### Exam 1 - RPI ECSE

```Name: ________Solutions___________
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Exam 1
Engineering Probability
ECSE-2500
February 20, 2014
Rules:
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Points
1
20
vastola@ecse.rpi.edu
2
30
vastola@ecse.rpi.edu
3
30
4
20
Total
100
Jie Chen
chenj24@rpi.edu
Kaan Duman
dumank@rpi.edu
Show all work.
You have 75 minutes to take this exam.
You are allowed one 8.5 x 11 inch piece of paper with nothing written on one side.
You may use a calculator, but no programmable features.
No communication devices of any kind. All cell phones off.
This exam has 6 pages including this cover page.
ECSE-2500, Exam 1
Solutions
2/20/2014
1. (20 points). Important: You must follow basic notational conventions: Curly brackets   are
for sets where the order is NOT important, e.g. a, b is the same as b, a. Parentheses   are
for ordered sets where order is important, e.g.  a, b  is NOT the same as  b, a  . Also, a is a
set, not an element. a,b is NOT the same as a, b.
1.a. (5 points) Three identical coins are tossed. We count the number of heads. What is the
simplest and smallest sample space we can use for this problem? Justify your answer.
Solution
All we need to keep track of here is the number of heads among the three coins so
S  0,1,2,3 is sufficient.
1.b. (5 points) What set A in your sample space in part (a) represents the event that either 2 or 3
Solution
A  2,3
1.c. (5 points) A nickel, a dime, and a quarter are each tossed. We record the heads or tails of each
coin separately. What is the simplest and smallest sample space we can use for this problem?
Solution
Here we need to keep track of each distinct coin, so we have to use vectors of length 3 where
the first position represents the outcome for the nickel, the 2nd position represents the
outcome for the dime, and the 3 rd represents the outcome for the quarter. So here we need
the 8 length-3 vectors with either h or t in each place:
S   hhh  ,  hht  ,  hth  , thh  ,  htt  , tht  , tth  , ttt 
1.d. (5 points) What set A in your sample space in part (c) represents the event that either 2 or 3
Solution
A   hhh  ,  hht  ,  hth  ,  thh 
Page 2 of 6
ECSE-2500, Exam 1
Solutions
2/20/2014
2. (30 points) A number x is selected at random in the interval  1,1. Define the events
A  x  0, B   x  0.5, and C  x  0.
2.a. (10) What are P  A, P B, and P C ? Clearly justify your answer for each.
Solution
For any interval a, b with 1  a  b  1, i.e. contained in the interval  1,1 , we have
ba
P a, b 
 0.5  b  a  regardless of whether the endpoints a or b are included in the
1   1
interval.
Thus,
P  A  P x  0  P  1,0   0.5  0   1   0.5,
P  B   P  x  0.5  P  0.5,0.5   0.5  0.5   0.5    0.5,
P C   P x  0  P  0,1  0.5 1  0   0.5.
2.b. (10) Compute P  A B, P  A C , P  AB, P  AC  , P BC . Justify your answer in each case.
Solution
P  A B  P  1,0.5  0.5  0.5   1   0.75
For A C we should use the disjoint union property (because A C is not an interval):
P  A C   P  A  P C   P  1,0   P  0,1  0.5  0   1   0.5 1  0   1.
P  AB   P  0.5,0   0.5  0   0.5    0.25.
P  AC   P    0.
P  BC   P  0,0.5   0.5  0.5   0    0.25.
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ECSE-2500, Exam 1
Solutions
2/20/2014
2.c. (10) Are A and B independent? Are A and C independent? Are B and C independent? Are A, B,
Solution
Using the results from parts (a) and (b) and the definition of independence, the answers are:
Yes. P  A P B  0.5  0.5  0.25  P  AB
No. P  A P C   0.5  0.5  0  P AC 
Yes. P B P C   0.5  0.5  0.25  P  BC 
No. Since A and C are not independent, A, B, and C cannot be independent by definition.
Page 4 of 6
ECSE-2500, Exam 1
Solutions
2/20/2014
3. (30 points) Recall the Type 1 Geometric PMF for a random variable X:
pX k   P X  k  1  p p for k  0,1,2,3,
k
Let p  0.3 and let X represent the number of packets arriving to a particular router in a
network during one minute.
3.a. (10 points) What is the probability that exactly 2 packets arrive during this minute? Justify
Solution
2
2
P X  2  pX2  1  p p   0.7 0.3  0.147.
3.b. (10 points) What is the probability that 3 or more packets arrive during this minute? Justify
Solution
P  X  3  1  P  X  3
 1  pX  0   pX 1  pX  2
 1  1  p  p  1  p  p  1  p  p
0
1
2
 1  0.3  0.21  0.147  0.343.
3.c. (10 points) What is the expected or average number of packets to arrive in one minute?
Solution
From the book and class notes, the formula for the expected value of a Type 1 Geometric
random variable X is

k
 1  p  1  p 0.7
E X   p k 1  p  p  2  

 2.3333
p
0.3
k 1
 p 
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ECSE-2500, Exam 1
Solutions
2/20/2014
4. (20 points) You are given two coins which look identical. One coin is fair (probability of heads is
one half) and the second coin comes up heads 90% of the time. You choose one of the coins at
it that the coin you tossed is the unfair one? That is, let
B0  coin we tossed is the fair one.
B1  coin we tossed is the unfair one.
A  the outcome of 4 flips was hhth.
and compute P  B1 A.
Solution
We are looking for P  B1 A. What we can compute easily is P  A B1  and P  A B0  using
the fact that coin tossing is Bernoulli Trials. So this is reversing a coniditonal probability, so
we should use Bayes Theorem.
Clearly B0 , B1  is a partition of the sample space S. And P  B0   P B1   0.5 since we are
picking the coin to toss “at random”.
P  A B1   ppqp   0.9  0.1  0.0729
3
P  A B0   ppqp   0.5  0.0625
4
Bayes Theorem tells us
P  B1 A 
P  A B1  P  B1 
1
 P  A B  P B 
i 0
i
i

0.0729  0.5
 0.5384
0.0625  0.5  0.0729  0.5
So it is more likely than not that coin we are tossing is the unfair one.
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