ISYE 6644 Homework week 2 Exam Questions and Answers.

ISYE 6644 Homework week 2 Exam Questions and Answers

Question 1
pts
(Lesson 2.5: Probability Basics.)
If P(A)=P(B)=P(C)=0.6P(A)=P(B)=P(C)=0.6 and A,B,A,B, and CC are independent, find
the probability that exactly one of A,B,A,B, and CC 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)=0.28
8.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=0.288P(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.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=0.288P(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 XX denote the number of times a "4" comes up.
Clearly, X∼Bin(3,16).ThusP(X=2)=(32)(16)2(56)3−2=572.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 XX denote the number of times a "4" comes up.
Clearly, X∼Bin(3,16).ThusP(X=2)=(32)(16)2(56)3−2=572.X∼Bin(3,16).ThusP(X=2)=(32)
(16)2(56)3−2=572.


Question 3
pts
(Lesson 2.7: Great Expectations.) Suppose that XX is a discrete random variable
having X=−1X=−1 with probability 0.2, and X=3X=3 with probability 0.8. Find E[X]E[X].


a. -1



b. 3



c. 1