Transcript Section 6.4

1.
2.
Probability of Equally Likely Outcomes
Complement Rule
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Let S be a sample space consisting of N equally
likely outcomes. Let E be any event. Then
number of outcomes in E 

Pr(E ) 
.
N
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Suppose
that a cruise ship returns to the US from
the Far East. Unknown to anyone, 4 of its 600
passengers have contracted a rare disease. Suppose
that the Public Health Service screens 20
passengers, selected at random, to see whether the
disease is present aboard ship. What is the
probability that the presence of the disease will
escape detection?
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The
sample space consists of samples of 20 drawn
from among the 600 passengers.
 600 
N 

 20 
There are 600 - 4 = 596 non infected passengers.
The disease is not detected if the 20 passengers
are chosen from this group.
Let E be the event that the disease is not detected.
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 596 
The number of outcomes in E is n( E ) 

.
 20 
 596 


20
  .87
Pr( E )  
 600 


 20 
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Complement
Rule
Let E be any event, E ' its
complement. Then
Pr(E) = 1 - Pr(E ').
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A
group of 5 people is to be selected at random.
What is the probability that 2 or more of them have
the same birthday?
For simplicity we will ignore leap years and
assume that each of the 365 days of the year are
equally likely.
Choosing a person at random is equivalent to
choosing a birthday at random.
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There
are 3655 ways of choosing 5 birthdays.
It will also be easier to first find the probability
that the 5 birthdays are different. Let this be E '.
The number of outcomes where the 5 birthdays
are different is 365364363362361.
365  364  363  362  361
Pr( E ) 
 .973
5
365
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Pr(E)
= 1- Pr(E ') = 1 - .973 = .027
This example could be worked for a general
group of r people.
In this case
365  364   (365  r  1)
Pr( E ) 
.
r
365
Probability that, in a randomly selected group of r
people, at least two will have the same birthday
r
5
10 15 20 22 23 25
30 40 50
Pr(E) .027 .117 .253 .411 .476 .507 .569 .706 .891 .970
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For a sample space with a finite number of
equally likely outcomes, the probability of an
event is the number of elements in the event
divided by the number of elements in the sample
space.
 The probability of the complement of an event is
1 minus the probability of the event.

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