MCB1007 Introduction to Probability and Statistics Fall 2017-2018
Midterm Exam November 9, 2017
No:
Name:
Section:
Justify your answers to get full credit − Exam Duration 90 Minutes 1) A multiple-choice test consists of 40 questions, each permitting a choice of three
alternatives, A, B and C, only one of which is correct. A student has to answer all questions. What is the number of ways this multiple choice exam can be answered if no two consecutive answers are the same?
Answer. A student can pick 1 out of 3 answers for the first question and then for the remaining 39 questions the student has 2 choices since the answer to the previous question is eliminate one of the choices so the answer is
3 1
239= 3 × 239.
2) The density function of the random variable X is given by
f (x) =
1 − cx, 0 < x < 2, 0, otherwise.
a) What is the value of the constant c?
Answer. 1 =R∞
−∞f (x)dx =R2
0(1 − cx)dx =
x − cx222
0 = 2 − 4c2 = 2 − 2c ⇒ c = 12.
1 / 10
2a / 8
2b / 8
3a / 5
3b / 5
3c / 5
4a / 7
4b / 7
5a / 10
5b / 7
6 / 9
7a / 10
7b / 9
P /100
b) Find the distribution function of the random variable X.
Answer. F (x) = Rx
−∞f (t)dt i) If x ≤ 0: F (x) =Rt
−∞f (t)
|{z}
0
dt = 0.
ii) If 0 < x < 2: F (x) =Rx
−∞f (t)dt =R0
−∞f (t)
|{z}0
dt +Rx 0 f (t)
|{z}
1−t/2
dt =
t − t42x
0 = x − x42. iii) If x ≥ 2: F (x) =Rx
−∞f (t)dt =R0
−∞f (t)
|{z}0
dt +R2 0 f (t)
|{z}
1−t/2
dt +R∞ 0 f (t)
|{z}0
dt =
t − t422 0 = 1
F (x) =
0, x ≤ 0,
x − x42, 0 < x < 2,
1, x ≤ 2.
3) In a dice game four 6-sided fair dice are rolled and a winning roll has two different pairs of numbers (for example 1144, 2323, 5665 etc.).
a) Find the probability of a winning.
Answer. Let A: winning a game. Then
P (A) = n N =
6 2
4
2
64 =
6!
2!4!
4!
2!2!
1296 = 15 × 6
1296 = 90 1296 = 5
72 = 0.069¯4 b) What is the probability of winning at most one of the 5 rolls?
Answer. The Binomial distribution with parameters n = 5, x = 0, 1 and θ = 725:
P (X ≤ 1) = b(0; 5, 5/72) + b(1; 5, 5/72)
=5 0
5 72
0 1 − 5
72
5−0
+5 1
5 72
1 1 − 5
72
5−1
= 67 72
5
+ 5 × 5 72
67 72
4
= 0.69777 + 0.26039 = 0.95816.
c) What is the probability of winning the game 4th times on the 10th roll?
Answer. The Negative Binomial distribution with parameters x = 10, k = 4 and θ = 725:
b∗(10; 4, 5/72) =9 3
5 72
4 1 − 5
72
6
= 0.0012685.
4) The probability distribution of the random variable X is given by f (x) = 35 25x−1
, x = 1, 2, 3, · · · . a) Find the moment generating function of X. (Hint: P∞
x=1arx−1 = 1−ra if |r| < 1) Answer.
MX(t) = E(etX) = X∞ n=1
etxf (x) = X∞ n=1
etx3 5
2 5
x−1
= X∞ n=1
3 5
5 2
2et 5
x
= X∞ n=1
3 2
2et 5
2et 5
x−1
=
3et 5
1 −2e5t = 3et
5 − 2et where
2et 5
< 1 ⇔ et< 5
2 ⇔ t < ln5 2.
b) Use the moment generating function of X to determine the mean of X.
Answer. Since MX(t) = 5−2e3ett ⇒ MX′ (t) = 3et(5−2e(5−2et)+2et)2 tet = (5−2e15ett)2, thus the mean is
µ = MX′ (0) = 15e0
(5 − 2e0)2 = 5 3.
5) The probability that an alarm system will work when there is a danger in a construction site is 0.99, the probability of the alarm system will work if there is no danger is 0.02, and the probability of occurrence of any danger in this construction site is 0.03.
a) What is the probability of a danger, given that the alarm system works?
Answer. Let A: alarm system works, D: there is a danger
P (D|A) = P (A ∩ D)
P (A) = P (D)P (A|D)
P (D)P (A|D) + P (D′)P (A|D′) = 0.03 × 0.99
0.03 × 0.99 + 0.97 × 0.02
= 297
491 = 0.60489.
b) What is the probability that the alarm system will not work in case of a danger?
Answer.
P (D ∩ A′) = P (D) × P (A′|D) = 0.03 × (1 − 0.99) = 3
10000 = 0.0003.
6) Six men and five women apply for an executive position in a small company. Two of the applicants are selected (at random) for interviews. How many women do you expect to be in the interview pool?
Answer. X: the number of women in the interview pool, x = 0, 1, 2. Since
f (0) =
6 2
11 2
= 3
11, f (1) =
6 1
5
1
11 2
= 6
11, f (2) =
5 2
11 2
= 2 11, then we find
E(X) = X2
x=0
xf (x) = 0 × 3
11 + 1 × 6
11+ 2 × 2 11 = 10
11.
7) We roll two dice. The first die has three ‘1’ faces and three ‘2’ faces. Each face is equally likely to come up. The second die has two ‘1’ faces, two ‘2’ faces, and two ‘3’ faces, also with equally weighted sides. Let X and Y be random variables:
X : how many ‘2’s rolled, Y : sum of these dice.
(a) Determine the joint probability distribution of X and Y .
Answer. X : how many ‘2’s rolled, x = 0, 1, 2; Y : sum of these dice, y = 2, 3, 4, 5
❍❍1 ❍❍❍
2
1 1 1 2 2 2
1 (0,2) (0,2) (0,2) (1,3) (1,3) (1,3)
1 (0,2) (0,2) (0,2) (1,3) (1,3) (1,3)
2 (1,3) (1,3) (1,3) (2,4) (2,4) (2,4)
2 (1,3) (1,3) (1,3) (2,4) (2,4) (2,4)
3 (0,4) (0,4) (0,4) (1,5) (1,5) (1,5)
3 (0,4) (0,4) (0,4) (1,5) (1,5) (1,5)
❅❅
❅❅
y
x 0 1 2 h(y)
2 6/36 6/36
3 12/36 12/36
4 6/36 6/36 12/36
5 6/36 6/36
g(x) 12/36 18/36 6/36 1 (b) Find the conditional variance of Y given X = 1.
Answer. w(y|1) = f (1, y)
g(1) = 3618f (1, y) ⇒ w(y|1) =
36
18f (1, 2) = 0, y = 2
36
18f (1, 3) = 1218, y = 3
36
18f (1, 4) = 0, y = 4
36
18f (1, 5) = 186 , y = 5 µY |1 = E(Y |1) =P5
y=2yw(y|1) = 2w(2|1) + 3w(3|1) + 4w(4|1) + 5w(5|1) = 3 × 1218+ 5 × 186 = 11 3 E(Y2|1) =P5
y=2y2w(y|1) = 22w(2|1) + 32w(3|1) + 42w(4|1) + 52w(5|1) = 32×1218 + 52× 186 = 43 3 σY |12 = E(Y2|1) − (E(Y |1))2 = 43
3 − 11 3
2
= 8
9 = 0.¯8.
