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```Sets, Combinatorics, Probability, and Number
Theory
Mathematical Structures
for Computer Science
Chapter 3
Copyright © 2006 W.H. Freeman & Co.
MSCS Slides
Probability
Introduction to Finite Probability


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
1
Introduction to Finite Probability



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) 



Section 3.5
E
S
For example, the probability of flipping a coin twice and having
both come up as heads is:
HH
E
1

P(E) 


S HH, HT, TH, TT 4
Probability involves finding the size of sets, either of the sample
space or of the event of interest.
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
2
Introduction to Finite Probability

Section 3.5
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:
Probability
3
Probability Distributions



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

2.
k
Because any probability value must fall within this range.
 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
4
Example
For a fair die, the probability of rolling a three is 1/6.


Sample space S = {1, 2, 3, 4, 5, 6}, event space E =
{3}
Suppose the die is loaded so that 4 comes up three
times more often than any of the other numbers. Now
the sample space S = {1, 2, 3, 41, 42, 43, 5, 6} so the
probability of rolling a 3 is only 1/8
The probability distribution for the loaded die is


xi
p(xi)

Section 3.5
1
1/8
2
1/8
3
1/8
4
3/8
5
1/8
6
1/8
 p(xi) for all xi above = 1, as we know it should be.
Probability
5
Conditional Probability





Section 3.5
Flip a coin twice, sample space = {HH, HT, TH, TT}. The
chance that we get two tails is 1/4.
Let E1 be the event that the first flip is a tail, so now
E1 = {TH, TT} and P(E1) = | E1 | / |S| = 1/2
Let E2 be the event that the second flip is a tail; E2 = {HT, TT}.
The event of interest, tossing two tails in a row, is E1 E2 or
{TT}. So …
P(E1  E2 ) = | E1  E2 | = 1
|S|
4
Now suppose we know that the first toss was a tail; does this
change the total probability? It does, because now S is limited:
S = {TH, TT}.
P(E2| E1) (probability of E2 given E1) = | E2 | / | S | = 1/2
Probability
6
Conditional Probability - Definition

Given events E1 and E2, the conditional probability
of E2 given E1, P(E2E1), is:
P E1  E 2 
PE 2 E1 
P E1 


For example, in a drug study of a group of patients,
17% responded positively to compound A, 34%
responded
 positively 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
Section 3.5
Probability
7
Conditional Probability

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.


Section 3.5
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
8
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
9
Average Case Analysis of Algorithms






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
10
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