İ s t a n b u l K ü l t ü r U n i v e r s i t y MCB1007
Introduction to Probability and Statistics First Midterm
Fall 2014-2015
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.
Arzu Yemişçi, PhD.
Question 1. Question 4.
Question 2. Question 5.
Question 3. Question 6.
TOTAL
8 + 8 points (a) How many odd positive integers less than 300 can be formed using the digits 0, 1, 2, 3, and 4, without repetition?
Answer. Since we only use the digits 0, 1, 2, 3, and 4 the number of 1-digit numbers is: 2 = 2
2-digit numbers is: 3 2 = 6
3-digit numbers is: if the last number is 1 then 1 3 1 = 3, if the last number is 3 then 2 3 1 = 6
thus, 2 + 6 + (3 + 6) = 17 different numbers can be formed under the given conditions.
(b) In İstanbul city, all vehicle license plates have 2 letters from the 23 letters of the alphabet followed by 4 one-digit numbers from0, 1, 2, · · · , 9. How many different license plates are possible for İstanbul if repetition is allowed?
Answer. There are 23 choices for the first letter. For each of these, there are 23 choices for the second letter. So, there are23 × 23 = 529 possible pairs of letters. On the other hand, there are 10 possibilities for each of the first, the second, the third and the fourth digits. This means that there are10×10×10×10 = 10.000 different numbers including the combination 0 0 0 0. Since there is no license plates which ends with the numbers 0 0 0 0, the total number of different pallets is
529 × (10.000 − 1) = 5.289.471.
7 + 10 points
(a) An urn contains six balls numbered 1 through 6. Three balls are randomly drawn from the urn in succession, without replacement. What is the probability that the smallest number in this sampling equal to 2?
Answer. There are 6
3
= 6!
3! · 3! = 20 different ways to choose 3 balls from 6. Since 2 is the smallest number in this sampling, we need to choose 2 extra numbers from 3, 4, 5 and 6, which are bigger than two. There are4
2
= 4!
4! · 4! = 6 different ways to do that.
So the probability that we are seeking for is n N = 6
20. (b) Find the constant term in the expansion of
1
x2 − 2x3
10 .
Answer. Since the constant term is the coefficient of the term x0 in the Binomial expansion of the given expression
1
x2 − 2x3
10
=10
r=0
10 r
1 x2
10−r
(−2x3)r=10
r=0
10 r
x2r−20(−2)rx3r
10 10
5 + 10 points The density function of the continuous random variable X is given by
f(x) =
c(x +√
x), for 0 < x < 1,
0, otherwise.
(a) What is the constant c?
Answer.
1 =
∞
−∞
f(x)dx =
0
−∞
f(x)dx +
1
0
f(x)dx +
∞
1
f(x)dx =
1
0
f(x)dx
=
1
0
c(x + x1/2)dx = c
x2
2 +x3/2 3/2
1 0
= c
1 2 +2
3
= c7
6 ⇒ c = 6 7. (b) Find the distribution function of the random variable X.
Answer.
If x≤ 0 then F(x) = x
−∞f(t)dt = 0 If0 < x < 1 then F (x) = x
−∞f(t)dt = 0
−∞f(t)dt + x
0 f(t)dt
= 6 7
x
0(t +√
t)dt = 6 7
t2
2 + t3/2 3/2
x 0 = 6
7
x2
2 +2x3/2 3
If x≥ 1 then F(x) = x
−∞f(t)dt = 1.
Thus
F(x) =
⎧⎪
⎪⎪
⎨
⎪⎪
⎪⎩
0, x≤ 0,
6 7
x2
2 + 2x3/2 3
, 0 < x < 1,
1, x≥ 1.
10 points Factories A, B and C produce a textile product. Factory A produces 2 times as many textile products as Factory B and Factory C. Factory A and Factory B produce defective products 5% of the time and Factory C produces defective products 1% of the time.
A textile product is selected at random and it is found to be defective. What is the probability it came from Factory C?
Answer. Let A be event that the textile product produced in Factory A, B be event that the textile product produced in Factory B, C be event that the textile product produced in Factory C, and D be event that the selected textile product is defective.
Since Factory A produces 2 times as many textile products as Factory B and Factory C, we have that P(A) = 0.50, P (B) = 0.25, P (C) = 0.25. Also it is given that P(D|A) = 0.05, P (D|B) = 0.05 and P (D|A) = 0.01. Using Bayes’ Theorem, we obtain that
P(C|D) = P(C ∩ D)
P(A ∩ D) + P (B ∩ D) + P (C ∩ D)
= 0.25 · 0.01
0.50 · 0.05 + 0.25 · 0.05 + 0.25 · 0.01 = 1
16 = 0.0625.
10 + 7 points
Suppose that a couple will continue having children until have a boy, under the assump- tion that they have ability to have children and theoretically number of births goes to infinity. So, if they have a female child they keep having more children until they have a boy. If they have a boy, they stop having children. Let X be the number of births.
Assume that outcomes of births are independent of each other, and boys and girls are equally likely.
(a) Find the probability distribution of the random variable X?
Answer. Since boys and girls are equally likely Number of Births
to First Boy Point Probability
1 B 12
2 GB 12 · 12 = 212
3 GGB 12 · 12 · 12 = 213
... ... ...
n GGG · · · G
(n−1) times
B 12 ·12 · · ·12 = 21n
... ... ...
thus the probability distribution of the random variable X is f(x) = 1 where x =
10 + 10 + 5 points 2 balls are selected at a random from an urn without replacement containing 3 blue and 5 white balls. Let the random variable X is the number of blue balls in the first draw and Y is the number of white balls in the second draw.
(a) Find the joint probability distribution as a table.
Answer. Since
P(X = 0, Y = 0) = f(0, 0) = 5 8 ·3
7 = 15
56, P(X = 1, Y = 0) = f(1, 0) = 3 8 ·2
7 = 6 56, P(X = 1, Y = 1) = f(1, 1) = 3
8 ·5 7 = 15
56, P(X = 0, Y = 1) = f(0, 1) = 5 8 ·4
7 = 20 56, thus
HHx HHHH
y 0 1 g(x)
0 15/56 20/56 35/56
1 6/56 15/56 21/56
h(y) 21/56 35/28 1 (b) Find the conditional distribution of Y given X = 1.
Answer. Conditional distribution of Y given X = 1 is w(y|1) = f(1, y)
g(1) = 56
21· f(1, y) which can be written as
w(y|1) =
56
21 · f(1, 0) = 5621· 566 = 216 for y = 0,
5621 · f(1, 1) = 5621· 1556 = 1521 for y = 1.
(c) Determine whether or not X and Y are independent.
Answer. X and Y are independent if the equality f(x, y) = g(x) · h(y) holds true for X = 0, 1 and Y = 0, 1. Let we consider the point where X = 0 and Y = 0. Since f(0, 0) = 1556, g(0) = 3556 and h(0) = 2156 it is easy to see that
f(0, 0) = 15 56 = 35
56· 21
56 = g(0) · h(0) thus X and Y are dependent.
