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Probability
Read: 4.6
Next Class: 5.1
Introduction to Finite Probability
1


If some action can produce Y different outcomes and X of those
Y outcomes are of special interest, we may want to know how
likely it is that one of the X outcomes will occur.
For example, what are the chances of:

Getting “heads” when a coin is tossed?
• The probability of getting heads is “one out of two,” or 1/2.

Getting a 3 with a roll of a die?
• A six-sided die has six possible results. The 3 is exactly one of
these possibilities, so the probability of rolling a 3 is 1/6.

Drawing either the ace of clubs or the queen of diamonds
from a standard deck of cards?
• A standard card deck contains 52 cards, two of which are
successful results, so the probability is 2/52 or 1/26.
Section 3.5
Probability
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Introduction to Finite Probability (cont’d)
The set of all possible outcomes of an action is called the
sample space S of the action.

Any subset of the sample space is called an event.

If S is a finite set of equally likely outcomes, then the
probability P(E) of event E is defined to be:

P(E) 
E
S

Section 3.5
Probability
2
Introduction to Finite Probability (cont’d)

For example, the probability of flipping a coin twice and having
both come up as heads is:
P(E) 

HH
E
1


S HH,HT,TH,TT 4
Probability involves finding the size of sets, either of the sample
space or of the event of interest.


Section 3.5
We may need to use the addition or multiplication principles,
the principle of inclusion and exclusion, or the formula for the
number of combinations of r things from n objects.
Probability
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2
Introduction to Finite Probability (cont’d)

Using the definition of P(E) as seen in the previous slide, we
can make some observations for any events E1 and E2 from a
sample space S of equally likely outcomes:
Section 3.5
Probability
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Probability Distributions


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A way to look at problems where not all outcomes are equally
likely is to assign a probability distribution to the sample
space.
Consider each distinct outcome in the original sample space as
an event and assign it a probability.
If there are k different outcomes in the sample space and each
outcome xi is assigned a probability p(xi), the following rules
apply:
1. 0  p(xi)  1

Because any probability value must fall within this range.
k
2.
 p(x )  1
i
i 1
•
The union of all of these k disjoint outcomes is the sample
space S, and the probability of S is 1.

Section 3.5
Probability
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3
Conditional Probability

Given events E1 and E2, the conditional probability of E2
given E1, P(E2E1), is:
P E 2 E1 


Section 3.5
P E1  E 2 
P E1 
For example, in a drug study of a group of patients, 17%
 positively to compound A, 34% responded positively
responded
to compound B, and 8% responded positively to both.
The probability that a patient responded positively to compound
B given that he or she responded positively to A is:
P(B|A) = P(A ∩ B) / P(A) = 0.08 / 0.17 @ 0.47
Probability
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Conditional Probability


Section 3.5
If P(E2E1) = P(E2), then E2 is just as likely to happen whether
E1 happens or not. In this case, E1 and E2 are said to be
independent events.

Then P(E1  E2) = P(E1) * P(E2)
This can be extended to any finite number of independent events
and can also be used to test whether events are independent.
Probability
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4
Expected Value
If the values in the sample space are not numerical, we may find
a function X: S  R that associates a numerical value with each
element in the sample space. Such a function is called a random
variable.

Given a sample space S to which a random variable X and a
probability distribution p have been assigned, the expected
value, or weighted average, of the random variable is:

n
E X    X(x i ) p(x i )
i1

Section 3.5
Probability
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Expected Value: Example
A fair coin is tossed three times. The sample space is
S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}
Let the random variable X assign to each outcome in S the
number of head in that outcome. Because the coin is fair, each
outcome in S occurs with equal opportunity which is the
multiplication of probability of each toss (total 3 tosses), ie.
½ * ½ * ½ = 1/8.

xi
HHH
X(xi)
3
2
2
1
2
1
1
0
p(xi)
1/8
1/8
1/8
1/8
1/8
1/8
1/8
1/8
Section 3.5
HHT HTH HTT THH THT TTH TTT
Probability
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Expected Value: Example (cont’d)
The expected value of X, that is, the expected number of heads
in three tosses, is:
8
E ( X )   X ( xi ) p( xi )
i 1
 3(1 / 8)  2(1 / 8)  2(1 / 8)  1(1 / 8)  2(1 / 8)  1(1 / 8)  1(1 / 8)  0(1 / 8)
 12(1 / 8)  1.5
Section 3.5
Probability
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Average Case Analysis of Algorithms

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
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Expected value may help give an average case analysis of an
algorithm, i.e., tell the expected “average” amount of work
performed by an algorithm.
Let the sample space S be the set of all possible inputs to the
algorithm.
We assume that S is finite.
Let the random variable X assign to each member of S the
number of work units required to execute the algorithm on that
input.
And let p be a probability distribution on S,
The expected number of work units is given as:
n
E X    X(x i ) p(x i )
i1
Section 3.5

Probability
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6
Class Exercise

A loaded die has the following probability distribution:
xi
1
2
3
4
5
6
p(xi)
0.2
0.05
0.1
0.2
0.3
0.15
When the die is rolled, let E1 be the event that the rolled
number is odd, let E2 be the event that the rolled number is 3 or
6, and let E3 be the event that the rolled number is 4 or more.
a. Find P(E1).
b. Find P(E2).
c. Find P(E3).
d. Find P(E2 ∩ E3 ).
e. Find P(E1 U E3 ).
Section 3.5
Probability
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