Week 2 Homework
Instructions
Please answer all the questions below.
This quiz was locked Jan 31 at 8:59pm.
Attempt History
Attempt Time Score
LATEST Attempt 1 6,702 minutes 10 out of 11
Score for this quiz: 10 out of 11
Submitted Jan 28 at 3:27pm
This attempt took 6,702 minutes.
Question 1
pts
(Lesson 2.5: Probability Basics.) If P(A)=P(B)=P(C)=0.6 and A,B, and C are
independent, find the probability that exactly one of A,B, and C occurs.
a. 0.144
b. 0.288
P(exactlyone)=P(A∩B¯∩C¯)+P(A¯∩B∩C¯)+P(A¯∩B¯∩C)=P(A)P(B¯)P(C¯)
+P(A¯)P(B)P(C¯)+P(A¯)P(B¯)P(C)(byindependence)=(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.,
P(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
P(exactlyone)=P(A∩B¯∩C¯)+P(A¯∩B∩C¯)+P(A¯∩B¯∩C)=P(A)P(B¯)P(C¯)
+P(A¯)P(B)P(C¯)+P(A¯)P(B¯)P(C)(byindependence)=(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.,
P(exactly one)=(31)(0.6)1(0.4)2=0.288
Question 2
pts
(Lesson 2.5: Probability Basics.) Toss 3 dice. What's the probability that a "4"
will come up exactly twice?
a. 5/72
Write out every possible outcome explicitly, or use the following binomial argument:
Let X denote the number of times a "4" comes up.
Clearly, X∼Bin(3,16).ThusP(X=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 X denote the number of times a "4" comes up.
Clearly, X∼Bin(3,16).ThusP(X=2)=(32)(16)2(56)3−2=572.
Question 3
pts
Instructions
Please answer all the questions below.
This quiz was locked Jan 31 at 8:59pm.
Attempt History
Attempt Time Score
LATEST Attempt 1 6,702 minutes 10 out of 11
Score for this quiz: 10 out of 11
Submitted Jan 28 at 3:27pm
This attempt took 6,702 minutes.
Question 1
pts
(Lesson 2.5: Probability Basics.) If P(A)=P(B)=P(C)=0.6 and A,B, and C are
independent, find the probability that exactly one of A,B, and C occurs.
a. 0.144
b. 0.288
P(exactlyone)=P(A∩B¯∩C¯)+P(A¯∩B∩C¯)+P(A¯∩B¯∩C)=P(A)P(B¯)P(C¯)
+P(A¯)P(B)P(C¯)+P(A¯)P(B¯)P(C)(byindependence)=(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.,
P(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
P(exactlyone)=P(A∩B¯∩C¯)+P(A¯∩B∩C¯)+P(A¯∩B¯∩C)=P(A)P(B¯)P(C¯)
+P(A¯)P(B)P(C¯)+P(A¯)P(B¯)P(C)(byindependence)=(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.,
P(exactly one)=(31)(0.6)1(0.4)2=0.288
Question 2
pts
(Lesson 2.5: Probability Basics.) Toss 3 dice. What's the probability that a "4"
will come up exactly twice?
a. 5/72
Write out every possible outcome explicitly, or use the following binomial argument:
Let X denote the number of times a "4" comes up.
Clearly, X∼Bin(3,16).ThusP(X=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 X denote the number of times a "4" comes up.
Clearly, X∼Bin(3,16).ThusP(X=2)=(32)(16)2(56)3−2=572.
Question 3
pts