İstanbul Kültür University
Faculty of Engineering
MCB1007
Introduction to Probability and Statistics
First Midterm
Fall 2013-2014
Solutions
– You have 90 minutes to complete the exam. Please do not leave the examination
room in the first 30 minutes of the exam. There are six questions, of varying credit
(100 points total). Indicate clearly your final answer to each question. You are
allowed to use a calculator. During the exam, please turn off your cell phone(s).
You cannot use the book or your notes. You have one page for “cheat-sheet” notes
at the end of the exam papers. The answer key to this exam will be posted on
Department of Mathematics and Computer Science board after the exam.
Good luck!
Emel Yavuz Duman, PhD.
M. Fatih Uçar, PhD.
Question 1.
Question 2.
Question 3.
Question 4.
Question 5.
Question 6.
TOTAL
15 points
There are n married couples in a party. All the participants shake each other’s hands
only once except his/her partner. What is the total number of handshakes at the party?
Answer. When two people shake hands, we can think of them as forming a temporary
“handshaking committee”. The total number of handshakes will be the same as the
number of ways of forming a committee of 2 people from 2n people (There are 2n
people at the party since the party consist of n married couples). As the 2 choices are
not ordered, we are counting combinations; thus the total number of handshakes is
2n
2
including partners ones. Since all the participants shake each other’s hands only once
except his/her partner, considering there are n partners in the party we obtain that
the total number of handshakes at the party is
2n
2n!
2n(2n − 1)
−n=
−n=
− n = 2n(n − 1)
2
(2n − 2)!2!
2!
5 + 10 points
A company decided to choose 6 of its employees by drawing and give them a weekend
holiday for every weekend during one year.
(a) What should be the minimum number of employees of this company if all holiday groups
are different then each other?
Answer. Let n denote the number of employees
working for the company. Since there are
n
52 weekends in a year, the inequality 6 ≥ 52 should be satisfied. Using the definition
of a combination, it is easy to see that n ≥ 6. Thus
for n = 6 then 66 = 1 < 52,
for n = 7 then 76 = 7 < 52,
8!
for n = 8 then 86 = 6!2!
= 28 < 52,
9
9!
for n = 9 then 6 = 6!3!
= 84 ≥ 52,
So, the minimum number of employees of this company is 9.
(b) It is given that the number of the employees of this company is equal to the minimum
number that you find in part (a). Also we know that two brothers are working for this
company. What is the probability of selecting their names consecutively in the first
drawing?
Answer. Let we define an event A = {Brother’s names are selected consecutively in
the first drawing}. Since the number of employees working for the company is 9, the
probability of selecting their names consecutively in the first drawing is
27
5!2!
P (A) = 2 4
=
9 P6
MCB1007 - Int. to Prob. and Statistics
2
7!
5!2!
4!3!
9!
3!
=
5
.
36
First Midterm
10 + 10 points
(a) Find the coefficient of
1
x4
in the expansion of
√1
x
+
1
√
3 2
x
7
.
Answer. Using the Binomial coefficient, we obtain
1
1
√ +√
3
x
x2
7
=
7 7 r=0
7 r
x−1/2
7−r −2/3 r
x
7 (r−7)/2 −2r/3
=
x
x
r
r=0
7 7 7 r−7 − 2r 7 − r+21
2
3
=
=
x
x 6 .
r
r
r=0
r=0
Since
r + 21
= 4,
6
thus we see that r = 3. So, the coefficient of x−4 is
7
7!
= 35.
=
3!4!
3
x−
r+21
6
= x−4 ⇒
(b) In a group of 6 married couple, 4 people are selected at random. What is the probability
that NOT married couple is selected?
Answer. Let we define an event A = {only one person from a couple is selected}. So,
the probability that not married couple selected is
6
24
16
15 · 16
= .
P (A) = 412 =
495
33
4
MCB1007 - Int. to Prob. and Statistics
3
First Midterm
15 points
Show that if events A and B are independent then events A and B are independent.
Answer. Since A and B are independent events then we know that A and B are also
independent. So, P (A ∩ B ) = P (A)P (B ). On the other hand, it is easy to see that the
B = (A ∩ B ) ∪ (A ∩ B ).
Since A ∩ B and A ∩ B are mutually exclusive, and A and B are independent by the
assumption, we have
P (B ) = P [(A ∩ B ) ∪ (A ∩ B )]
= P (A ∩ B ) + P (A ∩ B ) (by Postulate 3)
= P (A)P (B ) + P (A ∩ B ).
It follows that
P (A ∩ B ) = P (B ) − P (A)P (B )
= P (B )[1 − P (A)]
= P (B )P (A)
hence that A and B are independent.
7 + 8 points
A continuous random variable X has the following probability density function
kx−4 , x > 1,
f (x) =
0,
elsewhere.
(a) Find k.
Answer.
∞
1
∞
c
f (x)dx =
f (x)dx +
f (x)dx = lim
kx−4 dx
c→∞
−∞
−∞
1
1
c
k
x−3 k
k
= − lim (c−3 − 10 ) = − (0 − 1) = ⇒ k = 3.
= lim k
c→∞
−3 1
3 c→∞
3
3
1=
(b) Find the distribution function of the random variable X.
Answer.
If x ≤ 1 then F (x) =
x
f (u)du = 0
1
x
If x > 1 then F (x) = −∞ f (u)du = −∞ f (u)du + 1 f (u)du
x
x −4
3u−3 = 1 3u du = − 3 = 1 − x13
−∞
x
F (x) =
MCB1007 - Int. to Prob. and Statistics
4
1
0,
1−
1
,
x3
x ≤ 1,
x > 1.
First Midterm
10 + 7 + 3 points
Suppose that 3 calculators are randomly chosen without replacement from the following
group of 10 calculators: 7 new, 1 used (working) and 2 out of order (not working). Let
X denotes the number of new calculators chosen and Y denotes the number of used
calculators chosen.
(a) Find the joint probability distribution table.
Answer. Though X can take on values 0, 1, 2 and 3, and Y can take on values 0 and
1, when we consider them
X + Y ≤ 3. So, not all combinations of (X, Y ) are
jointly,
=
120
different ways to choose 3 out of 10, then
possible. Since there are 10
3
72
712
1
7
14
, f (1, 0) = 1 2 =
, f (1, 1) = 1 1 1 =
f (0, 1) = 1 2 =
120
120
120
120
120
120
72
71
7
42
21
35
f (2, 0) = 2 1 =
, f (2, 1) = 2 1 =
, f (3, 0) = 3 =
.
120
120
120
120
120
120
12
Therefore, we obtain the joint probability distribution P (X = x, Y = y) = f (x, y) for
(X, Y ):
HH
HH y
x
HH
0
1
2
3
h(y)
0
1
1/120
7/120 14/120
42/120 21/120
35/120
84/120 36/120
g(x)
1/120
21/120
63/120
35/120
1
(b) Find the conditional distribution of Y given X = 2.
Answer. Since the conditional distribution of Y given X = 2 is given by
w(y|2) =
f (2, y)
f (2, y)
=
,
g(2)
63/120
then
w(0|2) =
42/120
f (2, 1)
21/120
f (2, 0)
=
= 42/63, w(1|2) =
=
= 21/63.
63/120
63/120
63/120
63/120
(c) Determine whether or not X and Y are independent.
Answer. If X and Y are independent then f (x, y) = g(x)h(y) for all x = 0, 1, 2, 3 and
y = 0, 1. Let we consider (x, y) = (0, 0). Since
f (0, 0) = 0 =
83
1
·
= g(0) · h(0),
120 120
we see that X and Y are dependent.
MCB1007 - Int. to Prob. and Statistics
5
First Midterm