Download Chapter Ten: Introducing Probability The chance behavior is

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Chapter Ten: Introducing Probability
The chance behavior is unpredictable in the short
run but has a regular and predictable pattern in the
long run.
We call a phenomenon or experiment random if individual outcomes are uncertain but there is a nonetheless a regular distribution of outcomes in a large number of repetition. The probability of any outcome of
a random phenomenon is the proportion of times the
outcome would occur in a very long series of repetitions.
Example: Consider a fair coin tossing experiment.
There are two possible outcomes: head and tail. We
can not predict the outcome before tossing. However,
a very long series of tossing, the relative frequency
of head observations (= number of head observations
divided by total number of of tossing) will be very
close to 0.5 . So the probability
P({observe a head}) = 0.5
Definition: The sample space S of a random experiment or phenomenon is the set of all possible outcomes. An event is an outcome or a subset of the
sample space. A probability model is a mathematical
description of a random experiment consisting of two
parts: a sample space S and a way of assigning probabilities to events.
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Any probability must satisfy following Probability
Rules:
Rule 1 For any event A, 0 ≤ P(A) ≤ 1.
Rule 2 If S is the sample space in a probability model,
P(S) = 1.
Rule 3 (addition rule for disjoint events) Two events
A and B being disjoint, which means that they have
no outcomes in common, is denoted by {A and B} =
A ∩ B = ∅. If A ∩ B = ∅, we have
P(A ∪ B) = P(A) + P(B),
where A ∪ B = {A or B}.
Rule 4 For any event A, denote Ac := S − A, the complement of A. Then,
P(Ac) = 1 − P(A).
A probability model with a finite sample space is
called discrete. To assign probabilities in a discrete
model, list the probabilities of all the individual outcomes. The probability of any event is the sum of the
probabilities of the outcomes making up the event.
Example: Consider a random experiment of rolling
a fair die and observing its outcome number on the
upper face.
Solution: In this case, the sample space includes six
individual outcomes and S = {1, 2, 3, 4, 5, 6}. Since the
die is fair, all outcomes should be equally likely. So
for each 1 ≤ i ≤ 6 if we define
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P({i}) = ,
6
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we have a discrete probability model (S, P).
A continuous probability model assigns probabilities
as area under a density curve. The area under the
curve and above any range of values is the probability
of an outcome in that range. Normal distributions are
examples of continuous probability models.
A random variable is a variable whose value is a numerical outcome of a random experiment. The probability distribution of a random variable X tells us what
values X can take and how to assign probabilities to
those values.
Example: Consider the random experiment of rolling
one fair die and observe the number appearing on the
upper face. If we define that X is the number appearing on the upper face, then X is a random variable.
Find the probability distribution of X.
Solution: First, the sample space
S = {(1), (2), (3), (4), (5), (6)}
Since the die is fair, we should define that for each
possible outcome (i), where 1 ≤ i ≤ 6
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P({(i)}) = .
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Then, we can find the probability distribution of X as
follows.
The distribution table of r.v X
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X
1
2
3
4
5
6
P(X = i) 1/6 1/6 1/6 1/6 1/6 1/6
This is a discrete probability model.
Example: A couple plans to have three children. There
are 8 possible arrangements of girls and boys. For example, GGB means the first two children are girls and
the third child is a boy. All 8 arrangements are equally
likely.
(a) Write down all 8 arrangements of the sexes of three
children. What is the probability of any one these arrangements.
(b) Let X be the number of girls the couple has. What
is the probability that X = 2?
(c) Find the distribution of X.
Solution:
(a) All different 8 arrangements are as follows:
1
2
3
4
5
6
7
8
BBB BBG BGB BGG GBB GBG GGB GGG
They are equally likely. So each has probability 1/8.
Since X is the number of girls the couple has,
P(X = 2) = P({BGG, GBG, GGB})
= P({BGG}) + P({GBG}) + P({GGB}) = 3/8.
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(c) Similarly we can find that
P(X = 1) = P({BBG, BGB, GBB})
= P({BBG}) + P({BGB}) + P({GBB}) = 3/8,
P(X = 0) = P({BBB}) = 1/8,
and
P(X = 3) = P({GGG}) = 1/8.
Thus, we have the distribution table of X.
Value of X
0
1
2
3
Probability 1/8 3/8 3/8 1/8