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Probability
Computing Empirical Probability
The empirical probability of event E is
observed number of times E occurs
P( E ) 
total number of observed occurrences
Example
Hours of Sleep Number of Americans, in millions
4 or less
11
5
24.75
6
68.75
7
82.5
8
74.25
9
8.25
10 or more
5.5
Total 275
An American is randomly selected. Find the
probability of that person getting 6 hours sleep on
a typical night.
Example cont.
Solution:
P( six hours sleep ) 
number of Americans who sleep 6 hours
total number of Americans
68.75 275


 .25
275 1100
The empirical probability of randomly
selecting an American who gets eight hours
sleep in a typical night is 275/1100 or .25
Computing Theoretical
Probability
If an event E has n(E) equally-likely outcomes and its
Sample space S has n(S) equally-likely outcomes, the
Theoretical probability of event E, denoted by P(E), is
number of outcomes in event E
P( E ) 
number of outcomes in sample space S
The sum of the theoretical probabilities of all possible
Outcomes in the sample is 1.
Text Example
A die is rolled. Find the probability of getting a number less than 5.
Solution The sample space of equally likely outcomes is S  {1, 2, 3,
4, 5, 6}. There are six outcomes in the sample space, so n(S)  6.
We are interested in the probability of getting a number less
than 5. The event of getting a number less than 5 can be represented by
E  {1, 2, 3, 4}.
There are four outcomes in this event, so n(E)  4.
The probability of rolling a number less than 5 is
n(E) 4 2
P(E) 
 
n(S) 6 3
Example
What is the probability of getting at most 2
heads when a coin is tossed 3 times?
Solution:
Example cont.
What is the probability of getting at most 2
heads when a coin is tossed 3 times?
Solution:
Example cont.
What is the probability of getting at most 2
heads when a coin is tossed 3 times?
Solution:
E  {TTT , THT , HTT , TTH , HHT , HTH , THH }
n( E )  7
S  {TTT , THT , HTT , TTH , HHT , HTH , THH , HHH }
n( S )  8
n( E ) 7
P( E ) 

n( S ) 8
The probability of getting at most 2 heads
when a coin is tossed 3 times is 7/8
The Probability of an Event Not
Occurring
• The probability that an event E will not
occur is equal to one minus the probability
that it will occur.
P(not E) = 1 - P(E)
Or Probabilities with Mutually
Exclusive Events
If A and B are mutually exclusive events,
then
P(A or B)  P(A) + P(B).
Text Example
If one card is randomly selected from a deck of cards, what is the probability
of selecting a king or a queen?
Solution We find the probability that either of these mutually
exclusive events will occur by adding their individual probabilities.
4
4
8
2
+


P(king or queen)  P(king) + P(queen) 
52 52 52 13
The probability of selecting a king or a queen is 2/13 .
Or Probabilities with Events That
Are Not Mutually Exclusive
• If A and B are not mutually exclusive
events, then
• P(A or B)  P(A) + P(B) – P(A and B).
Text Example
The figure illustrates a spinner. It is
equally probable that the pointer will
land on any one of the eight regions,
numbered 1 through 8. If the pointer
lands on a borderline, spin again. Find
the probability that the pointer will
stop on an even number or a number
greater than 5.
Text Example cont.
Solution It is possible for the pointer to land on a number that is even
and greater than 5. Two of the numbers, 6 and 8, are even and greater
than 5. These events are not mutually exclusive. The probability of
landing on a number that is even and greater than 5 is
 even or

 even and

  P(even) + P(greater than 5)  P

P
greater
than
5
greater
than
5




4

8
Four of the eight numbers, 2,
4, 6, and 8, are even.
+
3
8

2
8
Three of the eight numbers, 6, 7,
and 8, are greater than 5.
4 +32 5


8
8
Two of the eight numbers, 6 and 8,
are even and greater than 5.
The probability that the pointer will stop on an even number or on a number
greater that 5 is 5/8.
And Probabilities with
Independent Events
• If events A and B are Independent, then the
probability of A and B is simply:
P(A and B) = P(A) · P(B)
Probability