ECEN 303: Assignment 6
Problems:
1. Two fair dice are rolled. Find the joint probability mass function of
X
and
Y
when
(a)
X
is the largest value obtained on any die and
Y
is the sum of the values;
In this first case, we have
p
X,Y
(
x,y
)
2
3
4
5
6
7
8
9
10
11
12
1
1
36
0
0
0
0
0
0
0
0
0
0
2
0
2
36
1
36
0
0
0
0
0
0
0
0
3
0
0
2
36
2
36
1
36
0
0
0
0
0
0
4
0
0
0
2
36
2
36
2
36
1
36
0
0
0
0
5
0
0
0
0
2
36
2
36
2
36
2
36
1
36
0
0
6
0
0
0
0
0
2
36
2
36
2
36
2
36
2
36
1
36
.
(b)
X
is the value on the first die and
Y
is the larger of the two values;
In this second case, we get
p
X,Y
(
x,y
)
1
2
3
4
5
6
1
1
36
1
36
1
36
1
36
1
36
1
36
2
0
2
36
1
36
1
36
1
36
1
36
3
0
0
3
36
1
36
1
36
1
36
4
0
0
0
4
36
1
36
1
36
5
0
0
0
0
5
36
1
36
6
0
0
0
0
0
6
36
.
(c)
X
is the smallest and
Y
is the largest value obtained on the dice.
Finally, in this third scenario, we obtain
p
X,Y
(
x,y
)
1
2
3
4
5
6
1
1
36
2
36
2
36
2
36
2
36
2
36
2
0
1
36
2
36
2
36
2
36
2
36
3
0
0
1
36
2
36
2
36
2
36
4
0
0
0
1
36
2
36
2
36
5
0
0
0
0
1
36
2
36
6
0
0
0
0
0
1
36
.
2. The number of people that enter a drugstore in a given hour is a Poisson random variable
with parameter
λ
= 10. Compute the conditional probability that at most 3 men entered the
drugstore, given that 10 women entered in that hour. What assumptions have you made?
Suppose that every customer that enters the store is a men with probability
p
or a women
with probability (1

p
), independently of other customers. Let
C
be the total number of
1
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customers,
M
be the number of men, and
W
be the number of women. From deriving the
splitting property of a Poisson random variable, we know that
p
M
(
m
) =
(10
p
)
m
m
!
e

10
p
m
= 0
,
1
,...
and similarly
p
W
(
w
) =
(10(1

p
))
w
w
!
e

10(1

p
)
w
= 0
,
1
,...
Furthermore, we know that
p
M

C
(
m

c
) =
parenleftbigg
c
m
parenrightbigg
p
m
(1

p
)
c

m
m
= 0
,
1
,...c.
We can therefore write
p
M

W
(
m

10) =
p
M

C
(
m

m
+ 10) =
parenleftbigg
m
+ 10
m
parenrightbigg
p
m
(1

p
)
10
m
= 0
,
1
,...
As such, the probability that at most 3 men entered the drugstore can be computed as
Pr(
M
≤
3

W
= 10) =
p
M

W
(0

10) +
p
M

W
(1

10) +
p
M

W
(2

10) +
p
M

W
(3

10)
=
parenleftbigg
10
0
parenrightbigg
(1

p
)
10
+
parenleftbigg
11
1
parenrightbigg
p
(1

p
)
10
+
parenleftbigg
12
2
parenrightbigg
p
2
(1

p
)
10
+
parenleftbigg
13
3
parenrightbigg
p
3
(1

p
)
10
.
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 Fall '07
 Chamberlain
 Probability theory, Yi, Poisson random variable

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