Transcript 03 probability distributions

Quantitative Methods
Varsha Varde
Quantitative Methods
Quantifying Uncertainty:
Basic Concepts of Probability
Quotes from You and Me
• Chances of your getting a handsome job
should improve if you obtain an MBA.
• Probably, collections will jump this month.
• Most probably, ERP will be on by June.
• Odds are great for my promotion this time.
• Winning cricket match against Australia is
not impossible, but is highly improbable
Defects from new machines are unlikely.
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Uncertainty
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Each Statement Involves Uncertainty.
Chances = Odds = Likelihood = Probability
Real Life is Usually Full of Uncertainty.
Almost Nothing is for Sure.
There are Chances of Something Happening
and Chances of Something Else Happening.
• In Such Situations, You can’t ‘Prove’ Anything.
• All You Can Do is to Assign a Probability to
Each of the Different Possible Outcomes.
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Quotes from You and Me
After This MBA
• Chances of your getting a handsome job
would be 90% if you obtain an MBA.
• I am 75% confident that collections will
jump this month.
• Odds are 80:20 for my promotion this time.
• . Winning cricket match against Australia
is not impossible but has only 10% chance
• New machines churn out good product 97
out of 100 times.
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Probability Theory
• How Do You Say 90% Chances, or 80:20
Odds, or 75% Confidence?
• Probability Theory Provides Tools to
Decision Makers to Quantify Uncertainties.
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Assigning Probabilities
• Classical Approach: Assumes equally likely
outcomes (card games ,dice games, tossing
coins and the like)
• Relative Frequency Approach: Uses relative
frequencies of past occurrences as probabilities
(Decision problems in area of management.
Delay in delivery of product)
• Subjective Approach :Guess based on past
experience or intuition.( At higher level of
managerial decisions for important ,specific and
unique decisions
Assume equally likely outcomes
Use Relative Frequencies
• Making use of relative frequencies
of past.
• Suppose an organisation knows from past
data that about 25 out of 300 employees
entering every year leave due to good
opportunities elsewhere
• then the organisation can predict the
probability of employee turnover for this
reason
• as 25/300=1/12=0.083
Subjective Probability
• Based on personal judgements
• Uses individual’s experience and familiarity
with facts
• An expert analyst of share prices may give
his judgement as follows on price of ACC
shares in next two months
• 20% probability of increase by Rs500or more
• 60% probability of increase by less than
Rs500
• 20%probability of remaining unchanged
Experiment
• Experiment: An experiment is some act, trial
or operation that results in a set of possible
outcomes.
-The roll of two dice to note the sum of spots
-The toss of a coin to see the face that turns
up.
- polling
- inspecting an assembly line
- counting arrivals at emergency room
- following a diet
Event
• Event: An event means any collection of
possible outcomes when an experiment is
performed. For example,
• When an unbiased die is rolled we may
get either spot 1, spot 2, spot 3, spot 4,
spot 5 or spot 6. Appearance of anyone of
the spots is an event.
• Appearance of an even spot is also an
event.
EVENT/OUTCOME
• -The roll of two dice (Appearance of
the sum of spots )
-The toss of a coin( the face that turns
up)
- polling (Win or lose)
- inspecting an assembly line(Number
of defectives)
- counting arrivals at emergency
room(Number of arrivals in one hour)
- following a diet (weight loss or gain)
Sample space
• Sample space: the set of all sample
points (simple events) for an experiment is
called a sample space; or set of all
possible outcomes for an experiment
• Venn diagram :It is a pictorial
representation of the sample space.It is
usually drawn as a rectangular figure
representing the sample space and circles
representing events in the sample space.
Venn Diagram
For Roll of a die
A:Odd spots
B:Even Spots
Equally Likely Events
• Equiprobable or Equally Likely Events:
Events are said to be equiprobable
when one does not occur more often
than the others.
• When an unbiased die is thrown any
one of the six spots may appear.
• When an unbiased coin is tossed either
a head or a tail appears
Exhaustive Events
• Exhaustive Events: Events are said to be
exhaustive when they include all possible
cases or outcomes. For example, in
tossing of fair coin, the two events
“appearance of a head” and “appearance
of a tail” are exhaustive events because
when a coin is tossed we would get either
a head or a tail.
Independent Events
• Independent Events: Two events A and
B are said to be independent if
occurrence of A does not affect and is
not affected by the occurrence of B.
• When a coin is tossed twice the result
of the first toss does not affect and is
not affected by the result of the second
toss. Thus, the result of the first toss
and the result of the second toss are
independent events.
Dependent Events
• Dependent Events: Two events A and B are
called dependent if the occurrence of A
affects or is affected by the occurrence of B.
• For example, there are four kings in a pack of
52 cards. The event of drawing a king at the
first draw and the event of drawing another
king at the second draw when the first drawn
king is not replaced, are two dependent
events. In the first event there are four kings
in a pack of 52 cards and in the second event
there are only three kings left in the pack of
remaining 51 cards
Mutually Exclusive Events
• Events are termed mutually exclusive if they cannot occur together
so that in any one trial of an experiment at most one of the events
would occur.
• Mutually Exclusive Events:
• “throwing even” and “throwing odd” with one die,
• “drawing the spade,” “drawing a diamond” and “drawing a club”
while drawing one card from a deck.
• purchase of a machine out of 3 brands available
• Not mutually exclusive
•
“drawing a spade” and “drawing a queen”
•
“even number” and “at least 3” with one die
• Selection of a candidate with post graduate qualification and over 3
years experience
• A particular easy way to obtain two mutually exclusive events is to
consider an event and its negative(Complement). Such as “even”
and “not even,” “spade”, “not spade” or in general ‘A’ and ‘not A’.
Notation.
• Sample space : S
• Sample point: E1, E2, . . . etc.
• Event: A,B,C,D,E etc. (any capital letter).
• Venn diagram:
Example.
• S = {E1, E2, . . ., E6}.
• That is S = {1, 2, 3, 4, 5, 6}. We may think
of S as representation of possible
outcomes of a throw of a die.
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Venn Diagram
A:Candidates over 3 years experience
B:Candidates with post graduate qualification
S
AB
A
B
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More definitions
• Union, Intersection and Complementation
Given A and B two events in a sample space S.
1. The union of A and B, AUB, is the event containing all
sample points in either A or B or both. Sometimes we
use A or B for union.
2. The intersection of A and B, A∩B, is the event
containing all sample points that are both in A and B.
Sometimes we use AB or A and B for intersection.
3. The complement of A, Ā the event containing all sample
points that are not in A. Sometimes we use not A or Ac
for complement.
Mutually Exclusive Events (Disjoint Events)
4 Two events are said to be mutually exclusive (or disjoint)
if their intersection is empty. (i.e. A ∩ B = ö).
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Example
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Suppose S = {E1, E2, . . ., E6}. Let
A = {E1, E3, E5};
B = {E1, E2, E3}. Then
(i)A U B = {E1, E2, E3, E5}.
(ii) A ∩ B = {E1, E3}.
(iii) Ā = {E2, E4, E6}; Bc ={E4, E5, E6};
(iv) A and B are not mutually exclusive
(why?)
• (v) Give two events in S that are mutually
exclusive.
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Probability of an event
• Relative Frequency Definition If an
experiment is repeated a large number, n,
of times and the event A is observed nA
times, the probability of A is
• P(A) = nA / n
• Interpretation
• n = # of trials of an experiment
• nA = frequency of the event A
• nA/n = relative frequency of A
• P(A) = nA /n , if n is large enough.
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Basic Formula of Probability
• Probability of an Event A:
No. of Outcomes Favourable to Event A
= ---------------------------------------------------Total Number of All Possible Outcomes
• Probability is a Ratio. (A Distribution Ratio)
It varies from 0 to 1.
• Often, It is Expressed in Percentage
Terms Ranging from 0% to 100%.
• It is denoted as P(A) and termed as
marginal or unconditional probability
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Rules of Probability:
Multiplication Rule
• It is for Probability of Simultaneous
Occurrence of Two Events
• If A and B are two independent events,
P(A & B) = P(A) x P(B)
• Example: Experiment: Toss Two Coins
A: Getting Head on Coin No. 1
B: Getting Head on Coin No. 2
P(A)= ½, P(B)= ½, P(A&B)= ¼ =0.25
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Rules of Probability:
General Multiplication Rule
• If A and B are two dependent events,
• P(A & B) = P(A) x P(B|A)
• P(B|A) The conditional probability of the event B
given that event A has occurred
• Example: Draw Two Cards from a Deck
A: First Card a King
B: Second Card also a King
P(A)=4/52=1/13, P(B|A)=3/51
P(A & B)=1/13 x 3/51=3/204=0.015=1.5%
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Rules of Probability:
Addition Rule
• It is for Probability of Occurrence of Either of the
Two Events
• If A and B are two mutually exclusive events,
P(A or B) = P(A) + P(B)
• Example: Experiment: Roll a Die
A: Getting the No. 5
B: Getting the No. 6
P(A)=1/6, P(B)=1/6, P(A or B)=1/3=0.33=33%
• Note: Two Events are Mutually Exclusive if They
Cannot Occur Together
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Rules of Probability:
General Addition Rule
• If A and B are any two events,
P(A or B) = P(A) + P(B) – P(A & B)
• Example: Toss Two Coins
A: Getting Head on Coin No. 1
B: Getting Head on Coin No. 2
P(A)= ½, P(B)= ½, P(A & B)= ¼
So, P(A or B)= ½ + ½ - ¼ = ¾ =0.75=75%
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Exercise
If 80% Company guests visit the HO,
70% visit the Plant, and 60% visit both,
what is the chance that a guest will visit
HO or Plant or both?
What is the probability that he will visit
neither the HO nor the Plant, but meet
Company Executives at the Taj?
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Solution
• P(A)=0.8
P(B)=0.7
P(A&B)=0.6
• Prob that a guest will visit HO or Plant or
both = P(A&B)=0.8 + 0.7 – 0.6=0.9 = 90%
• Prob that he will visit neither the HO nor
the Plant, but meet Company Executives
at the Taj = 1 - Prob that a guest will visit
HO or Plant or both = 1 – 0.9 = 0.1 = 10%
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Conceptual Definition of Probability
• Consider a random experiment whose
sample space is S with sample points E1,
E2, . . . ,.
• For each event Ei of the sample space S
let P(Ei) be the probability of Ei
(i) 0 ≤ P(Ei) ≤ 1 for all i
(ii) P(S) = 1
(iii)∑P(Ei) = 1,where the summation is
over all sample points in S.
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Example
• Definition The probability of any event A is
equal to the sum of the probabilities of the
sample points in A.
• Example. Let S = {E1, . . ., E10}.
• Ei
E1
E2
E3
E4
E5 E6 E7 E8 E9 E10
• P(Ei) 1/20 1/20 1/20 1/20 1/20 1/20 1/5 1/5 1/5 1/10
• Question: Calculate P(A) where A = {Ei, i≥6}.
• P(A) = P(E6) + P(E7) + P(E8) + P(E9) + P(E10)
= 1/20 + 1/5 + 1/5 + 1/5 + 1/10 = 0.75
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Steps in calculating probabilities of events
1. Define the experiment
2. List all simple events
3. Assign probabilities to simple events
4. Determine the simple events that constitute the
given event
5. Add up the simple events’ probabilities to obtain
the probability of the given event
• Example Calculate the probability of observing
one H in a toss of two fair coins.
• Solution.
• S = {HH,HT,TH, TT}
• A = {HT, TH}
• P(A) = 0.5
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Example.
Example. Toss a fair coin 3 times.
• (i) List all the sample points in the sample
space
• Solution: S = {HHH, · · ·TTT} (Complete
this)
• (ii) Find the probability of observing exactly
two heads and at most one head.
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Probability Laws
Complementation law:
P(A) = 1 - P(Ā)
Additive law:
P(A U B) = P(A) + P(B) - P(A ∩ B)
Moreover, if A and B are mutually exclusive,
then P(A∩B) = 0 and
P(A U B) = P(A) + P(B)
Multiplicative law (Product rule)
P(A ∩ B) = P(A|B)P(B)
= P(B|A)P(A)
Moreover, if A and B are independent
P(A∩B) = P(A)P(B)
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Example
• Let S = {E1, E2, . . ., E6}; A = {E1, E3, E5}; B =
{E1, E2, E3}; C = {E2, E4, E6};D ={E6}. Suppose
that all elementary events are equally likely.
• (i) What does it mean that all elementary events
are equally likely?
• (ii) Use the complementation rule to find P(Ac).
• (iii) Find P(A|B) and P(B|A)
• (iv) Find P(D) and P(D|C)
• (v) Are A and B independent? Are C and D
independent?
• (vi) Find P(A∩ B) and P(A UB).
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Law of total probability
• Let A, Ac be complementary events
and let B denote an arbitrary event.
Then
• P(B) = P(B∩ A) + P(B ∩ Ac) ,
or
• P(B) = P(B/A)P(A) + P(B/Ac)P(Ac).
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Bayes’ Law
• Let A,Ac be complementary events and let B denote an arbitrary
event. Then
P(A|B)= P(AB)/P(B )
P(B/A)P(A)
• P(A|B) =- --------------------------------P(B/A)P(A) + P(B/Ac)P(Ac)
Remarks.
(i) The events of interest here are A, Ac,
(ii) P(A) and P (Ac) are called prior probabilities,
(iii) P(A|B) and P(Ac|B) are called posterior (revised) probabilities.
(iv) Bayes’ Law is important in several fields of applications.
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Bayesian Approach
English mathematician Thomas Bayes
(1702-61) set out his theory of probability
It is being revived now 250 years later
Step ahead from Subjective Prob Method
P(Ai)
Prior Probabilities
P(A)=0.30
PAc)=0.70
P(B/Ai)
Conditional
Probabilities
P(AiB)
Joint
Probabilitie
s
P(Ai/B)
Posterior
Probabilities
P(A/B)=.195/.37
=.527
P(B/A)=0.65
P(AB)=0.195
P(B/Ac)=0.25
P(Ac/B)=.175/.37=
P(AcB)=0.175
.473
Example .
• A laboratory blood test is 95 percent effective in
detecting a certain disease when it is, in fact, present.
However, the test also yields a “false positive” results for
1 percent of healthy persons tested. (That is, if a healthy
person is tested, then, with probability 0.01, the test
result will imply he or she has the disease.) If 0.5 percent
of the population actually has the disease, what is the
probability a person has the disease given that the test
result is positive?
• Solution Let D be the event that the tested person has
the disease and E the event that the test result is
positive. The desired probability P(D|E) is obtained by
• P(D/E) =P(D ∩ E)/P(E)
• =P(E/D)P(D)/P(E/D)P(D) + P(E/Dc)P(Dc)
• =(.95)(.005)/(.95)(.005) + (.01)(.995)
• =95/294 ≈0 .323.
• Thus only 32 percent of those persons whose test
results are positive actually have the disease.
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General Bayes’Theorom
• A1,A2,…..Ak are k mutually exclusive and
exhaustive events with known prior probabilities
P(A1),P(A2),….P(Ak)
• B is an event that follows or is caused by prior
events A1,A2, …Ak with
Conditional probabilities
P(B/A1),P(B/A2),…P(B/Ak) which are known
• Bayes’ formula allows us to calculate posterior
(revised) probabilities
P(A1/B),P(A2/B),….P(Ak/B)
• P(Ai/B)=P(Ai)P(B/Ai)/{P(A1)P(B/A1)+…+P(Ak)P(B/Ak)}
Counting Sample Points
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Is it always necessary to list all sample points in S?
Coin Tosses
Coins
sample-points
Coins sample-points
1
2
2
4
3
8
4
16
5
32
6
64
10
1024
20
1,048,576
30
≈ 109
40
≈ 10 12
50
≈ 1015
60
≈1019
Note that 230 ≈ 109 = one billion, 240 ≈ 1012 = one
thousand billion, 250 ≈1015=one trillion.
• RECALL: P(A) = nA/n , so for some applications we need
to find n, nA where n and nA are the number of points in
S and A respectively.
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Basic principle of counting: mn rule
• Suppose that two experiments are to be
performed. Then if experiment 1 can result
in any one of m possible outcomes and if,
for each outcome of experiment 1, there
are n possible outcomes of experiment 2,
then together there are mn possible
outcomes of the two experiments.
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Examples.
(i) Toss two coins: mn = 2×2 = 4
(ii) Throw two dice: mn = 6× 6 = 36
(iii) A small community consists of 10 men, each
of whom has 3 sons. If one man and one of his
sons are to be chosen as father and son of the
year, how many different choices are possible?
Solution: Let the choice of the man as the
outcome of the first experiment and the
subsequent choice of one of his sons as the
outcome of the second experiment, we see,from
the basic principle, that there are 10 × 3 = 30
possible choices.
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Generalized basic principle of counting
If r experiments that are to be performed are
such that the first one may result in any of
n1 possible outcomes, and if for each of
these n1 possible outcomes there are n2
possible outcomes of the second
experiment, and if for each of the possible
outcomes of the first two experiments
there are n3 possible outcomes of the third
experiment, and so on,. . ., then there are
a total of n1 x n2 · · xnr possible outcomes
of the r experiments.
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Examples
• (i) There are 5 routes available between A and
B; 4 between B and C; and 7 between C and D.
What is the total number of available routes
between A and D?
• Solution: The total number of available routes is
mnt = 5.4.7 = 140.
• (ii) A college planning committee consists of 3
freshmen, 4 parttimers, 5 juniors and 2 seniors.
A subcommittee of 4, consisting of 1 individual
from each class, is to be chosen. How many
different subcommittees are possible?
• Solution: It follows from the generalized principle
of counting that there are 3·4·5·2 = 120 possible
subcommittees.
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Examples
(iii) How many different 7-place license plates
are possible if the first 3 places are to be
occupied by letters and the final 4 by numbers?
Solution: It follows from the generalized principle
of counting that there are 26 · 26 ·26 · 10 · 10 ·
10 · 10 = 175, 760, 000 possible license plates.
(iv) In (iii), how many license plates would be
possible if repetition among letters or numbers
were prohibited?
Solution: In this case there would be 26 · 25 · 24
· 10 · 9 · 8 · 7 = 78, 624, 000 possible license
plates.
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Permutations: (Ordered arrangements)
• Permutations: (Ordered arrangements) The number of
ways of ordering n distinct objects taken r at a time
(order is important) is given by
• n! /(n - r)! = n(n - 1)(n - 2) · · ·(n - r + 1)
• Examples
• (i) In how many ways can you arrange the letters a, b
and c. List all arrangements.
• Answer: There are 3! = 6 arrangements or permutations.
• (ii) A box contains 10 balls. Balls are selected without
replacement one at a time. In how many different ways
can you select 3 balls?
• Solution: Note that n = 10, r = 3. Number of different
ways is
= 10! /7! = 10 · 9 ·8= 720,
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Combinations
• Combinations For r ≤ n, we define
• nCr =n! / (n - r)! r!
and say that n and r represents the number of
possible combinations of n objects taken r at a time
(with no regard to order).
• Examples
• (i) A committee of 3 is to be formed from a group of 20
people. How many different committees are possible?
• Solution: There are 20C3 = 20! /3!17! = 20.19.18/3.2.1 =
1140 possible committees.
• (ii) From a group of 5 men and 7 women, how many
different committees consisting of 2 men and 3 women
can be formed?
• Solution: 5C2 x 27C3 = 350 possible committees.
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Random Sampling
• Definition. A sample of size n is said to be a random
sample if the n elements are selected in such a way that
every possible combination of n elements has an equal
probability of being selected .In this case the sampling
process is called simple random sampling.
• Remarks. (i) If n is large, we say the random sample
provides an honest representation of the population.
• (ii) For finite populations the number of possible samples
of size n is NCn
• For instance the number of possible samples when N =
28 and n = 4 is 28C4=20475
• Tables of random numbers may be used to select
random samples.
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Frequency Distribution:
Number of Sales Orders Booked by 50 Sales Execs April 2006
Number of Orders
Number of SEs
00 – 04
05 - 09
10 – 14
15 – 19
14
19
07
04
20 – 24
25 – 29
30 – 34
02
01
02
35 – 39
40 – 44
TOTAL
00
01
50
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Probability Distribution
Number of Orders
Number of SEs
Probability
00 – 04
05 - 09
10 – 14
15 – 19
14
19
07
04
0.28
0.38
0.14
0.08
20 – 24
25 – 29
30 – 34
02
01
02
0.04
0.02
0.04
35 – 39
40 – 44
TOTAL
00
01
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0.00
0.02
1.00
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Standard Discrete Prob Distns
• Binomial Distribution: When a Situation
can have Only Two Possible Outcomes
e.g. PASS or FAIL, ACCEPT or REJECT.
• This distribution gives probability of an
outcome (say, ACCEPT) occurring exactly
m times out of n trials of the situation, i.e.
probability of 10 ACCEPTANCES out of
15 items tested.
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Standard Discrete Prob Distns
• Poisson Distribution: When a Situation
can have Only Two Possible Outcomes, &
When the Total Number of Observations is
Large (>20), Unknown or Innumerable.
• This distribution gives the probability of an
outcome (say, ACCEPT) occurring m
times, i.e. probability of say 150
ACCEPTANCES.
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Standard Continuous Prob Distn
• Normal Distribution: Useful & Important
• Several Variables Follow Normal Distn or
a Pattern Nearing It. (Weights, Heights)
• Skewed Distns Assume This Shape After
Getting Rid of Outliers
• For Large No. of Observations, Discrete
Distributions Tend to Follow Normal Distn
• It is Amenable to Mathematical Processes
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Features of Normal Distribution
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Symmetrical and Bell Shaped
Mean at the Centre of the Distribution
Mean = mode = Median
Probabilities Cluster Around the Middle
and Taper Off Gradually on Both Sides
• Very Few Values Beyond Three Times the
Standard Deviation from the Mean
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Probabilities in Normal Distn
• 68 % of Values Lie in the Span of Mean
Plus / Minus One Standard Deviation.
• 95 % of Values Lie in the Span of Mean
Plus / Minus Two Standard Deviation.
• 99 % of Values Lie in the Span of Mean
Plus / Minus Three Standard Deviation.
• Standard Normal Distn Tables Readily
Show Prob of Every Value.
• Use Them to Draw Inferences.
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