Conditional probability
I
P(B|A)
I
Conditional probability of B given A
I
Probability of B if it is known that A has occured
Conditional probability
I
Basic Rule (Product Rule)
I
P(A ∩ B) = P(A)P(B|A)
I
Since A ∩ B is same as B ∩ A,
P(A ∩ B) = P(B)P(A|B)
I
Warning: P(A|B) is not same as P(B|A)
Conditional probability
I
How to justify the product rule
I
Try out in simple examples.
I
Suppose we have a box with 3 Green and 2 Red balls.
Draw one at random, do not replace and draw one more.
I
Let RI , RII stand for Red in the first draw, Red in second
draw
I
Similarly GI , GII
I
Events of interest are RI ∩ RII , RI ∩ GII , GI ∩ RII , GI ∩ GII
First draw
Second Draw
G
R
G
P(GI ∩ GII )
P(GI ∩ RII )
P(GI )
R
P(RI ∩ GII )
P(RI ∩ RII )
P(RI )
P(GII )
P(RII )
1
G1 G2
G2 G1
G3 G1
R1 G1
R2 G1
G1 G3
G2 G3
G3 G2
R1 G2
R2 G2
P(GI ∩ GII ) =
6
20
G1 R1
G2 R1
G3 R1
R1 G3
R2 G3
G1 R2
G2 R2
G3 R2
R1 R2
R2 R1
P(RI ∩ GII ) =
6
20
Second Draw
G
R
First draw
I
G
P(GI ∩ GII )
P(GI ∩ RII )
P(GI )
R
P(RI ∩ GII )
P(RI ∩ RII )
P(RI )
P(GII )
P(RII )
1
First draw
I
Second Draw
G
R
G
6
20
6
20
12
20
R
6
20
2
20 )
8
20
using product rule
I
Suppose we have a box with 3 Green and 2 Red balls.
Draw one at random, do not replace and draw one more.
I
Interpret this as
P(GI ) = 35 , P(RI ) = 24 , P(GII |GI ) = 24 , P(GII |GI ) =
32
54
23
54
6
20
6
20
I
P(G1 ∩ GII ) =
I
P(R1 ∩ GII ) =
I
We get the same numbers as before
=
=
3
4
A study revealed that 45% of college freshmen are male and
that 18% of male freshmen earned college credits while still in
high school. Find the probability that a randomly chosen
college freshman will be male and have earned college credits
while still in high school.
I
Two events of interest
M = Person is a male,
H = Person earned college credit in school
I
We have P(M) = .45, P(H|M) = .18 Want P(M ∩ H)
I
So P(M ∩ H) = P(M)P(H|M) = (.45)(.18)
3.50
For two events A and B, P(A) = .4, P(B) = .2, P(A|B) = .6
I
Find P(A ∩ B)
I
= P(B)P(A|B) = .2 × .6 = .12
I
Find P(B|A)
I
P(B|A) =
P(A∩B)
P(A)
=
.12
0.4
Independence
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Two events A, B are said to be independent, if
P(A|B) = P(A)
I
i.e
I
i.e.
P(A ∩ B)
= P(A)
P(B)
P(A ∩ B) = P(A)P(B)
I
Note independence is symmetric , if A is independent of B,
then B is also independent of A
I
What does A and B are independent mean?
I
A has no information about B. Occurrence of A does not
affect the probability of occurrence of B
I
How do we know events are independent?
I
Sometimes specified in the problem,
I
Description of the experiment like successive tosses of a
coin, sample with replacement etc.
I
If A, B are mutually exclusive (disjoint) can they be
independent?
I
Being mutually exclusive P(A ∩ B) = 0
I
If A, B are independent
P(A ∩ B) = P(A)P(B) = 0 !!
I
If mutually exclusive they cannot be independent
Summary Formulas
I
P(A) = 1 − P(A)
I
P(A ∪ B) = P(A) + P(B) − P(A ∩ B)
I
P(A ∩ B) = P(A)P(B|A)
I
P(A ∩ B) = P(B)P(A|B)
I
If A, B are independent
P(A ∩ B) = P(A)P(B)
3.50
For two events A and B, P(A) = .4, P(B) = .2, P(A ∩ B) = .1
I
Find P(B|A)
I
P(B|A) =
I
Find P(A|B)
I
P(A|B) =
I
Are A and B independent?
I
is P(A ∩ B) = P(A)P(B) ? No
I
Not independent
P(A∩B)
P(A)
P(A∩B)
P(B)
=
0.1
0.4
=
0.1
0.2
3.58
Suppose that 15% of all full time workers exhibit arrogant
behavior on the job and that 10% of all the full time workers
receive a poor performance rating. Also, assume that 5% of all
full time workers exhibit arrogant behavior and receive poor
rating. Let A be the event that full time worker exhibits arrogant
behavior and B be the event that a full time worker receives a
poor performance rating.
I Are the events A and B mutually exclusive?
I
Find P(B|A)
I
Are A and B independent?
3.67
An ambulance station has one vehicle and two demand
locations, A and B. The probability that an ambulance can
travel to a demand location under8 minutes is .58 for location A
and .42 for location B. The probability that the ambulance is
busy at any point of time is .3.
I
Find the probability that EMS can meet the demand for
ambulance at location A
3.67
An ambulance station has one vehicle and two demand
locations, A and B. The probability that an ambulance can
travel to a demand location under8 minutes is .58 for location A
and .42 for location B. The probability that the ambulance is
busy at any point of time is .3.
I
Find the probability that EMS can meet the demand for
ambulance at location B
3.69
A computer intrusion detection system (IDS)is designed to
provide an alarm whenever an intrusion is attempted into a
computer system. Consider a double IDS system with system
A and system B. If there is an intruder, system A sounds an
alarm with probability 0.9 and B sounds an alarm with
probability 0.95. If there is no intruder, system A sounds an
alarm ( false alarm) with probability 0.2 and system B sounds
an alarm with probability 0.1. Assume under a given condition (
intruder or not) system A and B operate independently.
I
Using symbols express the four probabilities given in the
example
I
If there is an intruder what is the probability that both
systems sound an alarm ?
I
If there is no intruder what is the probability that both
systems sound an alarm?
I
Given an intruder, what is the probability that at least one
of the systems sound an alarm?
3.69
A computer intrusion detection system (IDS)is designed to
provide an alarm whenever an intrusion is attempted into a
computer system. Consider a double IDS system with systems
A and systems B. If there is an intruder, system A sounds an
alarm with probability 0.9 and B sounds an alarm with
probability 0.95. If there is no intruder, system A sounds an
alarm ( false alarm) with probability 0.2 and system B sounds
an alarm with probability 0.1. Assume under a given condition (
intruder or not) system A and B operate independently.
I
Using symbols express the four probabilities given in the
example
3.69
I
If there is an intruder what is the probability that both
systems sound an alarm ?
I
If there is no intruder what is the probability that both
systems sound an alarm?
3.69
I
Given an intruder, what is the probability that at least one
of the systems sound an alarm?
Bayes Theorem
I
What is Bayes theorem all about?
I
You are given P(A), P(B|A), P(B|Ac ). Use these to find
P(A|B).
Bayes Theorem-Example
A study revealed that 40% of college freshmen are male and
that 15% of male freshmen earned college credits while still in
high school. Among the 60% female freshmen, 20% earned
college credit while in high school. If a freshman picked at
random has earned college credit in high school, find the
probability that the freshman is a male.
I
Two events of interest
M = Person is a male, M c = person is a female
H = Person earned college credit in school
I
We have P(M) = .4, P(H|M) = .15 ,P(H|M c ) = .2. Want
P(M|H)
I
Think in terms of percentages. What percentage of H is M?
H
Hc
M
45
Mc
55
I
We have P(M) = .4, P(H|M) = .15 ,P(H|M c ) = .2. Want
P(M|H)
H
Hc
M
.45
Mc
.55
B
Bc
P(A ∩ B)
P(A ∩ B c )
= P(A)P(B|A)
= P(A)P(B c |A)
P(Ac ∩ B)
P(Ac ∩ B c )
= P(Ac )P(B|Ac )
= P(Ac )P(B c |Ac )
P(B)
P(B c )
A
Ac
P(A|B) =
P(A)
P(A)
1
P(A ∩ B)
P(A)P(B|A)
=
P(B)
P(A)P(B|A) + P(Ac )P(B|Ac )
Bayes theorem - formal statement
If A1 , A2 , . . . , An are disjoint and A1 ∪ A2 , . . . ∪ An = S ,then
P(Ai |B) =
P(A ∩ B)
P(B)
P(Ai )P(B|Ai )
Pn
j=1 P(Aj )P(B|Aj )
(1)
(2)
3.88
A computer intrusion detection system (IDS)is designed to
provide an alarm whenever an intrusion is attempted into a
computer system. Consider a double IDS system with systems
A and systems B. If there is an intruder, system A sounds an
alarm with probability 0.9 and B sounds an alarm with
probability 0.95. If there is no intruder, system A sounds an
alarm ( false alarm) with probability 0.2 and system B sounds
an alarm with probability 0.1.Assume under a given condition (
intruder or not) system A and B operate independently. The
probability of an intrusion is 0.4.
I
Given both the system sound an alarm, what is the
probability that an intruder is detected?
I
Given at least one system sounds an alarm, what is the
probability that an intruder is detected?
3.88
A stands for A sounds alarm, B stands for B sounds alarm, I
there is an intruder
I
P(I) = .4
I
P(A ∩ B|I) = (.9)(.95)
I
P(Ac ∩ B|I) = (.1)(.95) P(Ac ∩ B|I c ) = (.8)(.1)
I
P(A ∩ B c |I) = (.9)(.05) P(A ∩ B c |I c ) = (.8)(.9)
I
P(Ac ∩ B c |I) = (.1)(.05)
P(A ∩ B|I c ) = (.2)(.1)
P(Ac ∩ B c |I c ) = (.8)(.9)
3.88
In terms of table, fill the following table
A∩B
I
Ic
Ac ∩ B
A ∩ Bc
Ac ∩ B c
problem 3.82
A company employs three sales engineers. Engineers 1,2 and
3 estimate the costs of 30%, 20% and 50% respectively of all
jobs bid by the company. For i = 1, 2, 3, let Ei b the event a job
is estimated by engineer i. The following table describes the
rates at which the engineers make serious errors in estimating
costs:
P( error |E1 ) = .01
P( error |E2 ) = .03
P( error |E3 ) =
.02
If a particular bid results in a serious error in estimating the
cost, what is the probability that it was made by
1. Engineer 1
2. Engineer 2
3. Engineer 3
Combinatorics
Suppose we have a box with 5 tickets with labels A,B,C,D,E and
we want to choose 3 out of this, one after the other. How many
ways can this be done?
There are 5 ways in which the first draw can be made
With each of the first draw there are 4 possibilities for the
second. There are 5 x 4 = 20 ways to choose two
1
2
1
3
3
4
2
4
5
5
1
1
2
2
4
3
4
3
5
5
1
2
5
3
4
Combinatorics
Proceeding,
The number of ways of choosing 3 out of 5 objects, keeping
track of order is
5×4×3=
5!
5×4×3×2×1
=
2×1
(5 − 3)!
In general: The number of ways of choosing r out of n objects,
keeping track of order is ( number of permutations of r out of n)
nPr = n(n − 1)(n − 2) . . . (n − (r − 1) =
n!
(n − r )!
Combinations
If we choose 3 out the 5 tickets how many distinct triplets are
possible? We do not keep track of order. For instance the triplet
{A, B, C} can arise in 6 ways. The first can be any of the three
the second any of the remaining 2 and the last one: i.e 3!.
That is 6 ordered triplets like
ABC,ACB,BAC,BCA,BAC,CAB,CBA will go to the same set
{A, B, C}.
5!
unordered triplets from 5 objects
So we can get 3!(5−3)!
Combinations
The number of ways of choosing 3 out of 5 objects, without
keeping track of order is
5×4×3=
5!
5×4×3×2×1
=
2×1
3!(5 − 3)!
In general: The number of ways of choosing r out of n objects,
without keeping track of order is ( number of combinations of r
out of n)
n
n!
=
r
r !(n − r )!
Problem
A company has four departments: manufacturing,distribution,
marketing and management. The number of people in each
department is 55,30,21 and 13. Each department is expected
to send one representative to a meeting. How many
possibilities are there?
55 × 30 × 21 × 13 = 450450
Problem
Nine sealed bids for oil drilling leases arrive at a regulatory
agency. In how many different orders can the nine bids be
opened.
9 × 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1 = 362880
Problem
fifteen locations in a given area are believed likely to have oil.
An oil company can only afford to dig 8 sites, sequentially
chosen. How many possibilities are there?
15!
= 15 × 14 × 13 × 12 × 11 × 10 × 9 × 8 = 259459200
(15 − 8)!
Problem
In a shipment of 14 computer parts, 3 are faulty and the
remaining 11 are in working order. Three elements are chosen
at random. what is the probability that all the faulty ones are
chosen
1
14
3
A more easy argument is
3 2 1
14 13 12
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