6.3 Assignment of Probabilities
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Transcript 6.3 Assignment of Probabilities
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Probability of an Outcome
Experimental Probability
Fundamental Properties of Probabilities
Addition Principle
Inclusion-Exclusion Principle
Odds
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Let a sample space S consist of a finite number of outcomes
s1, s2, … ,sN. To each outcome we associate a number,
called the probability of the outcome, which represents the
relative likelihood that the outcome will occur. A chart
showing the outcomes and the assigned probability is
called the probability distribution for the experiment.
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Toss
an unbiased coin and observe the side that
faces upward. Determine the probability
distribution for this experiment.
Since the coin is unbiased, each outcome is
equally likely to occur.
Outcome
Heads
Tails
Probability
½
½
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Let a sample space S consist of a finite number of
outcomes s1, s2, … ,sN. The relative frequency, or
experimental probability, of each outcome is
calculated after many trials.
The experimental probability could be different for
a different set of trials and different from the
probability of the events.
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Traffic engineers measure the volume of traffic
on a major highway during the rush hour.
Generate a probability distribution using the data
generated over 300 consecutive weekdays.
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We will use the
experimental
probability for
the distribution.
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Let an experiment have outcomes s1, s2, … , sN with probabilities p1, p2, … , pN.
Then the numbers p1, p2, … , pN must satisfy:
Fundamental Property 1 Each of the numbers p1, p2, … , pN is between 0 and
1;
Fundamental Property 2
p1 + p2 + … + pN = 1.
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Verify the fundamental properties for the
following distribution.
All
probabilities
are between
0 and 1
Total: 1.00
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Addition Principle
Suppose that an event E consists of the finite
number of outcomes s, t, u, … ,z. That is E = {s, t, u, … ,z }.
Then
Pr(E) = Pr(s) + Pr(t) + Pr(u) + … + Pr(z),
where Pr(E) is the probability of event E.
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Suppose that we toss a red die and a green die
and observe the numbers on the sides that face
upward.
a) Calculate the probabilities of the elementary
events.
b) Calculate the probability that the two dice
show the same number.
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a)As shown previously, the sample space
consists of 36 pairs of numbers
S = {(1,1), (1,2), … , (6,5), (6,6)}.
Each of these pairs is equally likely to occur.
The probability of each pair is 136 .
b) E = {(1,1), (2,2), (3,3), (4,4), (5,5), (6,6)}
1
1
1
1
1
1 1
Pr( E )
36 36 36 36 36 36 6
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Let E and F be any events. Then
Pr( E F ) Pr( E ) Pr( F ) Pr( E F ).
If E and F are mutually exclusive, then
Pr( E F ) Pr( E ) Pr( F ).
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A factory needs two raw materials. The
probability of not having an adequate supply of
material A is .05 and the probability of not
having an adequate supply of material B is .03. A
study determines that the probability of a
shortage of both materials is .01. What
proportion of the time will the factory not be able
to operate from lack of materials?
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Pr( A B) Pr( A) Pr( B) Pr( A B)
.05 .03
.01
.07
The factory will not operate 7% of the time.
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If the odds in favor of an event E are a to b, then
a
b
Pr( E )
and Pr( E )
.
ab
ab
On average, for every a + b trials, E will occur
a times and E will not occur b times.
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In the two dice problem, what are the odds of
rolling a pair with the same number on the faces?
The probability of obtaining a pair with the same
number on the faces is 1/6.
The probability of not obtaining a pair with the
same number on the faces is 5/6.
The odds are 1
6 1 or 1 to 5.
5
5
6
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A probability distribution for a finite sample space
associates a probability with each outcome of the
sample space. Each probability is a number
between 0 and 1, and the sum of the probabilities
is 1. The probability of an event is the sum of the
probabilities of outcomes in the event.
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The inclusion-exclusion principle states that the
probability of the union of two events is the sum of
the probabilities of the events minus the
probability of their intersection. If the two events
are mutually exclusive, the probability of the
union is just the sum of the probabilities of the
events.
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We say that the odds in favor of an event are a to
b if the probability of the event is a/(a + b).
Intuitively, the event is expected to occur a times
for every b times it does not occur.
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