Download Probability

What is Probability?
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Probability
Peter Lo
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CS218 © Peter Lo 2004
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Sample Spaces and Events
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CS218 © Peter Lo 2004
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Impossible
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Simple Event
Sample Space
u Complete Collection of Outcomes (the set S)
Simple Event (Sample Point)
u Outcome With 1 Characteristic; a particular outcome,
i.e, an element in S.
Joint Event
u 2 Events Occurring Simultaneously
Compound Event
u One or Another Event Occurring
Impossible Event
u The empty set ∅ in S.
CS218 © Peter Lo 2004
Probability is the numerical measure of
likelihood that the event will occur.
Certain
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u Simple Event
u Joint Event
u Compound Event
Lies between 0 & 1
.5
Sum of events is 1
A.
B.
C.
D.
E.
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Female
Under age 20, < 20
Has 3 credit cards
Red card from a deck of bridge cards
Ace card from a deck of bridge cards
CS218 © Peter Lo 2004
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Joint Event
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Compound Event
A and B, (A ∩ B):
u Female, Under age 20
D and E, (D ∩ E):
u Red, ace card from bridge deck
CS218 © Peter Lo 2004
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Example
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u
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Event that an odd number occurs, B = {1, 3, 5}
Event that a prime number occurs, C = {2, 3, 5}
Event that an even number or a prime number occurs,
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Let S be a sample space, let E be the class of events, and
let P be a real-value function defined on E. Then P is called
a Probability Function, and P(A) is called the probability
of the event A if the following hold:
u
u
A ∪ C = {2, 3, 4, 5, 6}
Event that an odd prime number occurs, B ∩ C = {3, 5}
Event that a prime number does not occurs, C’={1, 4, 6}
CS218 © Peter Lo 2004
CS218 © Peter Lo 2004
Theorems of Probability
The sample space S consist of the 6 possible numbers, S =
{1, 2, 3, 4, 5, 6}
u Event that an even number occurs, A = {2, 4, 6}
u
D or E, (D ∪ E):
u Ace or Red card from bridge deck
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u
u
For every event A, 0 ≤ P(A) ≤ 1
P(S) = 1
If A and B are mutually exclusive events, then
P(A ∪ B) = P(A) + P(B)
If A1, A2, … is a sequence of mutually exclusive events,
then P(A1 ∪A2 ∪ … ) = P(A1) + P(A2) + …
CS218 © Peter Lo 2004
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Theorems of Probability
Type of Sample Spaces
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For two events A and B,
u P (∅) = 0
u P(A’) = 1 – P(A)
u If A ⊆ B, then P(A) ≤ P(B)
u P(A\B) = P(A) – P(A∩B)
u P(A∪B) = P(A) + P(B) – P(A∩B)
CS218 © Peter Lo 2004
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Finite Probability Spaces
u A finite probability space is one in which the
set S contains a finite number of elements.
Finite Equiprobable Spaces
u A probability space is equiprobable if it is finite
and the probability mass are all equal.
CS218 © Peter Lo 2004
Example
Example
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Three horses A, B, and C are in a race; A is twice
as likely to win as B and B is twice as likely to
win as C. What are their respective probabilities of
winning, i.e. P(A), P(B) and P(C)?
CS218 © Peter Lo 2004
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Let a card be selected at random from an ordinary
pack of 52 cards. Let A= {the card is a spade} and
B = {the card is a face card}. What is P(A), P(B)
and P(A∩B)?
CS218 © Peter Lo 2004
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Probability of Simple Event
Conditional Probability
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P(Event) = X / T
u X = number of outcomes in which the event
occurs
u T = total number of possible events
P(B | A) =
n
100 Parts
Inspected, 2
Defects!
CS218 © Peter Lo 2004
The conditional probability that B occurs, given
that A occurred, is defined by
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Example
P( A ∩ B)
P( A)
Chain Rule for Conditional Probabilities
CS218 © Peter Lo 2004
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Multiplication Theorem
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For any events, A1, A2, … An,
P(A 1∩A2∩ … ∩An)
= P(A1)P(A2|A1)P(A3 |A1∩A2)… P(A n|A1∩A2∩ … ∩An-1)
CS218 © Peter Lo 2004
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CS218 © Peter Lo 2004
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Joint Probability Using
Contingency Table
Example
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A lot contains 12 items of which 4 are defective.
Three items are drawn at random from the lot one
after the other. Find the probability p that all three
are not defective.
Event
Event
B1
P(A 1 ∩ B1) P(A 1 ∩ B2) P(A 1)
A2
P(A 2 ∩ B1) P(A 2 ∩ B2) P(A 2)
Joint Probability
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Find the probability for a joint event that a female, and age
under 20 is drawn.
<20
Simple
Event
>20
Total
Female
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63
Male
45
22
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Total
92
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S = {F,<20; F,>20;
M,<20; M,>20}
CS218 © Peter Lo 2004
P(B 1)
P(B 2)
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Marginal (Simple) Probability
CS218 © Peter Lo 2004
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Stochastic Process & Tree
Diagrams
Example
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Total
A1
Total
CS218 © Peter Lo 2004
B2
Count
Total
Sample Space
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A (Finite) sequence of experiments in which each
experiment has a finite number of outcomes was
given probabilities is called a (finite) Stochastic
Process.
A convenient way of describing such a process
and computing the probability of any event is by a
Tree Diagram.
CS218 © Peter Lo 2004
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Example
Independence
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Find the probability of selecting 2 pens from 20
pens which 14 in blue and 6 in red and not replace.
P(R) = 6/20
P(B) = 14/20
CS218 © Peter Lo 2004
P(R|R) = 5/19
R
P(B|R) = 14/19
P(R|B) = 6/19
B
R
P(B|B) = 13/19
B
R
B
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Example
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An event B is said to be independent of an event A
if the probability that B occurs is not influenced
by whether A has or has not occurred.
u Event A and B are independent if
t P(A∩B) = P(A)P(B);
u Otherwise they are dependent.
CS218 © Peter Lo 2004
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Answer
Let a fair coin be tossed three times; we obtain the
equiprobable space S= {HHH, HHT, HTH, HTT,
THH, THT, TTH, TTT}. Consider the events A =
{first toss is heads}, B = {second toss is heads}, C
= {exactly two heads are tossed in a row}. Show
that A and B and A and C are independent events,
and that B and C are dependent events.
CS218 © Peter Lo 2004
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CS218 © Peter Lo 2004
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Repeated Trials
Example
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Let S be a finite probability space. By nIndependent or Repeated Trials, we mean the
probability space T consisting of ordered n-tuples
of elements of S with the probability of an n-tuple
defined to be the product of the probabilities of its
components:
u
P(S1, S2 , … , Sn) = P(S1)P(S2) … P( Sn)
CS218 © Peter Lo 2004
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Random Variables
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Whenever three horses a, b, and c race together, their
respective probabilities of winning are 1/2, 1/3 and 1/6. In
other words, S = {a, b, c} with P(a) = 1/2, P(b) = 1/3 and
P(c) = 1/6. If the horses race twice, what is the probability
of horse a winning the first race and horse c winning the
second race?
Random Variables
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Distribution
A Random Variable X is a rule that assigns a numerical
value to each outcome in a sample space S.
Experiment
CS218 © Peter Lo 2004
Event: Toss 2 Coins. Count No. of Tails.
Probability Distribution
Values, Xi
Probabilities, P(Xi)
Possible Values
Make 100 Sales Calls
No. of Sales
0, 1, 2, ... , 100
0
1/4 = .25
Inspect 70 Radios
No. of Defective
0, 1, 2, ... , 70
1
2/4 = .50
2
1/4 = .25
Answer 33 Questions
No. of Correct
0, 1, 2, ... , 33
Count Cars at Toll
between 11:00 & 1:00
No. of Car Arriving
0, 1, 2, ... , n
CS218 © Peter Lo 2004
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CS218 © Peter Lo 2004
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Expectation
Example
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Expected Value
u Mean of Probability Distribution
u Weighted Average of All Possible Values
u µ = E(X) = ΣXi P(Xi )
Variance
u Weighted Average Squared Deviation about
Mean
u σ2 = E[ (Xi − µ)2 ] = Σ (Xi − µ)2 P(Xi )
CS218 © Peter Lo 2004
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Answer
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In a game a player is paid $5 if he gets all heads or
all tails when three coins are tossed, and he will
pay out $3 if either one or two heads show. What
is his expected gain (or loss) per game?
CS218 © Peter Lo 2004
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Example
Three coins are tossed, the sample space
u S = {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}.
Assume unbiased coins, so any one of these eight
outcomes have equal chance of, say,
u P(HHT) = P(H)P(H)P(T) = (1/2)*(1/2)*(1/2) = 1/8.
Let E1 = {HHH, TTT}, and E2 = {the other 6 outcomes}.
u P{E1} = 1/8 + 1/8 = 1/4, and P{E2} = 1- 1/4 = 3/4.
It follows that the expected value is
u µ = $5*(1/4) - $3*(3/4) = -$1.
So the player is expected to lose one dollar per game.
CS218 © Peter Lo 2004
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CS218 © Peter Lo 2004
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Answer
Example
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CS218 © Peter Lo 2004
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Answer
CS218 © Peter Lo 2004
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Binomial Distribution Properties
f ( x) =
n
C x4C 33−x
C 37
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CS218 © Peter Lo 2004
A lot containing 7 components is sampled by a
quality inspector; the lot contains 4 good
components and 3 defective ones. A sample of 3 is
taken by the inspector. Find the expected value of
the number of good components in this sample.
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Two Different Sampling Methods
u Infinite Population Without Replacement
u Finite Population With Replacement
Sequence of n Identical Trials
Each Trial has 2 Outcomes
u ‘Success’(Desired Outcome) or ‘Failure ’
Constant Trial Probability
Trials are Independent
CS218 © Peter Lo 2004
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Binomial Probability Distribution
Function n
Probability of x successes
Pick x from n
( k)
P( X = k | n, p ) =
P(X=
n =
p =
k =
n!
k !( n − k )!
n
p k (1 − p )n − k
A fair die is tossed 180 times. Find the expected
number of sixes, and the standard deviation.
Probability
of (n-k)
failures
k | n,p) = probability that X = k given n & p
sample size
probability of ‘success’
number of ‘successes’in sample
(X = 0, 1, 2, ..., n)
CS218 © Peter Lo 2004
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Example
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Bayes’Theorem Formula
Toss 1 Coin 4 times in a row. What’s the
Probability of 3 tails?
P( X = x | n, p ) =
P( X = 3 | 4,.5 ) =
CS218 © Peter Lo 2004
CS218 © Peter Lo 2004
n!
x !(n − x )!
4!
3 !( 4 − 3 )!
= .25
P(Bi | A) =
p x (1 − p ) n − x
=
.5 3 (1−.5 )4 − 3
P(A | Bi ) ⋅ P(Bi )
P(A | B1 ) ⋅ P(B1) + L + P(A | Bk ) ⋅ P(Bk )
P(Bi ∩ A)
P(A)
All Bi ’s are the
same event
(e.g. B 2)!
Same
Event
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CS218 © Peter Lo 2004
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Example
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A box contains 12 items of which 4 are in red color. Three
items are drawn at random from the box one after the other.
Find the probability that all three are not in red color.
Probability = (8/12)(7/11)(6/10) = 14/55
CS218 © Peter Lo 2004
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