Chapter 5: A Survey of Probability Concepts
Homework #5 (Week 7): Chapter 5, Exercises 8, 12,
20, 26, 28, 30 and 32.
Homework #6 (Week 8): Chapter 5, Exercises 36, 38,
40, 70 & 74.
Permutations and Combinations
Our standard approach to probability questions:
P(A) =
#S =
#N
Number of successful outcomes
Number of possible outcomes
Example: Event A = even number on throw of die
S=3
N=6
{2, 4, 6}
{1, 2, 3, 4, 5, 6}
P(A) = 3/6 = 0.5
This approach, without some revision, is often very
difficult to implement, e.g. N (and S) increases.
Example: The Irish National Lottery works as
follows. Six numbers from the range 1 to 42 are
chosen at random. If you have correctly guessed all
six you win/share the first prize. What are your
chances of winning/sharing if you are allowed to
choose six numbers?
Useful Mathematical Notation
Factorial Notation: n! = n.(n-1).(n-2). ... .2.1
Examples: 5! = 5x4x3x2x1 = 120; 1! = 1; (Aside: 0! = 1
by convention)
Examples Leading to General Principles
Given the letters {a, b, c, d}
(1) How many 2-letter "words" can be formed,
allowing for the repeated use of letters?
aa ab ac ad ba bb bc bd ca ...
4.4 = (4 possible entries in first cell).(4 possible entries
in second cell) = 42 = 16
(2) How many 4-letter "words" can be formed,
allowing for the repeated use of letters?
aaaa aaab aaac aaad aaba aabb aabc ...
4x4x4x4 = (4 possible entries in the first cell). … .(4
possible entries in the fourth cell) = 44 = 256
General Principle
(1) and (2) There are an "words" of length n from an
alphabet of a letters, allowing for the repeated use of
letters.
(3) How many 2-letter "words" can be formed, not
allowing for the repeated use of letters?
ab ac ad ba bc bd ca cb cd ... [Note: ab is regarded as
different from ba]
4x3 = (four possible entries in the first cell).(three
possible entries in the second cell) = 12
{ = 4!/2! = 4!/(4-2)! }
(4) How many 4-letter "words" can be formed, not
allowing for the repeated use of letters?
abcd abdc acbd acdb adbc adcb bacd badc bcad bcda
... [Note: abcd is regarded as different from abdc]
4x3x2x1 = (four possible entries in the first cell). …
.(1 possible entry in the fourth cell) = 24 = 4!
{ = 4!/1 = 4!/0! = 4!/(4-4)! }
General Principle
(3) and (4) There are n!/(n-r)! "words" of length r
from an alphabet of n letters if the repeated use of
letters is not allowed, i.e. letters can only be used
once.
“Proof”/Intuition: n.(n-1). ... .(n-r+1) = n!/(n-r)!
(3) and (4) There are n!/(n-r)! “permutations” of n
distinct objects taken r at a time.
(nPr = nPr = n!/(n-r)!)
(5) How many 2-letter "words" can be formed not
allowing for the repeated use of letters and with the
extra stipulation that order becomes unimportant,
e.g. ab = ba and hence should only be
included/counted once?
ab ac ad bc bd cd [Note: ab is regarded as the same as
ba]
(4x3)/2 = 6 (i.e. divide the answer to part (3) by 2 (=
2!) in order to avoid double-counting)
General Principle
(5) There are n!/[r!.(n-r)!] words of length r from an
alphabet of n letters if the repeated use of letters is
not allowed and order is disregarded.
(5) There are n!/[r!.(n-r)!] “combinations” of n
distinct objects taken r at a time.
(nCr = nCr = n!/[r!.(n-r)!])
“Proof”/Intuition: As we move from (3) and (4) to (5)
order must be disregarded, i.e. as we move from
“permutations” to “combinations” order must be
disregarded. In how many ways can we permute r
objects? The answer is r!. Therefore we must divide
our formula from (3) and (4) by r!.
Example: The Irish National Lottery works as
follows. Six numbers from the range 1 to 42 are
chosen at random. If you have correctly guessed all
six you win/share the first prize. What are your
chances of winning/sharing if you are allowed to
choose six numbers?
How many 6 number "sets" can be formed from a 42
number “alphabet”, not allowing for the repeated use
of numbers and with the extra stipulation that order
is disregarded, e.g. 9, 12, 18, 23, 34, 42 = 18, 12, 34,
23, 42, 9 and should only be included/counted once?
42C6 = 42C6 = 42!/[6!(42-6)!] = 5,245,786
P(success) = 1/5,245,786 i.e. approximately 1 in 5
million
Aside:
36C6 = 1,947,792 and P(success) is approximately 1 in
2 million
39C6 = 3,262,623 and P(success) is approximately 1 in
3 million
Probability: Subjective Approach
“Probability is the degree of belief that someone holds
about the likelihood of an event occurring.”
Problem: Should the belief be rational? What does
rational mean?
Bayesian Statistics
Prior subjective “beliefs”, evidence and rational
posterior “beliefs”.
BAYES' THEOREM
Intuition: Prior Probabilities + Information =
Posterior Probabilities
Example:
Prior Probabilities:
P(Faulty) = P(F) = .3, P(Not Faulty) = P(NF) = .7
Information: Mechanic’s Record
P(“Guess” FaultyFaulty) = P(GFF) = .9
P(“Guess” Not FaultyFaulty) = P(GNFF) = .1
and
P(“Guess” FaultyNot Faulty) = P(GFNF) = .2
P(“Guess” Not FaultyNot Faulty) = P(GNFNF) =
.8
Posterior Probabilities?
(a) P(Faulty “Guess” Faulty) = P(FGF) = ?
Answer > .3 surely
(b) P(Faulty “Guess” Not Faulty) = P(FGNF) = ?
Answer < .3 surely
Method: Bayes' Formula
P(AB) = P(A and B)/P(B)
(i) P(A and B)?
P(BA) = P(B and A)/P(A) = P(A and B)/P(A)
 P(A and B) = P(BA).P(A)
(ii) P(B)?
P(B) = P(B and A) + P(B and Not A)
P(B) = P(BA).P(A) + P(BNot A).P(Not A)
P(AB) = P(A and B)/P(B)
P(AB) =
P(BA).P(A)
P(BA).P(A) + P(BNot A).P(Not A)
(a) P(Faulty “Guess” Faulty) = P(FGF) = ?
P(FGF) =
P(FGF) =
P(GFF).P(F)
P(GFF).P(F) + P(GFNot F).P(Not F)
(.9).(.3)
(.9).(.3) + (.2).(.7)
P(FGF) = (.27)/(.41) = .66 > .3 (as expected)
(b) P(Faulty “Guess” Not Faulty) = P(FGNF) = ?
P(FGNF) =
P(FGNF) =
P(GNFF).P(F)
P(GNFF).P(F) + P(GNFNot F).P(Not F)
(.1).(.3)
(.1).(.3) + (.8).(.7)
P(FGNF) = (.03)/(.59) = .05 < .3 (as expected)
Example: There are six male candidates and four
female candidates for four positions. All ten
candidates are equally qualified. The four selected
are male.
(a) Can this outcome be explained by chance, i.e.
what is the probability of the observed evidence
occurring given no discrimination?
P(EvidenceNo Discrimination) = P(END)
6C4/10C4 = 6C4/10C4 = 15/210 = .0714285
=
P(EvidenceDiscrimination) = P(ED) = 1.0
(b) Given the above information/evidence, calculate
the posterior probabilities of discrimination given the
following prior probabilities of discrimination:
(i) P(Discrimination) = P(D) = .8
P(No Discrimination) = P(ND) = .2
(ii) P(Discrimination) = P(D) = .5
P(No Discrimination) = P(ND) = .5
(iii) P(Discrimination) = P(D) = .2
P(No Discrimination) = P(ND) = .8
Prior Probabilities
Probabilities
+
Information
= Posterior
(i) Prior Probabilities:
P(Discrimination) = P(D) = .8
P(No Discrimination) = P(ND) = .2
Information:
P(EvidenceNo Discrimination) = P(END)
.0714285
P(EvidenceDiscrimination) = P(ED) = 1.0
=
Posterior Probabilities: P(DE)?
P(DE) =
P(ED).P(D)
P(ED).P(D) + P(END).P(ND)
P(DE) =
(1.0).(.8)
(1.0).(.8) + (.0714285).(.2)
Aside: P(NDE) = .0175439
= .9824561
(ii) Prior Probabilities:
P(Discrimination) = P(D) = .5
P(No Discrimination) = P(ND) = .5
Information:
P(EvidenceNo Discrimination) = P(END)
.0714285
P(EvidenceDiscrimination) = P(ED) = 1.0
Posterior Probabilities: P(DE)?
P(DE) =
P(ED).P(D)
P(ED).P(D) + P(END).P(ND)
P(DE) =
(1.0).(.5)
(1.0).(.5) + (.0714285).(.5)
Aside: P(NDE) = .0667
= .9333
=
(iii) Prior Probabilities:
P(Discrimination) = P(D) = .2
P(No Discrimination) = P(ND) = .8
Information:
P(EvidenceNo Discrimination) = P(END)
.0714285
P(EvidenceDiscrimination) = P(ED) = 1.0
Posterior Probabilities: P(DE)?
P(DE) =
P(DE) =
P(ED).P(D)
P(ED).P(D) + P(END).P(ND)
(1.0).(.2)
(1.0).(.2) + (.0714285).(.8)
Aside: P(NDE) = .223
= .777
=