Transcript Counting Sample Points

PROBABILITY
and
IMPORTANCE
to RISK
MANAGEMENT
What is PROBABILITY?
It is a quantitative
measure of uncertainty.
ETYMOLOGY
The word PROBABILTY
derives from PROBITY
HISTORY
OF
PROBABILITY
 MIDDLE OF 17 CENTURY
TH
RICHARD JEFFRY
term PROBABLE-means APPROVABLE
1774
PIERRE-SIMON LAPLACE
Made the first attempt to deduce a rule for the
combinations of observations.
INTERPRETATIONS
 FREQUENTISTS
-dealing with experiments that are random and welldefined.
BAYESIANS
-individual’s degree of belief
APPLICATIONS OF PROBABILITY
RISK ASSESSMENT
TRADE ON COMMODITY
MARKETS
RELATION OF PROBABILITY ON
RISK MANAGEMENT
Probability of Occurrence
0 - 10%orVery unlikely to occur
11 - 40%orUnlikely to occur
41 - 60%orMay occur about half of the
time
61 - 90%orLikely to occur
91 - 100%orVery likely to occur
Definition of Terms
Experiment
Is any action or process that generates observations or
set of data
Random Experiment
Is an experiment whose outcome cannot be predicted
with certainty,but the experiment is such a nature that
the collection of every possible outcomes can be
described prior to its performance.
Sample Space
Is the set of all possible outcomes of a statistical
experiment.It is collections of all conceivable outcome
of an experiment.
Element
Each outcome in sample space also called as sample
point.
Event
Is any collection of outcomes contained in the sample
space.
Simple Event
It consist exactly one outcome.
Compound Event
It consist more than one outcome.
Complement
The complement of an event A, A’ ,with respect to the
sample space,S,is the subset of all elements of S that
are not in A.
Example:
Let S={1,2,3,4,5,6,7,8,9,0} and let R={2,4,6,8,0}
R’=(1,3,4,7,9)
Intersection
The intersection of two events A and B,AΩB,is the event
containing all elements that are common to both A
and B.
Example:
Let S={1,2,3,4,5,6,7,8,9,0} and let R={2,4,6,8,0}
AΩB= (2,4,6,8,0)
Mutually Exclusive or Disjoint
Two events, A and B are mutually exclusive or disjoint if
A ΩB =ø,that is,if A and B have no elements in
common.
Example:
Let S={1,3,5,7,9} and let R={2,4,6,8,0}
A ΩB =ø
Union
The union of the two events A and B,A U B, is the event
containing all the elements that belong to A or B or
both .
Example:
Let S={1,2,3,4,5,6,7,8,9,0} and let R={2,4,6,8,0}
A U B=(1,2,3,4,5,6,7,8,9,0)
Counting Sample Points
Techniques in counting sample points
Multiplication rule
Generalized Multiplication Rule
Permutation
Circular Permutation
Distinct Permutation
Combination
Multiplication Rule
Is the fundamental principle of counting
Theorem #1
If an operation in n1 ways,and if for each of these
operation can be performed in n 2 ways,then the two
operations can be performed in n1 n 2 ways.
Example:
How many sample points are in the sample space when a
fair coin is tossed twice?
Solution:
n1 n 2 =6(6)ways= 36 possible ways
Generalized Multiplication Rule
It is used when involving operations is more than two.
Theorem #2
If an operation in n1 ways,and if for each of these
operation can be performed in n 2 ways,and for each of
the first two a third operation can be performed in n 3
ways and so forth then the sequence of k operations
can be performed in n1 n 2 n 3 …….ways.
Example:
How many 3 digits can be formed from the digits 1, 2, 5,
6, and 9 if each no. can only used once?
Solution:
n1 n 2 n 3 =5(4)(3)ways =60 possible ways
Permutation
It is the arrangement of all or part of a set of objects in a
definite order.
Order is important aspect of permutation.
Theorem #3
The number of permutation of n distinct objects is n!
Example:
Four people are to be arrange in a row to have their
picture taken. In how many ways can this done?
Solution:
n= 4 people to be arrange in definite order
n!= 4!=4(3)(2)(1)=24 ways
Permutation
Theorem #4
The number of permutation of n distinct objects taken r
at a time is
n Pr = n!
(n-r)!
Example:
Five applicants enter an interview room in which there
are seven seats.In how many ways may they be seated?
Solution:
n= 7;r=5
= 2520 ways
7P5 = 7!
(7-5)!
Circular Permutation
Permutations that occur by arranging objects in a circle
are called circular permutation .
Theorem # 5
The number of permutation of n distinct objects
arranged in a circle is (n-1)!.
Example:
Seven children join hands and form a circle. In how
many ways could the circle be formed?
Solution:
n=7
(n-1)!= (7-1)!= 6! = 720 ways
Distinct Permutation
Theorem # 6
The number of distinct permutation of n things of which n 1 are the same
kind, n 2 of a second kind……., n k of a second kind,…..n k of a kth kind
is
n!
n 1 ! n 2!....... n k
Example:
Eight books are to be arrange on a shelf. There are 2 math identical
books,3 identical English books and 3 identical Physics books. How
many distinct arrangement are possible?
Solution:
n= 8 books
n 1 =2 math books
n 2 =3 english books
n 3 =3 physics books
P= 8! = 560 arrangements
2! 3! 3!
Combination
It is arrangement of number of ways of selecting r objects from n
without regard to order.
Theorem # 7
The number of combination of n distinct objects taken r at a time
is
n!
n C r=
r!(n-1)!
Example:
In how many ways a committee of 3 can be chosen from a group of
8?
Solution:
n= 8 ; r= 3
8!
= 56 different ways
8 C 3=
3!(8-1)!
Probability of an event
P(A) = The Number Of Ways Event A Can Occur
The Total Number Of Possible Outcomes
P(A)= n__
S
Basic Properties of Probability
 Certainty- this property states that the probability
that the event is on the sample space is one.
P(S)=1
 Non-negativity- this property states that every
event has a non-negativity probability. The lowest
probability of an event is 0 or no chance and the
highest probability of an event is 1 or certain.
P(E)≤0, for all E c S
 Union- the probability of the union of mutually
exclusive events is the sum of each event’s
probability.
P (A U B)= P (A) + P (B)
Example 1:
A fair dice is tossed once. What is the probability that
a. ) 5 will appear
b.) an even number will appear?
Solution:
S={1,2,3,4,5,6}= 6
a.) If A is the event that a 5 will appear
A= {5}
n(A)= 1
P(A)= n(A) = 1
S
6
b.) If B is the event that an even number will appear
B= {2,4,6}
n(B)= 3
P(A)= n(A) =3 = 1
S
6 2
Example 2:
Suppose that a card is drawn at random from an ordinary deck of playing
cards,
What is the probability of drawing
a. ) a heart
b.) a red card
Solution:
S=52
P(heart)= n(heart) = 13 = 1
S
52
4
P(red card)= n(red card) = 26 = 1
S
52 2
ADDITIVE RULES
ADDITIVE RULES
• It is applicable to union of events
THEOREM
• If A and B are any two events, then P(AυB) = P(A)+P(B)-P(AΩ
B)
•For three events A, B, and C, P(AυBυC) = P(A)+P(B)+P(C)-P(A
Ω B)-P(A Ω C)-P(B Ω C)+P(AυBυC)
•If A and B are complementary events, then P(A)+P(A’)=1
EXAMPLES:
ADDITIVE RULE
1.
2.
The probability that Mel passes Calculus is 2/3 and the
probability that she passes statistics is 4/9. If the
probability of passing both courses is ¼, what is the
probability that Mel will pass at least one of these
subjects?
If the probabilities are 0.9, 0.15, 0.21, and 0.23 that a
person purchasing a new automobile will choose the color
green, white, red and blue, what is the probability that a
given buyer will purchase a new automobile that comes in
one of those colors? (mutually exclusive)
Solution #1
• Let A be the event that Mel passes Calculus
• Let B the event that Mel passes Statistics
• P(A U B)=P(A) + P(B) -P(A Ω B)
2 + 4 -1
3
9 4
=31
36
CONDITIONAL
PROBABILITY
CONDITIONAL
PROBABILITY
• The probability of an event B occurring when it is known
that some event A has occurred is called conditional
probability and is denoted by P(B/A).
• The conditional probability of A, given B, denoted by P(A/B),
P(A/B) = P(B Ω A)
P(B)
If P(B)>ø
• Two events A and B are independent if and only if P(B/A) =
P(B) and P(A/B) = P(A). Otherwise, A and B are dependent.
EXAMPLES:
CONDITIONAL PROBABILITY
1.
2.
A fair die is tossed once. Find the probability that a 4
appear, when it is known that a number greater than 2
results in the toss of the die.
The probability that a regularly scheduled flight departs
on time is P(A)= 0.83, the probability that it arrives on
time is P(B) = 0.82; and the probability that it departs
and arrive on time is P(A ΩB) = 0.78. Find the probability
that a plane
a) arrives on time given that it departed on time.
b) departed on time given that it has arrived on time.
Solution #1
•
•
•
•
•
•
S=(1,2,3,4,5,6)
Let A be the event that 4 appear in a single toss.
A=(4)
Let B the event that a number is greater that 2 appear.
B=(3,4,5,6)
A ΩB= (4)
•
1
• P(A/B)= 6
•
4
•
6
=1
4
RANDOM VARIABLES
 (r.v) is a function whose value is a real number
determined by each element in a sample space.
DISCRETE SAMPLE SPACE
 If the sample space contains a finite number of
possibilities or a n unending sequence with the
elements as there are whole numbers.
CONTINUOUS SAMPLE SPACE
• If a sample space contains an infinite number
possibilities equal to the number of points on a line
segment.
DISCRETE RANDOM VARIABLE
 If the set of possible outcome is countable and it
refers to the count data.
CONTINUOUS RANDOM
VARIABLE
• if it takes the values of on a continuous scale and it
refers to measured data.
Example:
1. A shipment of 8 similar microcomputers to a retail
outlet contains 3 that are defective. If a school makes
a random purchase of 2 of these computers, find the
probability distribution for the number of defectives.
Solution:
 Let X be a random variable whose values are the possible numbers of







defective computers purchased by the school. Then X can be any of the
numbers 0,1, and 2.
f(0)= P(X=0) = 3 C 0 * 5 C 2 = 5
14
8 C 2
f(1)= P(X=1) = 3 C 1 * 5 C 1 = 15
28
8 C 2
f(2)= P(X=2) = 3 C 2 * 5 C 0 = 3
28
8 C 2
Therefore,the probability distribution of X is
X
0
1
2
P(X=x)
5/14
15/28
3/28
 We can write P(X=x)= f(x)
Mathematical Expectation
Experiment
No Heads
-4
One Head
-7
Two Heads
-5
16
Average no. of Heads =
•
0(4)+1(7)+2(5)
16
Average Value is referred to as the Mean or Expected Value or the Mean of the Random
variable of x or the Mean of the probability of distribution of x.
Mean or Expected Value of X is
µ=E(x)= ∑xf(x)
x
∞
if x is discrete
µ=E(x)=∫ xf(x)dx if x is continuous
-∞
Variance of the Random Variable
Let x =be the random variable
Then the variance of x is
σ 2 = ∑(x-µ) 2 f(x),
x
σ2
if x is discrete
∞
= ∫ (x-µ) 2 f(x)dx, if x is continuous
-∞
Mean of the Function on Random Variable
Let g(x) =be the function of random variable
Then the expected value of x is
E (g(x))= ∑ g(x) f(x),
x
∞
E (g(x))= ∫ g(x) f(x) dx,
-∞
if x is discrete
if x is continuous
Example 1
If the probability mass function of x is given by:
f(x)=
1/3,
for x=1,2,3
0,
elsewhere
Find the mean and variance of x.
Solution:
µ=E(x)= ∑xf(x)
= [1(1/3)+2(2/3)+3(1/3)
µ=2
σ 2 = ∑(x-µ)2 f(x)
=(1-2)2(1/3)+(2-2)2(1/3)+(3-2)2(1/3)
=1/3+0+1/3
σ 2 =2/3
Example 2
Suppose that the probability density function of x is given by:
f(x)=
1/2,
for -1<x<1
0,
elsewhere
Find the mean and variance of x.
Solution:
∞
µ=E(x)= ∫ xf(x)dx
-∞
1
= ∫-1 x(1/2)dx
1
= x2/4]
-1
µ=0
∞
σ 2 = ∫ (x-µ) 2 f(x)dx
-∞
-1
1
∞
-1
1
= ∫ (x-0)2(0)dx + ∫ (x-0)2(1/2)dx + ∫ (x-0)2(0)dx
-∞
= x3/6]
1
-1
=1/6-(-1/6)
σ =1/3
2
Example 3
If the probability mass function of x is given by:
f(x)=
Let g(x)= x2
1/6,
0,
for x=1,2,3
elsewhere
Find the mean of g(x).
Solution:
E (g(x))= ∑ g(x) f(x)
= ∑ x2 (x/6)
=[(1)2(1/6)+(2)2(2/6)+(3)2(3/6)]
=1/6+8/6+27/6
E (g(x))= 6
Example 4
Given the probability density function of x :
f(x)=
2(1-X),
0,
Let g(x)= x2
Find the mean of g(x).
Solution:
∞
E (g(x))= ∫ g(x) f(x) dx
-∞
1
E (x2) = ∫ x 2 [2(1-x)]dx
0
1
= ∫ (2x 2-2x 3)dx
0
=2x 3-2x4
3
4
= 2/3-1/2
E (x2)
=1/6
1
0
for 0<x<1
elsewhere