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1
H. Introduction to Probability
1. Experiments and Probability
Text problems 4.1, 4.2.
2. The Venn Diagram and the Addition Rule.
Downing & Clark, pg. 96 (pg 85 in 3 rd ed) Basics 1, Application 2, 13. H1 (H0A), Text problems 4.3, 4.4, 4.10, 4.11, 4.8!, 4.9!.[4.3,
4.4, 4.8*, 4.9*]., (4.3,4.4, 4.8*, 4.9*)., H4, H5 (H1, H2).
3. Conditional and Joint Probability, Bayes’ Rule.
Text 4.16a-c, 4.18 [4.14a-c, 4.16] (4.13a-c, 4.15). H2, H3 (H0B, H0C), D&C pg 113 (pg 103 in 3 rd ed) 14 (Note error in text - 5/6 of
the people in the city support Jones, 5/9 of the people in the country support Jones), 15, 16. H6 (H3).
4. Statistical Independence.
Text 4.16d, 4.22, 4.21!, 4.24, 4.30, 4.31, 4.33 [4.14d*, 4.19*, 4.20*, 4.22*, 4.28, 4.29, 4.31, 4.68*] (4.18*, 4.19*, 4.21*, 4.26, 4.27,
4.29). H8, H9 (H5, H6). pg. 97( pg. 85 in 3 rd ed) Applications 4, 5, 8, 9, 10, 11, 44. H7(H4).
5. Review.
Section 3 is in this document.
----------------------------------------------------------------------------------------------------.
Exercise 4.16 [4.14a-c in 9th] (4.13 a-c in 8th edition): The text gives the following contingency table
B B
A 10 20 
. It asks for the following probabilities: (a) P A B ; (b) P A B and (c) P A B .


A 20 40 
 
B

A 10
Solution: Add the rows and columns to get

A 20
30
B B
10
20 

A
2090 40 90
A  90
90


30
90
60
90
 
B
20  30
. To get the probabilities, divide by 90.

40  60
60 90
B B

A .1111 .2222  .3333
90
or
. Note that this table consists of


60
A .2222 .4444  .6667
90
.3333 .6667 1.0000
1
30
B
B

A P  A  B  P A  B  P  A


A P A  B P A  B  P A


P B 
PB
1
According to the Instructor’s Solutions Manual
(Edited)
(a) P(A |B) = 10/30 = 1/3 = 0.3333
(b) P(A | B ) = 20/60 = 1/3 = 0.3333
(c) P( A | B ) = 40/60 = 2/3 = 0.6667

 

 
 


 
But if we use the multiplication rule:
10
P A  B 
1
(a) PA B  
 90 
30
P B 
3
90
20
P A B
1
(b) P A B 
 90 
60
3
PB
90
40
P AB
2
(c) P A B 
 90 
60
3
PB
90
   
  
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 
Exercise 4.18 [4.16 in 9th] (4.15 in 8th edition): If P A and B   .4 and PB   .8 , find P A B .
 
Solution: We can write the Multiplication rule as P A B 
PB   .8 . PA B  
P A  B  .4
  .5
PB 
.8
P A  B 
. We know that P A  B   .4 and
P B 
PROBLEM H2 (Old H0B): Using the same definitions as H1 (H0A)
Define Event A as rolling an even number (i.e. the sum of both faces is even)
Define Event B as rolling a 5.
Define Event C as a 1 on at least one of the dice.
a. Find what points are in the events and add the probabilities. P A B , P B A , P A C , P C A ,
   
   
PB C  , PC B , P B C , P C B
   
 
b. Show that the multiplication rule works for A and B . i.e. that P A  B  P A B PB and that
P A  B  PB AP A


c. Do the same for P A  C  , PB  C  and P C  B .
Solution: Define Event A as rolling an even number (i.e. the sum of both faces is even) P A  1 .
2
Define Event B as rolling a 5. PB   4  1 .
36
9
Define Event C as a 1 on at least one of the dice. PC   11 . P B  8  1  PB  .
36
9
y
 
1
2
Diagram for dice problems.
x 3
4
5
6
1
2
3
4
5
6
7
2
3
4
5
6
7
8
 
3
4
5
6
4 5 6 7
5 6 7  8
6 7 8 9
7  8  9  10 
8  9  10  11 
9  10  11  12 
a) P A B : This is a conditional probability. This is the probability of rolling an even number, given that
 
we roll a 5. There are 4 points in B . None of them are in A . Thus P A B  0.
PB A : This is the probability that we roll a 5, given that we roll an even number. There are 18 points in
B . None of then are in A . Thus PB A  0.
PA C  This is the probability of rolling an even number if we have a 1 on at least one of the dice. There
are 11 points in C . These are (1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 1), (3, 1), (4, 1), (5, 1) and (6, 1).
Only 5 of these add to an even number. PA C   5 .
11
PC A : There are 18 points in A . Of these 5, (1, 1), (1, 3), (1, 5), (3, 1) and (5, 1), are in C .
PC A  5
18
.
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PB C  : Since C is “a 1 on at least one of the dice” and B is “rolling a 5,” our conditional probability of
rolling a 5, assuming that we know there is a one on at least one of the dice, is taken out of the 11 points in
C . Only 2 of these points give us a 5, so PB C   2 .
11
PC B : There are only 4 points in B . Of these points only two contain a 1. PC B   2 4  1 2 .
 
P B C : This is the probability of not getting a 5 when you get a 1 on at least one of the dice and is actually
 
 
the complement of P B C . 9 of the 11 points in C do not give us a 5, so P B C  9
 
11
.
P C B : We are picking points with a one in them from the 32 points in B. We have already found that
 
there are 9 of them. P C B  9
32
.
 
b) We must show that the multiplication rule works for A and B ; i.e. that P A  B  P A B PB and
 
shown that PA B  0 and PB A  0. So P A  B   PA B PB   01   0
9
P A  B   PB AP A  01   0.
2
that P A  B  P B A P A . We already know that P A  B   0 from Problem H0A and we have just
and
 
c) We must show that the multiplication rule works for A and C ; i.e. that P A  C   P C A P A and
P A  C   PA C PC  . We already know P A  C   5 36 . This is the joint probability of A and C .
We have also found the conditional probabilities PA C   5
probabilities P A  1
2
and PC   11
P A  C   P C AP A  5
18
36
11
and PC A  5
18
and the total
. So we can say P A  C   P A C PC   5
11
1 2   365 .
11 36   365
and
Similarly, since we already know P B  C  
1
, we can use probabilities from part a) to show that
18
2
1
1
PB  C   PB C PC   2 11


and PB  C   PC B PB   1 1 
.
11 36 36 18
2 9 18
 

 

 
Also, since we already know P C  B  1 and PB   4  1 , P B  1  PB   1  1  8 , we can
4
36
9
9
9
9
8
1
use probabilities from part a) to show that P C  B  P C B P B 
and

32 9
4
P C  B  P B C PC   9 11
1
11 36
4

  
 

   
 
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 
   
PROBLEM H3 (Old H0C):. In H2(H0B), demonstrate Bayes’ rule for P A B and P B A , P A C and
 
 
PC A , PB C  and PC B , P B C and P C B .
 
Solution: P A  1 , PB   1 , PC   11 , P B  8 .
2
9
36
9
PB A  0. PA B  0. Bayes’ Rule says PB A 
PA B PB 
P  A

 9
01
1
0.
2
5
11
PA C PC 
 11 36  5 .
PC A  5 . PA C   5 . PC A 
18
18
11
1
P  A
2
1
PB C PC  211 11 36

 18  1 .
PC B   1 . PB C   2 . PC B  
2
2
11
1
1
P B 
9
9
1
P B C PC  911 11 36

 4 9 .
PCB 9 . P BC 9 . PCB 
32
32
11
8
8
PB
9
9
Downing and Clark, pg. 103, Application 14: In the city 5 6 support Jones, but in the country 5 9
support Jones. Half of the people live in the country and half in the city. What is the probability that a
randomly picked Jones supporter lives in the country?
Solution: The most important part of this problem is defining events and interpreting probabilities.
Use the following definitions for events: J , a person supports Jones; S , a person supports Smith; C , a
person lives in the city and Co , a person lives in the country. The corrected problem says P J C  5 6 ,
 
 
 
 
    
 
PJ Co  5 9 and PC   PCo  
 

P A B PB 
PJ Co PCo 
we need Bayes' Rule. P B A 
, so PCo J  
. The hardest part of this is
P  A
PJ 
finding PJ   PJ  C   PJ  Co  PJ C PC   PJ CoPCo  5 6 1 2  5 9 1 2  512  518
PJ CoPCo 5 9 12 10 2
10
25
 15




.
So
P
Co
J



 .
36
36
36
25
 
1
2

and asks us for P Co J . Since this is a condition-reversing problem,
PJ
36
25
5
Downing and Clark, pg. 103, Application 15: If 5% of people with type O blood are left-handed, 10% of
people with other blood types are left-handed and 40% have blood type O, what is the probability that a
randomly selected left-handed person has blood type O?
Solution: Use the following definitions for events: L left-handed, O type O. The problem says
 
 
PL O  .05 , P L O  .10 , PO  .40 and implies that P O  .60
If we do this as a 'box' problem, we can say that, out of 100 people, 40 are type O and 60 not type O. Of the
40 who are type O, 5% or 2 must be left-handed and the remaining 38 are right-handed. Of the 60 who are
not type O 10% or 6 are left-handed. Thus out of the 100 people there must be 8 left-handed people of
L L
40
O 2 38
whom 2 are type O. so the answer is 2 out of 8 or .25. If we put these in the box we get 
.
60
O 6 54
  
More formally, PL   PL O PO   P L O P O  .05 .40   .10 .60   .02  .06  .08
PO L  
PL O PO 
P L 

.05 .40   .02  .25
.08
.08
8 92
100
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Downing and Clark, pg. 103, Application 16: If 70% of brown-eyed people have brown hair, 20% of
green-eyed people have brown hair and 5% of blue-eyed people have brown hair; and, out of the entire
population, 75% of people have brown eyes, 5% of people have green eyes and 20% of people have blue
eyes, what is the probability that a randomly selected brown-haired person has green eyes?
Solution: Use the following definitions for events: BrH Brown-haired, BrE Brown-eyed, GE Greeneyed, BlE Blue-eyed.. The problem says P BrH BrE  .70 , P BrH GE  .20 , P BrH BlE  .05 ,




PBrE   75 , PGE   .05 and PBlE   .20 . It is asking for PGE BrH 


If we do this as a 'box' problem, we can say that, out of 100 people, 75 are brown-eyed, 5 are green-eyed
and 20 are blue-eyed. Of the 75 who are brown-eyed, 70% or 52.5 have brown hair. Of the 5 who are greeneyed, 20% or 1 has brown hair. Of the 20 who are blue-eyed, 5% or 1 has brown hair. Thus a total
of 52.5 + 1 + 1 = 54.5 have brown hair. Of these
BrH BrH
54.5 people only one has green eyes. The
BrE 52 .5 22 .5 75
requested probability is one out of 54.5 and
1
GE
1
4
5.
 .01835 . If we put these in the box to the
54.5
BlE
right we get PGE BrH   1  .01835 .
54 .5
1
19
20
54 .5 45 .5 100
More formally,
PBrH   P BrH BrE PBrE  P BrH GE PGE  P BrH BlE PBlE  .70.75  .20.05  .05.20






PBrH GE PGE  .20 .05 

 .01835
 .525  .010  .010  .545 So PGE BrH  
PBrH 
.545
PROBLEM H6 (OldH3): Your firm produces 3 different VCRs, The Deluxe, Super Deluxe, and
Incredible models. You offer a 1 year warranty. 50% of your sales are Deluxe, 30% are Super Deluxe, and
20% are Incredible.
During the warranty period 80% of the Deluxe models fail, 50% of the Super Deluxe models fail, and 30%
of the Incredible models fail. Since the Deluxe and Super Deluxe models are not worth repairing, they are
replaced when they are returned after failure. To decide on how many repair people to hire, find out what
per cent of the returns are incredible.
Solution: Let D be the event that a VCR is Deluxe, SD be the event that a VCR is Super Deluxe and I be
the event that it is Incredible. Then PD  .50,PSD  .30 and PI   0.20 . Furthermore, if F represents




 
the event that the VCR fails, P F D  .80,P F SD  .50 and P F I  .30 . From the first set of probabilities
we can say that of 100 VCRs, 50 are Deluxe, 30 are Super Deluxe and 20 are Incredible. From the second
set of probabilities we can deduce that, of the 50 deluxe VCRs, 80% or 40 fail, of the 30
Super Deluxe VCRs, 50% or 15 fail and, of the
20 Incredible VCRs, 30% or 6 fail. We add these
SD
Total
D
I
up on the table at left and conclude that there are
40
15
6
61
F
a total of 61 failures in the 100 VCRs, of which 6
10
15
14
39
F
6
are incredible so that P I F 
 .0984 .
Total
50
30
20
100
61
P F I PI 
More formally, P I F  
. But PF   PF  I   PF  SD  PF  D
P F 
 
 PF I PI   PF SDPSD  PF DPD  .30 .20   .50 .30   .80 .50   .61 .
So that P I F  
P F I PI 
.06

 .0984 .
P F 
.61
Parts not copied ©2003 Roger Even Bove