
TTT
THT
d. The decimal of outcomes in that region.
HTT
TTH
Example:
Consider the experiment of flipping
three coins. The sample space for
this event is:
elements
S F
3
{HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}.
There are two events associated
with this:
F: The first coin is a head (H)
A: All coins are the same
S F
A
1
1
3
37.5% 12.5%
3
8
percentage
S F
A
1
8
1
8
fraction
12.5%
37.5%
number
S F
A
3
8
.375
A
.125
.125
.375
decimal
S
S
A
2
A
B
3
4
30%
S
B
15%
2
7
45%
1
A
B
1
7
3
7
Experiment 2
A
B
.1
.6
.2
1
7
10%
Experiment 1
S
Experiment 3
.1
Experiment 4
For each of the four experiments above fill in the table below. Do you notice any patterns?
P(A)
Experiment 1
Experiment 2
5
10
45%
P(B)
P(A∩B)
P(AB)
P(A')
P(B')
7
10
3
10
9
10
5
10
3
10
60%
15%
90%
55%
40%
Experiment 3
3
7
4
7
1
7
6
7
4
7
3
7
Experiment 4
.7
.8
.6
.9
.3
.2
P(A)+P(B)=P(A∩B)+P(AB)
P(A)+P(A')=1 (or 100%)
P(B)+P(B')=1 (or 100%)
The chance that A happens plus the chance B happens
is equal to the chance they both happen plus the
chance either happens.
The chance an event occurs plus the chance it will not
occur is 1 (or 100%).
In a certain experiment there are two events that can happen A and B. In this
experiment P(A) = .4 and P(B) = .7 and P(AB) = .9, use this to answer each of
the questions below.
Before we begin we attempt to draw the Venn
Diagram. But in order to do that we need to get
P(A∩B) but we were given the P(AB). We use
the previous relationship to find this.
P(A) + P(B) = P(A∩B) + P(AB)
.4 + .7 = P(A∩B) +.9
1.1 = P(A∩B) +.9
P(A∩B) = .2
P(A∩(B')) = .2
P((A')∩B) = .5
S
A
B
.2
.2
.5
.1
P((AB)') = .1
Mutually Exclusive Events
Two events are called mutually exclusive if they can not both happen at the
same time. For example if you flip a coin you can not get both a head and a tail,
so a head and tail are mutually exclusive. In terms of numbers:
P(A∩B) = 0
or
P(A) + P(B) = P(AB)
A researcher has 50 rats. He feeds 25 of them a high fat diet for six months and
the others he feeds normally. After six months he examines all the rats for signs of
cardiovascular disease and finds that 32 have the disease. Of the 32 rats with
cardiovascular disease he finds that 21 of them were given a high fat diet.
An experiment is conducted where a rat from this
group is picked at random. Let F be the event the
rat was fed a high fat diet and C be the event the
rat has cardiovascular disease. Find each of the
probabilities below.
S
F
4
C
21
11
14
Before we begin we draw a Venn Diagram
P(C) = The chance a rat develops cardiovascular disease =
32
 64 %
50
P(F∩C) = The chance a rat was fed a high fat diet and has disease =
21
 42 %
50
4
P(F∩(C')) = The chance a rat has a high fat diet and does not have disease = 50  8%
11
 22 %
50
14
 28 %
P((FC)') = The chance a rat was neither fed a high fat diet nor has disease =
50
P((F')∩C) = The chance a rat was not fed a high fat diet and gets the disease =
Another way the numerical information can be organized is in the form of a table. In
the problem below use the table of information to answer each of the questions
below.
The results of a survey in which 200
people were asked if they were
married (M) or unmarried (U) and if
they were a smoker (S) or nonsmoker (N) are given to the right. An
experiment is conducted were a
person is selected at random from
this group.
Married
(M)
Unmarried
(U)
Smoker
(S)
23
42
Non-Smoker
(N)
94
41
23  42 65


P
S


 32.5%
P(The person is a smoker) =
200
200
P(The person is unmarried) = PU  
42  41 83

 41.5%
200
200
P(The person is married and a non-smoker) = PM  N  
P(The person is either unmarried or a non-smoker) =
94
 47%
200
PU  N  
P(The person is neither married nor a smoker) = PM  S ' 
42  41  94 177

 88.5%
200
200
41
 20.5%
200
```