ISyE 3770 — Spring 2026
Homework #1 – Due Thursday, Jan. 29th
Only the 1st five problems will be graded, any additional problems are for practice.
1. Suppose I roll a four-sided die (like a standard die, but only has 4 sides), flip a
coin, and record the paired results as the outcome of my experiment. What is the
sample space of this experiment?
The experiment consists of rolling a four-sided die (outcomes 1,2,3,4) and flipping a coin
(outcomes H, T). The sample space is the set of all ordered pairs:
{(1,H),(1,T),(2,H),(2,T),(3,H),(3,T),(4,H),(4,T)}{(1,H),(1,T),(2,H),(2,T),(3,H),(3,T),(4
,H),(4,T)}
2. On any day that I have to proctor an exam, I go through the drive-through of a
popular donut chain and get myself a little treat. I always buy myself exactly one
beverage, either a hot chocolate or a sweet tea. I often also get myself something
to eat as well: either a donut, an order of hashbrowns, a bagel, or possibly even
some combination of the three.
Since I am obviously obsessed with probability, I leave what I order up to complete
chance (I’ll give you the full algorithm in a future assignment).
Let’s define the events as follows: C is the event I order a hot chocolate, T is the
event I order a sweet tea, D is the event I order a donut, H is the event I order
hashbrowns, and B is the event I order a bagel.
Given the above setup, if the probability I order a hot chocolate is 0.2, what is the
probability I order a sweet tea?
Since I always order exactly one beverage, either hot chocolate (C) or sweet tea (T),
these events are mutually exclusive and exhaustive. Thus,
P(C)+P(T)=1⇒P(T)=1−0.2=0.8P(C)+P(T)=1⇒P(T)=1−0.2=0.8
3. Using the setup from problem 2, regardless of what I order to drink, the probability
I order a donut is 0.35, the probability I order a bagel is 0.3, the probabilty I order
hashbrowns is 0.45, the probability I order a donut and hashbrowns is 0.2, the
probability I order a bagel and hashbrowns is 0.1, and I never order a bagel and
donut together.
, 2
Given this, what is the probability that I order a bagel and no hashbrowns?
We are given P(B)=0.3P(B)=0.3 and P(B∩H)=0.1P(B∩H)=0.1. The event “bagel
and no hashbrowns” is B∩HcB∩Hc. Therefore,
P(B∩Hc)=P(B)−P(B∩H)=0.3−0.1=0.2P(B∩Hc)=P(B)−P(B∩H)=0.3−0.1=0.2