İ s t a n b u l K ü l t ü r U n i v e r s i t y 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 4.
Question 2. Question 5.
Question 3. 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 2
− n = 2n!
(2n − 2)!2! − n = 2n(2n − 1)
2! − n = 2n(n − 1)
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 52 weekends in a year, the inequality n
6
≥ 52 should be satisfied. Using the definition of a combination, it is easy to see that n≥ 6. Thus
for n= 6 then 6
6
= 1 < 52, for n= 7 then 7
6
= 7 < 52, for n= 8 then 8
6
= 6!2!8! = 28 < 52, for n= 9 then 9
6
= 6!3!9! = 84 ≥ 52,
So, the minimum number of employees of this company is9.
(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
P(A) =
2
2
7
4
5!2!
9P6 = 4!3!7! 5!2!
9!3!
= 5 36.
10 + 10 points (a) Find the coefficient of x14 in the expansion of
√1 x + √31
x2
7 . Answer. Using the Binomial coefficient, we obtain
√1
x + 1
√3
x2
7
=
7 r=0
7 r
x−1/27−r
x−2/3r
=
7 r=0
7 r
x(r−7)/2x−2r/3
=7
r=0
7 r
xr−72 −2r3 =7
r=0
7 r
x−r+216 .
Since
x−r+216 = x−4 ⇒ r+ 21 6 = 4, thus we see that r= 3. So, the coefficient of x−4 is
7 3
= 7!
3!4! = 35.
(b) In a group of6 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
P(A) =
6
4
24
12
4
= 15 · 16 495 = 16
33.
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
f(x) =
kx−4, x >1, 0, elsewhere.
(a) Find k.
Answer.
1 = ∞
−∞
f(x)dx = 1
−∞
f(x)dx + ∞
1
f(x)dx = lim
c→∞
c
1
kx−4dx
= lim
c→∞kx−3
−3 c
1 = −k 3 lim
c→∞(c−3− 10) = −k
3(0 − 1) = k
3 ⇒ k = 3.
(b) Find the distribution function of the random variable X.
Answer.
If x≤ 1 then F (x) = x
−∞f(u)du = 0 If x >1 then F (x) = x
−∞f(u)du = 1
−∞f(u)du + x
1 f(u)du
= x
1 3u−4du= −3u3−3 x
1 = 1 − x13
F(x) =
0, x≤ 1, 1 −x13, x >1.
10 + 7 + 3 points Suppose that3 calculators are randomly chosen without replacement from the following group of10 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 jointly, X + Y ≤ 3. So, not all combinations of (X, Y ) are possible. Since there are 10
3
= 120 different ways to choose 3 out of 10, then
f(0, 1) =
1
1
2
2
120 = 1
120, f(1, 0) =
7
1
2
2
120 = 7
120, f(1, 1) =
7
1
1
1
2
1
120 = 14
120 f(2, 0) =
7
2
2
1
120 = 42
120, f(2, 1) =
7
2
1
1
120 = 21
120, f(3, 0) =
7
3
120 = 35 120.
Therefore, we obtain the joint probability distribution P(X = x, Y = y) = f(x, y) for (X, Y ):
HHx HHHH
y 0 1 g(x)
0 1/120 1/120
1 7/120 14/120 21/120 2 42/120 21/120 63/120
3 35/120 35/120
h(y) 84/120 36/120 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)
g(2) = f(2, y) 63/120, then
w(0|2) = f(2, 0)
63/120 = 42/120
63/120 = 42/63, w(1|2) = f(2, 1)
63/120 = 21/120
63/120 = 21/63.
(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 = 1 120 · 83
120 = g(0) · h(0), we see that X and Y are dependent.
