MIE236: Probability
Department of Mechanical and Industrial Engineering
Midterm 1
Thursday October 7 12:10 p.m. – 2 p.m. EDT
34 marks
Solutions
Question 1 (13 marks)
Anne is responsible for all the drinks and beverages for their upcoming class trip. For the
class of 20 students, there are 20 drinks in total, 8 of which are bottled water, 8 are iced tea,
and 4 are energy drinks. The drinks are distributed to the class randomly.
(a) Anne asks the class to line up to pick up their drinks. How many different arrangements of queues can Anne possibly see? (An expression is fine) (1 marks)
Solution: 20!
(b) Among all the classmates, Anne has 6 good friends. What is the probability that
exactly 2 of her good friends will receive iced tea? (2 marks)
Solution:
(82)(12
4)
(20
6)
= 0.357
(c) What is the probability that at least 1 of her good friends get an energy drink? (3
marks)
Solution:
Let 𝐸 be the event that at least one friend gets an energy drink.
𝑃(𝐸)̄ =
(16
6)
(20
6)
= 0.207
𝑃(𝐸) = 1 − 𝑃(𝐸)̄ = 0.793
(d) Among all 20 classmates, how many different distributions of drinks are possible?
(2 marks)
Solution:
20!
= 62, 355, 150
8!8!4!
While each person was given a random drink, everyone has exactly one preferred drink. If
they receive their preferred drink, they are satisfied. Anne surveys all 20 people (including
herself) about whether they are satisfied with their drinks or not.
25% of people who received bottled water were satisfied with their drink
50% of people who received iced tea were satisfied with their drink
75% of people who received an energy drink were satisfied with their drink
(e) What is the probability that a randomly selected person is satisfied with their drink?
(3 marks)
Solution:
Let 𝑆 be the event of being satisfied with the drink
Let 𝐵, 𝑇, 𝐸 be the event of receiving bottled water, iced tea and energy drink respectively.
𝑃(𝑆 ∣ 𝐵) = 0.25 𝑃(𝑆 ∣ 𝑇) = 0.5 𝑃(𝑆 ∣ 𝐸) = 0.75
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𝑃(𝐵) = 8/20
𝑃(𝑇) = 8/20
𝑃(𝐸) = 4/20
𝑃(𝑆) = 𝑃(𝑆 ∣ 𝐵)𝑃(𝐵) + 𝑃(𝑆 ∣ 𝑇)𝑃(𝑇) + 𝑃(𝑆 ∣ 𝐸)𝑃(𝐸)
= 0.25(8/20) + 0.5(8/20) + 0.75(4/20)
= 0.45
(f) Is the event of getting an iced tea independent from drink satisfaction? Why or why
not? (2 marks)
Solution:
𝑃(𝑆 ∣ 𝑇) = 0.5, and 𝑃(𝑆) = 0.45. Since 𝑃(𝑆 ∣ 𝑇) ≠ 𝑃(𝑆), drink choice satisfaction and
receiving iced tea are not independent events.
Question 2 (6 marks)
A box has two 1-dollar coins and four 2-dollar coins. Three coins are selected at random,
without replacement. Find the probability distribution of the total value 𝑇 of the three
selected coins. Express the probability distribution as a histogram or a probability mass
function plot.
Solution:
Let 1 and 2 be the event of selecting a 1-dollar coin and 2-dollar coin respectively.
𝑆 = {1 1 2, 1 2 1, 2 1 1, 1 2 2, 2 1 2, 2 2 1, 2 2 2}
Possible values of 𝑇 = {4, 5, 6}
𝑃(𝑇 = 4) =
𝑃(𝑇 = 5) =
𝑃(𝑇 = 6) =
t
𝑃(𝑇 = 𝑡)
4
0.2
5
0.6
(22)(41)
(63)
(21)(42)
(63)
(20)(43)
(63)
= 0.2
= 0.6
= 0.2
6
0.2
Question 3 (8 marks)
Metalmec is a company that manufactures aluminum casts. The time in hours that a die
cast takes to solidify and cool can be modelled as a random variable 𝑋 with the following
probability density function:
⎧
𝑘𝑥,
0≤𝑥<2
{
{
𝑥
𝑓 (𝑥) = ⎨1 − 4 , 2 ≤ 𝑥 < 4
{
{0
otherwise.
⎩
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(a) Find the value of 𝑘 so that 𝑓 (𝑥) is correctly defined. (5 marks)
Solution:
∫
∞
−∞
𝑓 (𝑡) 𝑑𝑡 = 1
2
4
𝑡
∫ 𝑘𝑡 𝑑𝑡 + ∫ (1 − ) 𝑑𝑡 = 1
4
0
2
2
4
𝑘𝑡2 ∣
𝑡2 ∣
∣∣ + (𝑡 −
)∣ = 1
2 ∣0
2(4) ∣∣2
2𝑘 +
1
=1
2
1
𝑘=
4
(b) Give an expression for the cumulative distribution function 𝐹(𝑥) (3 marks)
Solution:
For 𝑥 < 2
𝐹(𝑥) = ∫
𝑥
−∞
𝑥 1
=∫
=
0 4
𝑥
𝑡2
𝑓 (𝑡) 𝑑𝑡
𝑡 𝑑𝑡
𝑥2
∣
∣∣ =
8 ∣0
8
For 𝑥 ≤ 𝑥 < 4
2
𝑥
𝑡2 ∣
𝑡
𝐹(𝑥) = ∣∣ + ∫ (1 − ) 𝑑𝑡
8 ∣0
4
2
𝑥
1
𝑡2 ∣
= + (𝑡 − )∣∣
2
8 ∣2
𝑥2
=𝑥−
−1
8
⎧0
{ 2
{𝑥
{
𝐹(𝑥) = ⎨ 8
{𝑥 −
{
{1
⎩
𝑥<0
0≤𝑥<2
𝑥2
8
−1
2≤𝑥<4
𝑥≥4
Question 4 (7 marks)
A software company is developing a spam filter for their email service. After analyzing a
large number of emails, the company has found:
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20% of all emails are spam
the probability that an email comes from a source that the recipient knows is 0.85
if an email is from a known source, the probability that it is spam is 0.10
If you select an email at random,
(a) if the email is known to be spam, what is the probability that the email is from a
known source? (3 marks)
Solution:
Let 𝑆 be the event that an email is spam
Let 𝐾 be the event that an email comes from a known source.
𝑃(𝑆) = 0.2
𝑃(𝐾 ∣ 𝑆) =
𝑃(𝐾) = 0.85
𝑃(𝑆 ∣ 𝐾) = 0.1
𝑃(𝑆 ∣ 𝐾)𝑃(𝐾)
(0.1)(0.85)
=
= 0.425
𝑃(𝑆)
0.2
(b) what is the probability that the email is spam if it you know it comes from an unknown source? (2 marks)
Solution:
𝑃(𝑆 ∣ 𝐾)̄ =
𝑃(𝐾̄ ∣ 𝑆)𝑃(𝑆)
(1 − 0.425)(0.2)
=
= 0.767
0.15
𝑃(𝐾)̄
(c) what is the probability that the email comes from an unknown source if you know
it was not spam? (2 marks)
Solution:
𝑃(𝐾̄ ∣ 𝑆)̄ =
̄
𝑃(𝑆 ̄ ∣ 𝐾)𝑃(
𝐾)̄
(1 − 0.767)(0.15)
=
= 0.0437
̄
0.8
𝑃(𝑆)
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