ECON2222A – Midterm 1
Professor: Daniel Chaves
October 13, 2022
Name:
Student ID:
Instructions
• Show your work
• Write your calculations and answers in the exam booklet.
• Express numeric answers in either decimal form or simplified (“reduced”) fraction form. Non-reduced
fractions may be marked as incorrect.
• Use of only non-graphic calculators is allowed.
• All exam materials, INCLUDING THIS SHEET OF PAPER, must be turned in with the exam.
Useful Formulae
P
• E(X) = x xP (X = x)
P
• V ar(X) = x (x − E(X))2 P (X = x)
• short cut for the variance: V ar(X) = E(X 2 ) − (E(X))2
• If X is a Bernoulli Random Variable with probability of success p, then P (X = 1) = p and P (X =
0) = 1 − p. Furthermore, E(X) = p and V ar(X) = p(1 − p)
• If X is a Binomial RV with probability of success p for n independent Bernoulli trials, then P (X =
n!
x) = Cxn px (1 − p)n−x where Cxn = x!.(n−x)!
• If X is a Binomial RV with probability of success p for n independent Bernoulli trials, E[X] = np
• If X is a Poisson RV, then P (X = x) =
e−λ λx
x!
and E(X) = λ
Points Breakdown
Question
Points
1
20
2
10
3
10
1
4
15
5
25
6
10
7
10
Question 1
The random variable X may take the values 0,1,2,3,4. The probability distribution function for X is given by:
P (X = x) =
5−x
c
(a) (5 points) What value of c makes P (X = x) a legitimate probability distribution function? Use this
value for the remainder of the question. If you are not able to answer (a), use c = 20.
(b) (5 points) What is the expected value of X?
(c) (5 points) What is the variance of X?
(d) (5 points) What is the probability that X is an odd number?
Solution
P4
(5 − 0) + (5 − 1) + (5 − 2) + (5 − 3) + (5 − 4)
(a)
= 1 ⇒ c = 15
x=0 P (X = x) = 1 ⇒
c
P
4
(b) Using c = 15, E(X) = x xP (X = x) = ≈ 1.33
3
P
14
(c) Using c= 15, V ar(X) = x (x − E(X))2 P (X = x) =
≈ 1.56
9
2
(5 − 1) + (5 − 3)
= = 0.4
(d) P (X is an odd number) = P (X = 1) + P (X = 3) =
15
5
Question 2
The London Police Department keeps detailed information on drunk driving and car accidents. They estimated that in any given day, 20% of drivers are under the influence of alcohol. Furthermore, their records
show that when a driver consumes alcohol, there is a 10% chance the driver will be involved in a car accident.
When a driver does not consume alcohol, there is a 1% chance the driver will be involved in a car accident.
An officer is called to the scene of a car accident. What is the probability the driver was under the influence
of alcohol?
Solution
P (alcohol|accident) =
20% ∗ 10%
5
= ≈ 0.7142
20% ∗ 10% + 80% ∗ 1%
7
Question 3
Of 500 first year Social Sciences undergraduate students, 100 were chosen at random to receive a letter
containing detailed information about the advantages of pursuing an Economics major. It is known that
receiving the letter increases the chance of a student declaring Economics as its major: 30% of those that
do not receive the letter declare Economics as their major and 50% of those that receive the letter declare
economics as their major.
(a) (5 points) What is the probability that a randomly chosen student both received the letter and declared
Economics as its major?
(b) (5 points) What is the probability that a student who declared Economics as its major had received
the letter?
Solution
2
(a)
P (received the letter) =
100
= 20%
500
⇒ P (received the letter & declared Economics) = 20% ∗ 50% = 10%
(b) P (received the letter|declared Economics) =
Question 4
10%
5
=
≈ 0.2941
10% + 80% ∗ 30%
17
An assistant professor at Western University is about to submit five papers for peer review. For each paper,
it is estimated that the probability of acceptance is 0.3 and the probability of rejection is 0.7. The outcomes
of the five submissions are independent of one another.
(a) (5 points) What is the probability that all papers are accepted?
(b) (5 points) What is the expected number of rejections?
(c) (5 points) Assume that the assistant professor will be promoted if she publishes at least three papers and fired otherwise. What is the probability she is promoted?
Solution
(a) 0.35 = 0.00243
(b)
0.7 ∗ 5 = 3.5
(c)
P (promoted) = P (X ≥ 3) = 5 C3 ∗ 0.33 ∗ 0.72 + 5 C4 ∗ 0.34 ∗ 0.7 + 0.35 = 0.16308
Question 5
Let X = {1, 2, 3} denote the number of customers that enter an used car dealership in a given day. The
probability distribution function of X is such that P (X = 1) = P (X = 2) = P (X = 3) = 1/3.
Customers purchase one car or no car. The purchase decision is independent across customers. Also,
conditional on entering the dealership, the probability a customer will purchase a car is 0.4.
Let S be the random variable representing the number of cars sold in a given day.
(a) (5 points) What are the possible realizations of S?
(b) (7.5 points) What is the probability distribution function of the random variable S?
(c) (5 points) What is the expected number of customers that enter the dealership in any given day?
(d) (7.5 points) Assuming that the dealership sells each car for $10,000, then what is the expected daily
revenue?
Solution
(a) S = {0, 1, 2, 3}
1
1
1
∗ 0.6 + ∗ 0.62 + ∗ 0.63 = 0.392
3
3
3
1 2
1
∗ 0.4 + ∗ C1 ∗ 0.4 ∗ 0.6 + ∗ 3 C1 ∗ 0.4 ∗ 0.62 = 0.4373
3
3
1
∗ 0.42 + ∗ 3 C2 ∗ 0.42 ∗ 0.6 = 0.1493
3
(b) P (S = 0) =
1
3
1
P (S = 2) =
3
P (S = 1) =
3
1
∗ 0.43 ≈ 0.0213
3
1
E[X] = (1 + 2 + 3) = 2
3
P (S = 3) =
(c)
(d) E[S] = 1 ∗ 0.4373 + 2 ∗ 0.1493 + 3 ∗ 0.0213 = 0.808
⇒ E[Revenue] = 0.808 ∗ 10000 = $8080
Question 6
Suppose that we have a data set including race (white or non-white) and employment status (employed or
unemployed). Define W to be the event that our randomly chosen person is white and define E to be the
event that this person is employed. The joint probabilities are given in the following table:
White
Non-white
Total
Employed
a
0.4
0.7
Unemployed
0.10
b
0.3
Total
0.4
c
(a) (5 points)Considering the information on the table, what is a+b+c?
(b) (5 points) What is P (E|W )?
Solution
(a) a = 0.3; b = 0.2; c = 0.6 ⇒ a + b + c = 1.1
(b) P (E|W ) =
0.3
= 0.75
0.4
Question 7
We have the following list of numbers: {2, 3, 4, 5, 7, 9}
(a) (5 points) How many four-digit numbers can we form?
(b) (5 points) How many of these four-digit numbers are smaller than 3500?
Solution
• Assume numbers are taken with replacement:
(a) 64 = 1296
(b) Numbers smaller than 3000 (or numbers with 2 as the first digit): 63 = 216
Numbers greater than 3000 but smaller than 3500: 3 ∗ 62 = 108
Hence, there are 216 + 108 = 324 four-digit numbers smaller than 3500.
• Assume numbers are taken without replacement:
(a) 6P4 = 360
(b) Numbers that start with 2: 5P3 = 60
Numbers between 3000 and 3500: 2 ∗ 4P2 = 24
Hence, there are 84 four-digit numbers smaller than 3500.
4