probabilities
DESCRIPTION
Probability Theory and DistributionTRANSCRIPT
Probabilities(What are your chances?)
Presented by:
Sheryl B. Satorre([email protected])
---TM 705 – Operations Research
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Terminology:
• Experiment • A systematic investigation where the answer is
unknown• Trial• One specific instance of an experiment
• Outcome• Result of a single trial
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Terminology:
• Event• A selected outcome
• Sample Space• Set of all possible outcomes of an experiement
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Games of Chance Games of chance commonly involve the toss of a coin, the roll of a die, or the
use of a pack of cards. Examples: • Rolling a six-sided die will generate
• If two dice are rolled, the following will be the Sample Space{1,2,3,4,5,6}S
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Probability Concepts• Probability is the study of random events.• The probability, or chance, that an event will happen can be
described by a number between 0 and 1:
• A probability of 0, or 0%, means the event has no chance of happening.
• A probability of 1/2 , or 50%, means the event is just as likely to happen as not to happen.
• A probability of 1, or 100%, means the event is certain to happen.
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You can represent the probability of an event by marking it on a number line like this one
Impossible 0 = 0%
Certain 1 = 100%
50 – 50 Chance½ , .5, 50%
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General FormulaLet
S be a Sample SpaceE be an EventP(E) be the Probability of event E
P(E) = the number of ways event E will occur (/E/) Total number of possible outcomes (/S/)
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Any other probability is between 0 and 1.In symbol:
0 <= P(A) <= 1 for any event Aand that
P( f ) + P( s ) = 1 for any eventSo that,
P ( f ) = 1 – P( s ) f means failureP ( s ) = 1 – P( f ) s means success
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Probability ValuesA fair die will have each of the six outcomes equally
likely.
1(1) (2) (3) (4) (5) (6)
6P P P P P P
An example of a biased die would be one of which
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If two dice are thrown and each of the 36 outcomes is equally likely ( as will be the case two fair dice that are shaken properly), the probability value of each outcome will necessarily be 1/36.
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Basic Rules of Probability
1. If A is any event, then 0 <= P(A) <= 1.2. The probability of a null space P(Ø) = 0 while
the probability of the sample space P(S) = 1.3. If an experiment can result in any one of N
different equally likely outcomes and if exactly n of these outcomes correspond to event A, then the probability of event A is P(A) = n/N.
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EVENTS AND COMPLEMENTS 12
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Events
An event A is a subset of the sample space S. It collects outcomes of particular interest. The probability of an event is obtained by
summing the probabilities of the outcomes contained within the event A.
An event is said to occur if one of the outcomes contained within the event occurs.
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Complement
The event A’ , the complement of event A, is the event consisting of everything in the sample space S that is not contained within the event A. In all cases
• Events that consist of an individual outcome are sometimes referred to as elementary events or simple events
( ) ( ) 1P A P A
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Example: ComplementSuppose that two dice are thrown, what is the probability that at least one of the two dice is 6?
11( )
3611 25
( ) 136 36
P B
P B
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Properties of Events
•Mutually Exclusive Events• Non-Mutually Exclusive Events• Independent Events• Dependent Events
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Mutually Exclusive Events
Let A and B be Events.
Events A and B are said to be mutually exclusive if and only if they cannot happen at the same time.
Examples:• Movement: Turning left and turning right• Tossing a coin: Heads and Tails • Drawing a card: King and Queen 17
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Non-Mutually Exclusive Events
Let A and B be Events.
Events A and B are said to be non-mutually exclusive if and only if they can happen at the same time.
Examples:• Movement: Turning left and turning right• Drawing a Card: King and Heart
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Independent Events
Let A and B be events.
If event A cannot influence the outcome of event B, then two events are said to be independent.
Example: Flipping a coin 5 times• The chance that HEAD will occur does not affect the
occurrence of the TAIL from one trial to another.
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Dependent Events
Let A and B be events.
If event A can influence the occurrence of event B, then these two events are said to be dependent.
Example: Drawing 5 balls from the box• The chance of the next ball to be drawn is dependent on
the outcome of the previous draw.
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THE ADDITIVE RULE 21
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Mutually Exclusive Events
Let A and B be Mutually Exclusive events.
P (either A or B) = P (A) + P (B)
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Example: Mutually Exclusive Events
(1,1) (1,2) (1,3) (1,4) (1,5) (1,6)
(2,1) (2,2) (2,3) (2,4) (2,5) (2,6)
(3,1) (3,2) (3,3) (3,4) (3,5) (3,6)
(4,1) (4,2) (4,3) (4,4) (4,5) (4,6)
(5,1) (5,2) (5,3) (5,4) (5,5) (5,6)
(6,1) (6,2) (6,3) (6,4) (6,5) (6,6)
Let P(A) and P(B) be the probabilities of getting the sum of 6, and the sum of 8, respectively.
P(A) = 5/36P(B) = 5/36
P(A or B) = P(A) + P(B)P(A or B) = (5/36) + (5/36)P(A or B) = 5/18
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Non-Mutually Exclusive Events
Let A and B be Non-Mutually Exclusive events.
P (either A or B) = P (A) + P (B) – P(A and B)
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Example: Non-Mutually Exclusive EventsIf a card is drawn from an ordinary well-shuffled deck of cards, what is the probability that it is either a spade or a face card?
Let A be an event of drawing a SPADELet B be an event of drawing a FACE cardLet P(A or B) be the probability of getting a Spade or a Face card
P(A or B) = P(A) + P(B) – P(A and B)P(A or B) = (13/52) + (12/52) – (3/52)
P(A or B) = 11/26 25
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MULTIPLICATION OF EVENTS 26
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Independent Events
If two events A and B are independent, that is, P(B/A) = P(B) or P(A/B) = P(A) then the probability that both A and B occurs is
P(A and B) = P(A) x P(B)
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Example: Independent EventsIf a coin is tossed two times in succession, what is the probability that a head appears in the first toss and a tail in the second?
P(HT) = P(H) * P(T) 1
st toss 2
nd toss
P(HT) = (1/2) * (1/2)P(HT) = 1/4
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Dependent Events
Let A and B be dependent events.If events A and B can both occur then the probability that both A and B occur is
P(A and B) = P(A) x P(B/A)
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Example: Dependent EventsA box contains one dozen Casio calculators, 3 of which are defective and the rest are good. What is the probability that if 3 calculators are drawn at random without replacement, they will be defective?
Let D = 3 total defective Casio calculatorsLet D’ = 9 total non-defective Casio calculatorsLet P(3D) be the chance that the 3 calculators drawn are all defectiveP(3D) = P(1D) * P(2D/1D) * P(3D/2D)P(3D) = (3/12) * (2/12) * (1/12)
P(3D) = 1/288 30
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CONDITIONAL PROBABILITY 31
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Basic Concepts
Conditional probability refers to the probability of an event B occurring given the condition that some other event A has occurred.
This is denoted by the symbol P(B/A) which is usually read as “the probability that B occurs given that A occurs” or simply, “the probability of B, given A”.
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Conditional Probability The conditional probability of event A conditional on event B
is
for P(B)>0. It measures the probability that event A occurs
when it is known that event B occurs.
( )( | )
( )
P A BP A B
P B
( ) 0( | ) 0
( ) ( )
( ) ( )( | ) 1
( ) ( )
A B
P A BP A B
P B P B
B A
A B B
P A B P BP A B
P B P B
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Example: Conditional ProbabilityThe probability that the lady of the house is home when the Avon representative calls is 0.60. Given that the lady of the house is home, the probability that she makes a purchase is 0.40. Find the probability that the lady of the house is home and makes a purchase when the Avon representative calls.
Given:P(H) = 0.60 ; the chance that the lady of the house is home when Avon rep. callsP(P|H) = 0.40 ; the chance that the lady of the house makes a purchase given that she is home 34
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Given:P(H) = 0.60 ; the chance that the lady of the house is home when Avon rep. callsP(P|H) = 0.40 ; the chance that the lady of the house makes a purchase given that she is home
Required:P(H and P) = the chance that the lady of the house is home and makes purchase
P(P|H) = P(H and P) P(H)
0.40 = P(H and P) 0.60
P(H and P) = 0.2435
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Probability Tree
• a.k.a Tree Diagram• Is a decision-tool that helps visualize all
possible choices with their corresponding chances of outcomes in order to avoid mathematical errors
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Probability Tree: Car Warranties
A company sells a certain type of car, which it assembles in one of four possible locations. Plant I supplies 20%; plant II, 24%; plant III, 25%; and plant IV, 31%.
A customer buying a car does not know where the car has been assembled, and so the probabilities of a purchased car being from each of the four plants can be thought of as being 0.20, 0.24, 0.25, and 0.31.
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Probability Tree: Car Warranties (cont.)Each new car sold carries a 1-year bumper-to-bumper
warranty. P( claim | plant I ) = 0.05, P( claim | plant II ) = 0.11 P( claim | plant III ) = 0.03, P( claim | plant IV ) = 0.08For example, a car assembled in plant I has a probability of
0.05 of receiving a claim on its warranty. Notice that claims are clearly not independent of
assembly location because these four conditional probabilities are unequal
Problem: If a car was purchased from this company, what is the chance that it’s warranty will be claimed? 38
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P( claim ) = P( plant I, claim ) + P( plant II, claim ) + P( plant III, claim) + P( plant IV, claim )
P(claim) = 0.0687
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Probability Tree: Freeway
In a certain freeway, your chance of getting off at Exit I is 0.60. If you get off at Exit I, your chance of getting lost is 0.30. If you miss Exit I, and have to get off at Exit II, your chance of getting lost is 0.70. What is the probability of not getting lost?
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PROBABILITY DISTRIBUTIONS 41
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Basic ConceptsProbability distributions are theoretical models that describe the behavior of the population associated with statistical experiments. The concept of probability distributions can best be understood by means of the following experiment.
Let us consider the probability distributions of the number of heads which shows in three tosses of a balanced coin.
No. of Heads Probability
0 1/81 3/82 3/83 1/8 42
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Probability distribution may be defined as a listing of probabilities associated with all possible outcomes that could result if the experiment were done.
A probability distribution involves random variable.
A random variable is a variable whose numerical values are determined by chance.
The representation of the values of a random variable and the probabilities associated with those values is called a probability distribution
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Random Variable• Discrete Random Variable• This is a random variable that takes on a finite number of possible
values or an unending sequence with as many values as there are whole numbers. Examples include the number of heads or tails that occur when a coin is tossed a specific number of times and the number of patients coming in to a particular clinic in a given day.
• Continuous Random Variable• This is a random variable that takes on an infinite number of
possible values equal to the number of points on a line segment. Examples include measured quantities such as weight, height, length and distance.
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Types of Probability Distribution
1. Binomial2. Poisson3. Hypergeometric4. Trinomial
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Binomial Probability Distribution
The binomial distribution is a distribution that describes discrete data that results from an experiment called the Bernoulli Process. The Bernoulli Process is a sampling process that conforms to the following characteristics:
1. Each trial has only two possible outcomes. (Ex: head or tail, yes or no, success or failure)
2. The probability of success on any trial remains fixed or constant from trial to trial.
3. Each trial is statistically independent, that is, the outcome of one trial could not affect the outcomes of the other trials.
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Binomial FormulaThe formula for determining the probability of a designated number of X success in n independent trials in a binomial distribution is:
P(x) = C(n,x) px (1-p)n-x
Where:
x = 0, 1, 2, 3, …, np = constant probability of success for each trialC(n,x) = combination of n trials taken r at a time1 – p = probability of failuren = number of trials
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Poisson DistributionThe Poisson Distribution was named after the mathematician and physicist Simeon Poisson (1781-1840). The Poisson distribution can be used to determine the probability of a designated number of successes when the events occur continuously. It is similar to the Bernoulli process except that the events occurs continuously and not on fixed trials or observations. For instance, the number of cars arriving at a gasoline station for refill in a given 10-minute period and the number of commuters arriving at a bus waiting station within a given minute can be assumed to be Poisson variables.
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Poisson Distribution (cont.)
A Poisson experiment gives the number of outcomes occurring during a time interval independent of the number of outcomes. In determining the probability of a designated number of successes x in a Poisson process, the only value needed is the average number of successes for the specific time or space which is represented by λ(Greek letter lambda).
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The Poisson Formula:
P(x,λ) = λx * e –λx!
Where:e = 2.7183 (base of natural logarithm)λ = (Greek letter lambda) is the average or mean of the distributionx = 0, 1, 2, 3,…,n
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When the number of trials n is large and p is small, binomial probabilities are approximated by means of the Poisson distribution changing λ into np (λ = np).
P(x) = (np)x e –np
x! Where:
n = total number of trials
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Hypergeometric Distribution
The Hypergeometric Distribution is the appropriate distribution when sampling without replacement is used in a situation that could qualify as a Bernoulli process. The probability distribution of the hypergeometric variable x is the number of successes in a random sample size n selected from N items of which k are classified as successes, and N-k as failures. Its variables are denoted by P(x; N, n, k).
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In a hypergeometric distribution, the formula for determining the probability of a designated number of successes x is:
P(x;N,n,k) = C(k,x) C(N-k,n-x) C(N,n)
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Trinomial Distribution
A trinomial distribution is one in which: • A certain basic action is performed with n trials or
repetitions.• Each repetition involves three possible outcomes. A, B,
and C.• Each of the three outcomes has a fixed probability. • p = P(A) q = P(B) r = P(C)• The individual trials are independent.
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The trinomial formula:
P(x,y,z) = [n!/(x!y!z!)] [pxqyrz] Where: x+y+z = n and p+q+r = 1 The term, n!/(x!y!z!) is called a trinomial coefficient.
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Thank you.
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