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CSE 312 Homework 2, Due Friday October 10
Instructions:
• When asked for a short answer (such as a single number), also show and explain your work briefly.
Unless you are asked to, you can
leave your answer in terms of factorials, combinations, etc., for
instance 267 or 26!/7! or 26 · 26
7 . You can also leave answers in the form of a sum.
• Please write down on your homework the names of all people you discussed the homework with.
Problems
1. Consider the question: Suppose a team of 10 players is selected at random out of eighteen possible
people, where 15 are women and 3 are men. Each team (unordered) is equally likely. What is the
probability that at least one of the players is a man?
Each of the 18
10 teams is equally likely, so that is the size of the sample space. Let E be the
event that the team selected has at least one man. Then
P r(E) =
|E|
18
10
To count |E|, first pick one man out of the 3 possible, and then pick the rest of the team
from the remaining people. So
17
3
3
17
1 · 9
|E| =
·
and hence
P r(E) =
18
1
9
10
Explain what is wrong with this solution. If there is over-counting in the calculation of |E|, characterize
all teams that are counted more than once, and how many times each such team is counted using the
above solution.
What is the correct answer for P r(E)?
2. Suppose 10 players are selected at random out of eighteen possible people, where 15 are women and 3
are men, with each player assigned to one of the 10 distinct positions. Each outcome is equally likely.
What is the probability that at least one of the positions is assigned to a man? What is the probability
that exactly one of the positions is assigned to a man?
3. Consider the question: what is the probability of getting a 6-card poker hand that contains a card
from all four suits (order doesn’t matter):
Each of the 52
6 hands is equally likely. Let E be the event that the hand selected contains
a card from all four suits. Then
|E|
P r(E) = 52
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For |E|, first pick one heart, then one spade, then one diamond, then one clubs, and, finally,
two additional cards. Therefore
134 · 48
48
4
2
.
|E| = 13 ·
and hence
P r(E) =
52
2
6
Explain what is wrong with this solution. If there is over-counting in |E|, characterize all hands that
are counted more than once, and how many times each such hand is counted.
What is the correct answer for P r(E)?
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4. There are 30 psychiatrists and 24 psychologists attending a certain conference. Three of these 54
people are randomly chosen to take part in a panel discussion. What is the probability that at least
one psychologist is chosen? What is the probability that exactly three psychologists are chosen?
5. We flip a fair coin 10 times. Each outcome is equally likely. Find the probability of the following
events:
• The number of heads and the number of tails are equal.
• There are more heads than tails.
• The i-th flip and the 11 − i-th flip are the same for i = 1, . . . , 5.
• We flip at least five consecutive heads.
6. Consider the following variant of the Monty Hall Problem: There are 5 doors and the host promises
to open 2 of the doors with goats behind them after the contestant has chosen a door. What is the
probability that the contestant wins a car by switching? State all your assumptions.
7. Consider two probability spaces: In the first, the outcomes are all unordered subsets of size k out of a
set of n distinct elements. In the second, the outcomes are all ordered subsets of size k out of a set of
n distinct elements. Both probability distributions are uniform.
Let E be any event in the first probability space. Let F be the event in the second probability space
defined as follows:
F = {All ordered sequences (i1 , i2 , . . . , ik ) such that the unordered subset{i1 , . . . , ik } is an element of E}.
Show that the probability of E in the first probability space is equal to the probability of F in the
second probability space.
8. Consider a weighted die such that
• P r(1) = P r(2),
• P r(3) = P r(4) = P r(5) = P r(6), and
• P r(1) = 2P r(3).
What is the probability that the outcome is 5 or 6?
9. Let E, F, G be events in a certain probability space (Ω, p), where P r(ω) is the probability of outcome
ω ∈ Ω. (Note that this probability space is not necessarily uniform). Prove that
P r(E ∪ F ∪ G) =P r(E) + P r(F ) + P r(G)
− P r(E ∩ F ) − P r(F ∩ G) − P r(E ∩ G)
+ P r(E ∩ F ∩ G).
10. I have a fair coin and a two-headed coin. I choose one of the two coins randomly with equal probability
and flip it. Given that the flip was heads, what is the probability that I flipped the two-headed coin?
11. A medical company touts its new test for a certain genetic disorder. The false negative rate is small:
if you have the disorder, the probability that the test returns a positive results is 0.999. The positive
rate is also small: if you do not have the disorder, the probability that the test returns a positive result
is only 0.005. Assume that 2% of the population has the disorder. If a person chosen uniformly from
the population is tested and the result comes back positive, what is the probability that the person
has the disorder?
12. A certain number of dollars is placed in each of two envelopes. The amounts differ from one another,
but you do not know what the values of the two amounts are. You do know that they lie between 2
positive integers s and b, where 0 < s < b (where b > s + 2). You choose an envelope randomly. After
inspecting its contents, you may switch envelopes. What is the probability of winning the larger sum
of money if you use the following procedure:
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• Choose an envelope and look to see how much cash is inside.
• Pick a random integer between s and b.
• If the number you drew is greater than the amount of cash in your envelope, you exchange the
envelope. Otherwise, you keep the envelope you have.
13. The names of 100 people are placed into 100 closed bags, one name per bag, and the bags are lined up
on a table in a room. One by one the people are led into the room; each may look in at most 50 bags,
but must leave the room exactly as they found it. From the moment a person is led into the room,
that person is not allowed any further communication with the others.
The people have a chance to plot their strategy in advance (and come up with a coordinated strategy if
they so desire) and they are going to need it, because unless every single person finds their own name
among the 50 bags they look in, all of them will be shot!
(a) What is the probability they are not shot if they all look in the same 50 boxes?
(b) What is the probability they are not shot if each person looks in a random set of 50 boxes?
(c) Extra Credit: (hard without a hint) Come up with a strategy for the people that guarantees
that they won’t be shot with probability at least 0.3. (In next week’s homework, I will walk you
through the development of the answer to this question, but I thought I’d first give you a chance
to ponder it.)
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