Презентация Discrete mathematics. Probability онлайн
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№1 слайд
Содержание слайда: Discrete Mathematics
PROBABILITY-1
Adil M. Khan
Professor of Computer Science
Innopolis University
“Information: The Negative Reciprocal Value of Probability!”
- Claude Shannon -
№2 слайд
Содержание слайда: Probability---Introduction
One of the most important disciplines in Computer Science (CS).
Algorithm Design and Game Theory
Information Theory
Signal Processing
Cryptography
№3 слайд
Содержание слайда: Probability---Introduction---Cont.
But it is also probably the least well understood
Human intuition and Random events
Goal: To try our best to teach you how to easily and confidently solve problems involving probability
“What is the probability that … ?”
№4 слайд
Содержание слайда: Probability
Contents
Basic definitions and an elementary 4-step process
Counting
Conditional probability and the concept of independence
Random Variable
Expected value and Standard Deviation
№5 слайд
Содержание слайда: Probability
Let’s Make a Deal
The famous game show (you might have seen this problem in your books)
Participant is given a choice of three doors. Behind one door is a car, behind the others, useless stuff. The participant picks a door (say door 1). The host, who knows what is behind the doors, opens another door (say door 3) which has the useless stuff. He then asks the participant if he would like to switch (pick door 2)?
Is it to participant’s advantage to switch or not?
№6 слайд
Содержание слайда: Probability
Precise Description
The car is equally likely to be hidden behind the three doors.
№7 слайд
Содержание слайда: Probability
Precise Description
The car is equally likely to be hidden behind the three doors.
The player is equally likely to pick each of the doors.
After the player picks a door, the host must open a different door (with the useless thing behind it) and offer the player to switch.
When a host has a choice of which door to pick, he is equally likely to pick each of them.
Now here comes the question:
“What is the probability that a player who switches wins the car?”
№8 слайд
Содержание слайда: Probability
Solving Problems Involving Probability
Model the situation mathematically
Solve the resulting mathematical problem
№9 слайд
Содержание слайда: Probability
Solving Problems Involving Probability
Step 1: Finding the sample space
Set of all possible outcomes of a random process
№10 слайд
Содержание слайда: Probability
Solving Problems Involving Probability
Step 1: Finding the sample space
Set of all possible outcomes of a random process
№11 слайд
Содержание слайда: Probability
Solving Problems Involving Probability
Step 1: Finding the sample space
Set of all possible outcomes of a random process
To find this, we must understand the quantities involve in the random process
№12 слайд
Содержание слайда: Probability
Solving Problems Involving Probability
Step 1: Finding the sample space
Set of all possible outcomes of a random process
To find this, we must understand the quantities involve in the random process
Quantities in the above problem:
The door concealing the car
The door initially chosen by the player
The door that host opens to reveal the useless thing
№13 слайд
Содержание слайда: Probability
Finding the Sample Space
Every possible value of these quantities is called an outcome.
And (as said earlier) the set of all possible outcomes is called the sample space
№14 слайд
Содержание слайда: Probability
Finding the Sample Space
Every possible value of these quantities is called an outcome.
And (as said earlier) the set of all possible outcomes is called the sample space
A tree structure (Possibility tree) is a useful tool for keeping track of all outcomes
When the number of possible outcomes is not too large
№15 слайд
Содержание слайда: Probability
Possibility Tree
The first quantity in our example is the door concealing the car
Represent this as a root of tree with three branches (three doors)
№16 слайд
Содержание слайда: Probability
Possibility Tree --- Cont.
The car can be at any of these three locations
№17 слайд
Содержание слайда: Probability
Possibility Tree --- Cont.
The car can be at any of these three locations
For each possible location of the car, the player can choose any of the three doors
№18 слайд
Содержание слайда: Probability
Possibility Tree --- Cont.
The car can be at any of these three locations
For each possible location of the car, the player can choose any of the three doors
Then the final possibility is regarding the host opening a door to reveal the useless thing
Overall tree turns out to be
№19 слайд
Содержание слайда: Probability
Possibility Tree --- Cont.
№20 слайд
Содержание слайда: Probability
Finding The Sample Space
The leaves of the possibility tree represent the outcomes of a random process
The set of all leaves represent the sample space
№21 слайд
Содержание слайда: Probability
Finding The Sample Space
In our example, if we represent the leaves as a sequence of “labels” of intermediate nodes including the leaf node then,
№22 слайд
Содержание слайда: Probability
Solving Problems Involving Probability
Step 2: Defining the Events of Interest:
№23 слайд
Содержание слайда: Probability
Solving Problems Involving Probability
Step 2: Defining the Events of Interest:
Remember, we want to answer the questions of type:
“What is the probability that … ?”
№24 слайд
Содержание слайда: Probability
Solving Problems Involving Probability
Step 2: Defining the Events of Interest:
Remember, we want to answer the questions of type:
“What is the probability that … ?”
Replacing the “…” with some specific event. For example,
№25 слайд
Содержание слайда: Probability
Solving Problems Involving Probability
Step 2: Defining the Events of Interest:
Remember, we want to answer the questions of type:
“What is the probability that … ?”
Replacing the “…” with some specific event. For example,
“What is the probability that the car is behind door C?”
Doing this reduces S to some specific outcomes, called event of interest.
№26 слайд
Содержание слайда: Probability
Event of Interest
For the event,
“What is the probability that the car is behind door C?”
The set of possible outcomes reduces to
№27 слайд
Содержание слайда: Probability
Event of Interest
For the event,
“What is the probability that the car is behind door C?”
The set of possible outcomes reduces to
Simply speaking, an event is a subset of S
№28 слайд
Содержание слайда: Probability
Solving Problems Involving Probability
Coming back to our example
We want to know:
“What is the probability that the player will win by switching?”
This event can be represented as the following set
№29 слайд
Содержание слайда: Probability
Solving Problems Involving Probability---Cont.
Notice: Half of the outcomes are checked. Does this mean that the player wins by switching in half of all outcomes?
№30 слайд
Содержание слайда: Probability
Solving Problems Involving Probability---Cont.
Step 3: Determining Outcome Probability
Assign Edge Probabilities
Compute Outcome Probabilities
№31 слайд
Содержание слайда: Probability
Equally likely probability formula
E: the equally likely event
S: the sample space
№32 слайд
Содержание слайда: Probability
Solving Problems Involving Probability---Cont.
Step 3: Determining Outcome Probability
Assign Edge Probabilities
Compute Outcome Probabilities
№33 слайд
Содержание слайда: Probability
Edge Probabilities
To understand, let’s analyze the path leading to the leaf node (A, A, B)!
№34 слайд
Содержание слайда: Probability
Multiplication Rule
The probability that Events A and B both occur is equal to the probability that Event A occurs times the probability that Event B occurs, given that A has occurred.
№35 слайд
Содержание слайда: Probability
Outcome Probabilities
To understand, let’s analyze the probability of the outcome (A, A, B).
№36 слайд
Содержание слайда: Probability
Solving Problems Involving Probability---Cont.
Step 4: Compute Event Probability
№37 слайд
Содержание слайда: Probability
Summary
To solve problems involving probability, that is, “what is the probability that … ?”
Perform the following four steps:
Find the sample space
Define event of interest
Compute outcome probabilities
Compute event probability
№38 слайд
Содержание слайда: Probability
Uniform Sample Space
Strange Dice
If we picked dices (a) and (b), rolled them once, what is the probability that (a) beats (b) (has a higher value)?
№39 слайд
Содержание слайда: Probability
Applying Four-Step Method
When the probability of every outcome is the same, we say such a sample space is uniform
№40 слайд
Содержание слайда: Probability
Applying Four-Step Method
Example--- Cont.
So what is the probability that (a) beats (b)?
Which in this case =
(a) Beats (b) more than half of the time.
№41 слайд
Содержание слайда: Probability
Applying Four-Step Method
Example--- Cont.
What about the following:
(a) vs. (c)
(b) vs. (c)
Homework!
№42 слайд
Содержание слайда: Set Theory and Probability
Sample Space S : A nonempty countable set.
An element is called an outcome.
A subset of S is called an event to which a probability is assigned.
If you look closely, you will realize that to calculate this probability we first have to count the elements in these sets.
№43 слайд
Содержание слайда: Probability
Counting
Rules of counting the elements in a set
№44 слайд
Содержание слайда: Probability
The Addition Rule
The basic rule underlying the calculation of the number of elements in a union or difference or intersection is the addition rule.
This rule states that the number of elements in a union of mutually disjoint finite sets equals the sum of the number of elements in each of the component sets.
Theorem 9.3.1:
Suppose a finite set A equals the union of k distinct mutually disjoint subsets A1, A2, …., Ak. Then
N(A) = N(A1)+N(A2)+…+ N(Ak)
№45 слайд
Содержание слайда: Probability
The Addition Rule---Cont.
Example: A computer access password consists of from one to three letters chosen from the 26 in the alphabet with repetitions allowed. How many different passwords are possible?
Solution: The set of all passwords can be partitioned into subsets consisting of those of length 1, those of length 2, and those of length 3 as shown in the figure below.
№46 слайд
Содержание слайда: Probability
The Addition Rule---Cont.
By the addition rule, the total number of passwords equals
the number of passwords of length 1, plus the number of
passwords of length 2, plus the number of passwords of
length 3.
Now the,
Number of passwords of length 1= 26
Number of passwords of length 2 =262
№47 слайд
Содержание слайда: Probability
The Addition Rule---Cont.
Number of passwords of length 3 =263
Hence the total number of passwords= 261+262+263=18,278
№48 слайд
Содержание слайда: Probability
The Difference Rule
An important consequence of the addition rule is the fact that if the number of elements in a set A and the number in a subset B of A are both known, then the number of elements that are in A and not in B can be computed.
Theorem 9.3.2: The Difference Rule:
If A is finite set and B is a subset of A, then
N(A-B) = N(A) – N(B)
№49 слайд
Содержание слайда: Probability
The Difference Rule---Cont.
The difference rule is illustrated below.
№50 слайд
Содержание слайда: Probability
The Difference Rule---Cont.
The difference rule holds for the following reason: If B is a subset of A, then the two sets B and A – B have no elements in common and B (A – B) = A. Hence, by the addition rule,
N(B) + N(A – B) = N(A).
Subtracting N(B) from both sides gives the equation
N(A – B) = N(A) – N(B).
№51 слайд
Содержание слайда: Probability
The Difference Rule---Cont.
Example:
A typical PIN (personal identification number) is a sequence of any four symbols chosen from the 26 letters in the alphabet and the ten digits, with repetition allowed.
a. How many PINs contain repeated symbols?
b. If all PINs are equally likely, what is the probability that a randomly chosen PIN contains a repeated symbol?
№52 слайд
Содержание слайда: Probability
The Difference Rule---Cont.
a. How many PINs contain repeated symbols?
Let’s use the board to intuitively explain why the Difference Rule will work here!
№53 слайд
Содержание слайда: Probability
The Difference Rule---Cont.
Example --- Cont.:
There are 364 = 1,679,616 PINs when repetition is allowed,
and there are 36 35 34 33 = 1,413,720
PINs when repetition is not allowed.
№54 слайд
Содержание слайда: Probability
The Difference Rule---Cont.
Example --- Cont.:
There are 364 = 1,679,616 PINs when repetition is allowed,
and there are 36 35 34 33 = 1,413,720
PINs when repetition is not allowed.
Thus, by the difference rule, there are
1,679,616 – 1,413,720 = 265,896
PINs that contain at least one repeated symbol.
№55 слайд
Содержание слайда: Probability
The Difference Rule---Cont.
b. If all PINs are equally likely, what is the probability that a randomly chosen PIN contains a repeated symbol?
So, how would you figure this out?
№56 слайд
Содержание слайда: Probability
The Difference Rule---Cont.
Example --- Cont.:
There are 1,679,616 PINs in all, and by part (a) 265,896 of these contain at least one repeated symbol.
Thus, by the equally likely probability formula, the probability that a randomly chosen PIN contains a repeated
symbol is
№57 слайд
Содержание слайда: Probability
The Difference Rule---Cont.
An alternative solution to Example 3(b) is based on the observation that if S is the set of all PINs and A is the set of all PINs with no repeated symbol, then S – A is the set of all PINs with at least one repeated symbol.
№58 слайд
Содержание слайда: Probability
The Difference Rule---Cont.
An alternative solution to Example 3(b) is based on the observation that if S is the set of all PINs and A is the set of all PINs with no repeated symbol, then S – A is the set of all PINs with at least one repeated symbol.
It follows that
№59 слайд
Содержание слайда: Probability
The Difference Rule---Cont.
An alternative solution to Example 3(b) is based on the observation that if S is the set of all PINs and A is the set of all PINs with no repeated symbol, then S – A is the set of all PINs with at least one repeated symbol.
It follows that
№60 слайд
Содержание слайда: Probability
The Difference Rule---Cont.
An alternative solution to Example 3(b) is based on the observation that if S is the set of all PINs and A is the set of all PINs with no repeated symbol, then S – A is the set of all PINs with at least one repeated symbol.
It follows that
№61 слайд
Содержание слайда: Probability
The Difference Rule---Cont.
An alternative solution to Example 3(b) is based on the observation that if S is the set of all PINs and A is the set of all PINs with no repeated symbol, then S – A is the set of all PINs with at least one repeated symbol.
It follows that
№62 слайд
Содержание слайда: Probability
The Difference Rule---Cont.
We know that the probability that a PIN chosen at random contains no repeated symbol is
№63 слайд
Содержание слайда: Probability
The Difference Rule---Cont.
We know that the probability that a PIN chosen at random contains no repeated symbol is
And hence
№64 слайд
Содержание слайда: Probability
The Difference Rule---Cont.
This solution illustrates a more general property of probabilities: that the probability of the complement of an event is obtained by subtracting the probability of the event from the number 1.
Formula for the Probability of the Complement of an event!
If S is a finite sample space and A is an event in S, then
P(Ac) = 1- P(A).
№65 слайд
Содержание слайда: Probability
The Inclusion/Exclusion Rule
The addition rule says how many elements are in a union of sets if the sets are mutually disjoint. Now consider the question of how to determine the number of elements in a union of sets when some of the sets overlap.
For simplicity, begin by looking at a union of two sets A and B, as shown below.
№66 слайд
Содержание слайда: Probability
The Inclusion/Exclusion Rule--- Cont.
To get an accurate count of the elements in , it is necessary to subtract the number of elements that are in both A and B. Because these are the elements in .
№67 слайд
Содержание слайда: Probability
Counting Rules in terms of Probabilities
If {E0, E1, ….} is collection of disjoint events, then
№68 слайд
Содержание слайда: Probability
Counting Rules in terms of Probabilities---Cont.
Complement Rule:
)
№69 слайд
Содержание слайда: Probability
Counting Rules in terms of Probabilities---Cont.
)
)
№70 слайд
Содержание слайда: Further Counting
Counting Subsets of a Set: Combinations:
Look at these examples:
In how many ways, can I select 5 books from my collection of 100 to take on vacation?
How many different ways 13-card Bridge hands can be dealt from a 52-card deck?
In how many ways, can I select 5 toppings for my pizza if there are 14 available?
What is common in all these questions?
№71 слайд
Содержание слайда: Further Counting
Counting Subsets of a Set: Combinations:
Look at these examples:
In how many ways, can I select 5 books from my collection of 100 to take on vacation?
How many different ways 13-card Bridge hands can be dealt from a 52-card deck?
In how many ways, can I select 5 toppings for my pizza if there are 14 available?
What is common in all these questions?
Each is trying to find “how many k-element subsets of an n-element set are there?”
№72 слайд
Содержание слайда: Counting Subsets of a Set: Combinations---Cont.
Is read as “n choose k”
№73 слайд
Содержание слайда: Why Count Subsets of Set?
Example:
Suppose we select 5 cards at random from a deck of 52 cards.
What is the probability that we will end up having a full house?
Doing this using the possibility tree will take some effort.
№74 слайд
Содержание слайда: Counting Subsets of a Set: Combinations---Cont.
How to calculate “n choose k”??
Permutations
Division rule
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