Week 2 Homework
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School
Georgia Institute Of Technology *
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Course
6242
Subject
Mathematics
Date
Feb 20, 2024
Type
docx
Pages
8
Uploaded by BailiffNeutron10026
Week 2 Homework
Due
Jan 26 at 8:59pm
Points
11
Questions
11
Available
Jan 19 at 5am - Jan 29 at 8:59pm
Time Limit
None
Instructions
Please answer all the questions below.
This quiz was locked Jan 29 at 8:59pm.
Attempt History
Attempt
Time
Score
LATEST
Attempt 1
8,663 minutes
11 out of 11
Score for this quiz:
11
out of 11
Submitted Jan 25 at 2:02pm
This attempt took 8,663 minutes.
Question 1
1
/ 1
pts
(Lesson 2.5: Probability Basics.) If �(�)=�(�)=�(�)=0.6 and �,�, and � are independent, find the probability that exactly one of �,�, and � occurs.
a. 0.144
Correct!
b. 0.288
�(����������)=�(�∩�¯∩�¯)+�(�¯∩�∩�¯)
+�(�¯∩�¯∩�)=�(�)�(�¯)�(�¯)+�(�¯)�(�)�(�¯)+�(�¯)�(�¯)�(�)
(��������������)=(0.6)(0.4)(0.4)+(0.4)(0.6)(0.4)+(0.4)(0.4)
(0.6)=0.288.
You could also have used a binomial distribution argument to solve this problem,
i.e.,
�(exactly one)=(31)(0.6)1(0.4)2=0.288
c. 0.576
d. 0.6
e. I'm from The University Of Georgia. Is the answer -3?
The answer is (b). To see why, note that
�(����������)=�(�∩�¯∩�¯)+�(�¯∩�∩�¯)
+�(�¯∩�¯∩�)=�(�)�(�¯)�(�¯)+�(�¯)�(�)�(�¯)
+�(�¯)�(�¯)�(�)(��������������)=(0.6)(0.4)(0.4)+(0.4)
(0.6)(0.4)+(0.4)(0.4)(0.6)=0.288.
You could also have used a binomial distribution argument to solve this problem,
i.e.,
�(exactly one)=(31)(0.6)1(0.4)2=0.288
Question 2
1
/ 1
pts
(Lesson 2.5: Probability Basics.) Toss 3 dice. What's the probability that a "4"
will come up exactly twice?
Correct!
a. 5/72
Write out every possible outcome explicitly, or use the following binomial argument:
Let � denote the number of times a "4" comes up. Clearly, �
∼
Bin(3,16).ThusP(�=2)=(32)(16)2(56)3−2=572.
b. 1/2
c. 13/16
d. 1/8
(a). Write out every possible outcome explicitly, or use the following binomial
argument: Let � denote the number of times a "4" comes up. Clearly, �
∼
Bin(3,16).ThusP(�=2)=(32)(16)2(56)3−2=572.
Question 3
1
/ 1
pts
(Lesson 2.7: Great Expectations.) Suppose that � is a discrete random variable having �=−1 with probability 0.2, and �=3 with probability 0.8. Find E[�].
a. -1
b. 3
c. 1
Correct!
d. 2.2
E[�]=∑���(�)=(−1)(0.2)+(3)(0.8)=2.2 So the answer is (d).
E[�]=∑���(�)=(−1)(0.2)+(3)(0.8)=2.2. So the answer is (d).
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Question 4
1
/ 1
pts
(Lesson 2.7: Great Expectations.) Suppose that � is a discrete random variable having �=−1 with probability 0.2, and �=3 with probability 0.8. Find Var[�].
a. -1
b. 1
Correct!
c. 2.56
In addition to the above work,
E[�2]=∑��2�(�)=((−1)2)(0.2)+(32)(0.8)=7.4,
so that we have Var(�)=E[�2]−(E[�])2=2.56. So the answer is (c).
d. 5.12
In addition to the above work,
E[�2]=∑��2�(�)=((−1)2)(0.2)+(32)(0.8)=7.4,
so that we have Var(�)=E[�2]−(E[�])2=2.56. So the answer is (c).
Question 5
1
/ 1
pts
(Lesson 2.7: Great Expectations.) Suppose that � is a discrete random variable having �=−1 with probability 0.2, and �=3 with probability 0.8. Find E[3−1�].
a. 3
b.
∞
c. -2
Correct!
d. 44/15
Finally, by LOTUS,
E[1/�]=∑�(1/�)�(�)=0.2/(−1)+0.8/3=1/15,
so that E[3−1�]=3−E[1�]=4415. So the answer is (d).
Finally, by LOTUS,
E[1/�]=∑�(1/�)�(�)=0.2/(−1)+0.8/3=1/15,
so that E[3−1�]=3−E[1�]=4415. So the answer is (d).
Question 6
1
/ 1
pts
(Lesson 2.7: Great Expectations.) Suppose X is a continuous random variable
with p.d.f. �(�)=4�3 for 0≤�≤1. Find E[1/�2].
a. 2/3
b. 1
c. 3/2
Correct!
d. 2
By LOTUS,
E[1/�2]=∫�(1/�2)�(�)��=∫014���=2.
(d) By LOTUS,
E[1/�2]=∫�(1/�2)�(�)��=∫014���=2.
Question 7
1
/ 1
pts
(Lesson 2.8: Functions of a Random Variable.) Suppose � is the result of a 5-
sided die toss having sides numbered −2,−1,0,1,2. Find the probability mass
function of �=�2.
a.
P(�=1)=P(�=4)=1/2
b.
P(�=1)=P(�=2)=1/2
Correct!
c.
�(�=0)=15,����(�=1)=�(�=4)=25
This follows because
P(�=0)=P(�2=0)=P(�=0)=1/5,P(�=1)=P(�2=1)=P(�=−1)+P(�=1)=2/5,andP(
�=4)=P(�2=4)=P(�=−2)+P(�=2)=2/5.
No other possible values for �=�2.
d.
P(�=−2)=P(�=−1)=P(�=0)=P(�=1)=P(�=2)=1/5
(c). This follows because
P(�=0)=P(�2=0)=P(�=0)=1/5,P(�=1)=P(�2=1)=P(�=−1)+P(�=1)=2/5
,andP(�=4)=P(�2=4)=P(�=−2)+P(�=2)=2/5.
No other possible values for �=�2.
Question 8
1
/ 1
pts
(Lesson 2.8: Functions of a Random Variable.) Suppose � is a continuous random variable with p.d.f. �(�)=2� for 0<�<1. Find the p.d.f. �(�)of�=�2. (This may be easier than you think.)
Correct!
a.
�(�)=1
, for
0<�<1
Note that the c.d.f. of � is �(�)=�2 (you can do this in your head). So by the Inverse Transform Theorem, we immediately have that
�(�)=�2=� is Unif(0,1), with the p.d.f. �(�)=1.
b.
�(�)=�
, for
0<�<1
c.
�(�)=�2
, for
−1<�<1
d.
�(�)=�2
, for
0<�<1
(a). Note that the c.d.f. of � is �(�)=�2 (you can do this in your head). So by the Inverse Transform Theorem, we immediately have that �(�)=�2=� is Unif(0,1), with the p.d.f. �(�)=1.
Question 9
1
/ 1
pts
(Lesson 2.9: Jointly Distributed RVs.) Suppose that �(�,�)=6� for 0≤�≤�≤1. Find P(�<1/2and�<1/2).
a. 1
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b. 1/2
c. 1/4
Correct!
d. 1/8
P(�<1/2and�<1/2)=∫01/2∫0��(�,�)����=∫01/2∫0�6�����=1/8.
So the answer is (d).
P(�<1/2and�<1/2)=∫01/2∫0��(�,�)����=∫01/2∫0�6�����=1/8.
So the answer is (d).
Question 10
1
/ 1
pts
(Lesson 2.9: Jointly Distributed RVs.) Suppose that �(�,�)=6� for 0≤�≤�≤1. Find the marginal p.d.f. ��(�) of �.
Correct!
a.
6�(1−�)
, for
0≤�≤1
��(�)=∫−∞∞�(�,�)��=∫�16���=6�(1−�), for 0≤�≤1. So the answer is (a).
b.
6�
, for
0≤�≤1
c.
6�
, for
0≤�≤1
d.
6�(1−�)
, for
0≤�≤1
��(�)=∫−∞∞�(�,�)��=∫�16���=6�(1−�), for 0≤�≤1. So the answer is (a).
Question 11
1
/ 1
pts
(Lesson 2.9: Jointly Distributed RVs.) YES or NO? Suppose � and � have joint
p.d.f. �(�,�)=���/(1+�2+�2) for 0<�<1, 0<�<1, and whatever constant � makes the nasty mess integrate to 1.
Are � and � independent?
a. Yes
Correct!
b. No
NO! The lesson has a theorem that says that
�,� are independent if and only if you
can write �(�,�)=�(�)�(�) with no funny limits for some functions �(�) and �(�). Can't do such a factorization, so � and � ain't indep.
NO! The lesson has a theorem that says that �,� are independent if and only if you can write �(�,�)=�(�)�(�) with no funny limits for some functions �(�) and �(�). Can't do such a factorization, so � and � ain't indep.
Quiz Score:
11
out of 11