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ECSE-2500 Engineering Probability HW#6 Solutions
Due 9/18/14
1. (12 points) Two distinct fair dice are tossed independently. We record the number on die 1
followed by the number on die 2 as an ordered pair, so that our sample space is
S     s1 , s2  si1,2,3,4,5,6, i  1,2.
1.a. (3 points) Clearly define the probability law that corresponds to this problem given that the
dice are fair and independent. Specifically, what is P  i, j  i  j ? Justify your answer.
Solution
Since the dice are independent, for any pair    i, j   S with i, j 1,2,3,4,5,6, we have
1 1 1
P  i, j   P i P  j   
6 6 36
Then because S is a discrete set, for any event A  S, we have
P  A   P  
 A
Finally,
P  i, j  i  j  P 1,1 ,  2,2  , 3,3  ,  4, 4  , 5,5  , 6,6 
  P i, j  6 
i j
1 1
 .
36 6
1.b. (3 points) Define the random variable X1 on S by
0 if i  j
X1   i, j    
1 if i  j
Clearly identify range SX1  x1 , x2 , x3 ,

for this random variable and the sets
A0    S X1    0 and A1    S X1    1. Show that A0 and A1 form a partition of
S.
Solution
Here SX1  0,1 ,
Page 1 of 4
ECSE-2500 Engineering Probability HW#6 Solutions
Due 9/18/14
A0    S X1    0   i, j   S i  j

1,2 , 1,3  , 1, 4  , 1,5  , 1,6  ,  2,3  , 2, 4  , 2,5  , 2,6  , 



3, 4  , 3,5  , 3,6  ,  4,5  ,  4,6  ,  5,6 



and A1  A0C . We know that A0 and A1 form a partition of S since we know that any set and
its complement form a partition of the space.
1.c. (3 points) For the random variable X1 defined in (b), compute the probability mass function
(PMF) pX1  xk  for each xk  SX1 , directly from the Ak as shown in class and the book. Which
of the well-known random variables covered in class and in the book is X1 ? Identify the value
of any parameters associated with this well-known random variable. Justify your answer.
Solution
Using pX1  xk   P X1  xk   P  Ak  and the fact that each element of Ak has probability
1
36
, we have pX1  xk   P  Ak  
So, we have pX1  0   P  A0  
Ak
, where Ak is the number of elements in Ak .
36
15
21
and pX1 1  P  A1   .
36
36
This is a Bernoulli random variable with p 
21
.
36
1.d. (3 points) For the random variable X1 defined in (b), compute E X1  using the definition
given in Eq. (3.8) on page 104 in the book. Show your work.
Solution

1
 15   21  21
E X1    xk pX1  xk    kpX1  k   0    1    .
 36   36  36
k 1
k 0
Page 2 of 4
ECSE-2500 Engineering Probability HW#6 Solutions
Due 9/18/14
2. (14 points) For the sample space defined in problem 1, define a new random variable
X2   i, j    i  j.
2.a. (2 points) What is the range SX2  x1 , x2 , x3 ,

for this random variable?
Solution
SX2  2,3,4,5,6,7,8,9,10,11,12
2.b. (4 points) For this random variable X2 , list the elements of the set Ak    S X2    xk 
for each xk  SX2 , and show that the Ak together form a partition of S.
Solution
Here, for k  2,3,


,12, Ak   i, j   S X2   i, j    i  j  k , so
A2  1,1,
A3  1,2  ,  2,1,
A4  1,3  , 3,1 2,2 ,
A5  1, 4  ,  4,1 ,  2,3  , 3,2 ,
A6  1,5  ,  5,1 ,  2, 4  ,  4,2  , 3,3 ,
A7  1,6  ,  6,1 ,  2,5  ,  5,2  , 3, 4  ,  4,3 ,
A8   2,6  ,  6,2  , 3,5  ,  5,3  ,  4, 4 ,
A9  3,6  ,  6,3  ,  4,5  ,  5, 4 ,
A10   4,6  ,  6, 4  ,  5,5 ,
A11   5,6  ,  6,5 ,
A12   6,6 .
We can see that they are pairwise disjoint and their union is S by inspection.
2.c. (4 points) For the random variable X2 , compute the probability mass function (PMF) pX2  xk 
for each xk  SX2 , directly from the Ak as shown in class and the book. Justify your answer.
Page 3 of 4
ECSE-2500 Engineering Probability HW#6 Solutions
Due 9/18/14
Solution
As in problem 1, using pX  xk   P X  xk   P  Ak  and the fact that each element of Ak has
probability
1
36
, we have pX2  xk   P  Ak  
So, we have
pX2  2  P  A2   P 1,1 
1
36
Ak
, where Ak is the number of elements in Ak .
36
,
pX2 3   P  A3   P 1,2  ,  2,1 
2
36
,
pX2  4   P  A4   P 1,3  , 3,1 2,2  
3
36
,
pX2  5   P  A5   P 1, 4  ,  4,1 ,  2,3  , 3,2  
4
36
,
pX2  6   P  A6   P 1,5  ,  5,1 ,  2, 4  ,  4,2  , 3,3  
5
36
,
pX2  7   P  A7   P 1,6  ,  6,1 ,  2,5  , 5,2  , 3, 4  ,  4,3  
pX2  8   P  A8   P  2,6  ,  6,2  , 3,5  ,  5,3  ,  4, 4  
pX2  9   P  A9   P  3,6  ,  6,3  ,  4,5  ,  5, 4  
pX2 10   P  A10   P  4,6  ,  6, 4  ,  5,5  
pX2 11  P  A11   P  5,6  ,  6,5  
pX2 12  P  A12   P  6,6  
1
36
2
36
3
36
4
36
5
36
6
36
,
,
,
,
,
.
2.d. (4 points) For the random variable X2 , compute E X2  using the definition given in Eq. (3.8)
on page 104 in the book. Show your work.
Solution

12
k 1
k 2
E X2    xk pX1  xk    kpX2  k 
 1   2
 3   4 
 5
 6 
 2   3   4    5   6    7 
 36   36 
 36   36 
 36 
 36 
 5
 4 
 3 
 2
 1 
 8    9    10    11    12  
 36 
 36 
 36 
 36 
 36 


 2  12 1  3  11 2   4  10  3  5  9  4  6  8  5   7  6
36
14  1  2  3  4  5   7  6
36

14  15   7  6
36

252
 7.
36
Page 4 of 4