pg. 1
SS2657a Midterm Exam, October 27th 2008
SS257a Midterm Exam
Monday Oct 27th 2008, 6:30-9:30 PM
Talbot College 342 and 343
You may use simple, non-programmable scientific calculators.
This exam has 15 questions worth a total of 125 marks on 16
pages, including this cover page. Attempt All questions.
Good luck! :+)
Name:
Student Number:
For Marker’s Use only:
Question
Mark
Question
1
9
2
10
3
11
4
12
5
13
6
14
7
15
8
Total
Mark
SS2657a Midterm Exam, October 27th 2008
pg. 2
Q1 (6 marks): A poll conducted by an election candidate asked
two questions: (1) Do you support the candidate’s position on
environmental issues and (2): Do you support the candidate’s
position on income tax cuts?
A total of 1200 responses were received; 600 said ‘yes’ to the
first question and 400 said ‘yes’ to the second. If 300
respondents said ‘no’ to the environmental question and ‘yes’
to the tax cut question, how many said ‘yes’ to the
environmental question but ‘no’ to the tax cut question?
Solutions:
A = yes to q1, #(A) = 600
B = yes to q2, #(B) = 400
#( Ac  B)  300
#( A  B)  #( B)  #( Ac  B)  400  300  100
#( A  B c )  #( A)  #( A  B)  600  100  500
So there are 500 people said ‘yes’ to the environment question
but ‘no’ to the tax cut question.
SS2657a Midterm Exam, October 27th 2008
pg. 3
Q2 (5 marks). Write down Kolmogorov’s Axioms for
characterizing the probability function P on a finite sample
space S.
P1: for all A  S , P( A)  0
P 2 : P( S )  1
P3: If A and B are mutually exclusive, ie A  B  , Then
P( A  B)  P( A)  P( B)
SS2657a Midterm Exam, October 27th 2008
pg. 4
Q3 (6 marks) Three events: A, B, and C – are defined on a
sample space S. Given that P(A) = 0.2, P(B) = 0.1 and P(C) = 0.3,
what is the smallest possible value for P[(A U B U C)C]?
Solutions:
If we want
P[(A U B U C)C] to be as small as possible then we want
P[(A U B U C)] to be as big as possible. That means we don’t
want any overlap. ]
Therefore,
P[(A U B U C)] =0.2 + 0.1 + 0.3 = 0.6
And P[(A U B U C)C] = 0.4
SS2657a Midterm Exam, October 27th 2008
pg. 5
Q4 (6 marks) Two fair dice are rolled. What is the probability
that the number on the first die was at least as large as 4 given
that the sum of the two dice was 8?
Solutions:
Possible outcomes:
(2, 6), (3, 5), (4, 4), (5, 3), (6, 6)
5 equally likely outcome have sum of two dice = 8, only 3 of the
5 have first die at least as large as 4.
So P(first die ≥ 4\sum=8) is 3/5.
SS2657a Midterm Exam, October 27th 2008
pg. 6
Q5 (10 marks): When Xiaohong wants to contact her boyfriend
and she knows he is not at home, she is twice as likely to send
him an email as she is to leave a message on his answering
machine. The probability that he responds to her email within
3 hours is 80%; his chance of responding to a phone message
within 3 hours is 90%. Suppose he responded to the message
she left this morning within 2 hours. What is the probability
that Xiaohong was communicating with him via email?
Solutions:
P(email )  2 / 3
P(phone)  1/ 3


 P(response email )  0.8, P(response email )  0.9
B  event she responded


A1  event she sent him an email

 A  Event she email him a phone message
 2
P( A1 \ B) 

P( B \ A1 ) P( A1 )
P( B \ A1 ) P( A1 )  P( B \ A2 ) P( A2 )
(0.8)(2 / 3)
1.6
1.6


 64%
(0,8)(2 / 3)  (0.9)(1/ 3) 1.6  0.9 2.5
Therefore, the probability that she contact him visa email was
64% [make sense: This is slightly less than base rate of 2/3 bince
phone is more successful response setting method].
SS2657a Midterm Exam, October 27th 2008
pg. 7
Q6 (6 marks): A desk has four drawers. Drawer 1 contains two
gold coins, Drawers 2 and 3 each contain two silver coins, and
Drawer 4 contains one silver coin and one gold coin. A coin is
drawn at random from a drawer selected at random. Suppose
the coin selected was silver. What is the probability that the
other coin in the drawer was gold?
Solutions:
B  event coin drawn was silver
Ai  event drawer i was picked
Only was for other coin to be gold is if drawer 4 was picked.
P( B \ A4 ) P( A4 )
P( A4 \ B) 
P( B \ A1 ) P( A1 )  P( B \ A2 ) P( A2 )  P( B \ A3 ) P( A3 )  P( B \ A4 ) P( A4 )
1
1 1
( )( )
1
2 4

8
1
1
1
1 1
5 5
(0)( )  (1)( )  (1)( )  ( )( )
4
4
4
2 4
8
Check: 5 silver coins could be picked only one of them is with a
gold coin.
SS2657a Midterm Exam, October 27th 2008
pg. 8
Q7 (10 marks): A computer is instructed to generate a random
sequence using the digits 0 through 9; repetitions are
permissible. What is the shortest length the sequence can be
and still have at least a 70% probability of containing at least
one 4?
Solutions:
1

each
digit
is
4
with
with
probability
,

10

9

or not 4 with probability
.

10
With N digits
P(at least 1 4) = 1- P(no 4s)
N
 1    (0.1)0 (0.9) N 0
0
 1  (0.9) N
Find smallest N  (0.9) N  0.7 or (0.9) N  0.3
N  12
pg. 9
SS2657a Midterm Exam, October 27th 2008
Q8 (10 marks) Experience has shown that only 1/3 of all
patients having a certain disease will recover if given the
standard treatment. A new drug is to be tested on a group of
12 volunteers. If Health Canada, which regulates new drugs,
requires that at least 7 of these patients recover before it will
licence the new drug, what is the probability that the
treatment will be rejected even though it increases the
recovery rate to ½?
Solutions:
0
1
6
P( recover)  P( recover)    P( recover)
12
12
12
12 
12 
12 
=   P 0 (1  P)120    P1 (1  P)121      P 6 (1  P)126
0
1
6
1
(here P  1  P  )
2
1
1
 12  11  1 
    12    
   
2
2
 12  2 
12
12
12
12 11 12 11 10 12 11 10  9 12 11 10  9  8 12 11 10  9  8  7 
1 
   1  12 




2
1

2
1

2

3
1

2

3

4
1

2

3

4

5
1  2  3  4  5  6 
  
12
12
2510
1
   1  12  66  220  55.9  99.8  11 12  7  
 61.3%
4096
2
SS2657a Midterm Exam, October 27th 2008
pg. 10
Q9 (6 marks): Suppose that f(x) is the probability distribution
function corresponding to a continuous random variable which
takes values only in the interval [2,3].
Which of the following sketches of f(x) cannot be valid? Why?
Check areas covered in[2,3] :
1
a ) Area = , not valid
2
b) Area is not possible to be 1, not valid
1
c) Area  , not valid
2
d ) Area  1, valid
SS2657a Midterm Exam, October 27th 2008
Q10 (10 marks): Suppose that the random variable X has
probability density function
f(x) = K, c ≤ x ≤ d,
with c < d and for K > 0.
a) In terms of c and d, what is the value of K?
Solutions:
k
1
d c
b) Find E[X]
Solutions:
1
1 1 2 d
1
d c
2
2
xdx

x
|

(
d

c
)

c
d  c c
d c 2
2(d  c)
2
d
c) Find Var[X]
Solutions:
1
d c
2
x
dx



d  c c
 2 
d
2
1 1 3 d d c

x |c  

d c 3
 2 
d 3  c 3 d 2  2cd  c 2


3(d  c)
4
2
d 2  cd  c 2 d 2  2cd  c 2


3
4
1
1
1
1
 d 2  cd  c 2  (d  c) 2
12
6
12
12
pg. 11
SS2657a Midterm Exam, October 27th 2008
pg. 12
Q11 (10 marks) Persons suffering from purple spot fever have
purple spots on their face for a period of time Y, where Y is a
random variable described by the continuous probability
density function fY(y) = 3y2/8, 0 ≤ y ≤2, (fY(y) = 0, otherwise)
where Y is measured in weeks. What is the probability that
Maxwell, who began suffering from purple spot fever today,
will still have purple spots 1 week from now, when his class
photos are taken?
Solutions:
1
3y2
1 
dy
8
0
y3 1
 1  |0
8
1 7 7
 1    
8 8 8
SS2657a Midterm Exam, October 27th 2008
pg. 13
Q12 (10 marks) If fY(y) = 2y/k2, 0 ≤ y ≤k, for what value of k
does Var(Y) = 2?
Solutions:
2y
y 2 k k 2  02
0 k 2 dy  k 2 |0  k 2  1
k
k
2 y2
2 y 3 k 2k
E Y    2 dy 
| 
2 0
k
3
k
3
0
k
2 y3
1 y4 k k 2
E Y    2 dy 
| 
2 0
k
2
k
2
0
2
2
k 2  2k 
k 2 4k 2 k 2
Var Y  
  


2  3 
2
9
18
k2
Var Y  
 2  k 2  36  k  6
18
pg. 14
SS2657a Midterm Exam, October 27th 2008
Q13 (10 marks) Suppose that the random variable Y has
probability density function fY(y) = 2/y3, y ≥ 1, (fY(y) = 0, y < 1)
a) Show that this is a valid probability density function.
Solutions:


1
2
1 
dy


|1  1
y3
y2
b) Find the expected value of Y.
Solutions:


1
 2
2y
1 
dy

dy


2
|1  2
3
2

1
y
y
y
c) However, show that Y does not have a finite variance.
Solutions:
Var ( y )  E ( y )  ( E ( y ))  
2
2

1
 lim 
R
R  1
Not finite
 2
2 y2
dy   dy
1 y
y3
2
dy  lim 2 ln R
R 
y
pg. 15
SS2657a Midterm Exam, October 27th 2008
Q14 (10 marks): Assume that the number of hits X that a
baseball teams makes in an inning of a baseball game has a
Poisson distribution. If the probability that a team makes zero
hits in an inning is 1/3:
a) what are their chances of getting two or more hits in the
first inning?
Solutions:
P( X
P( X
P( X
P( X
e  0
1
 0) 
 e     ln 3
0!
3
 1
e 
1
 1) 
  e    ln 3
1!
3
1
1
 0)  P( X  1)  ln 3  ln 3  (1  ln 3)
3
3
1
2  ln 3
 1)  1   P( X  0)  P( X  1)   1  (1  ln 3) 
 30.05%
3
3
b) What is the probability that they get 10 hits in a regular
9 inning game?
Solutions:
game  9 ln 3
9
1
10
  (9 ln 3)
9ln 3
10
e
(9 ln 3)
(3ln 3)10 ln 3
3

P(10 Hits) 


 0.12503  12.50%
10!
10!
10!
SS2657a Midterm Exam, October 27th 2008
pg. 16
Q15 (10 marks)
An experiment is performed independently 10 times.
The results are 1,1,0,0,0,0,1,1,0,1.
If the same experiment were performed again 5 times
(independent of the above results), estimate the probability
that exactly three of the five new experiments result in a 1?
Solutions:
5 successes out of 10 trials:
Estimate
P( success ) 
3
1
2
 5  1   1 
    
 3  2   2 
5 3
5  4  1  10
  
 31.25%
2  2  32
5
