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Marginal and Conditional PMF, Exercises of Probability and Statistics

A construction site is needed to be excavated and prepared for the next stage of a project. That work requires haul trucks to transport the excavated materials. X is the number of haul trucks working at the construction site. Y is the number of the days the haul units can finish the job.

Typology: Exercises

2019/2020

Uploaded on 11/07/2021

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Homework (1)
A construction site is needed to be excavated and prepared for the next stage of a project.
That work requires haul trucks to transport the excavated materials.
X is the number of haul trucks working at the construction site.
Y is the number of the days the haul units can finish the job.
Marginal probability mass function of X and Y:
p
(
x
)
=P
(
X=x
)
=
yj=5
10
p(x , y j)
p
(
y
)
=P
(
Y=y
)
=
xi=1
6
p(xi, y )
Marginal PMF of X:
p
(
1
)
=P
(
X=1
)
=
yj=5
10
p
(
1, y j
)
=0+0+0.009+0.005 +0.0021+0.075=0.0911
p
(
2
)
=P
(
X=2
)
=
yj=5
10
p
(
2, y j
)
=0+0.002+0.0035 +0.0058 +0.004 +0.0032=0.0185
p
(
2
)
=P
(
X=3
)
=0.1300
p
(
4
)
=P
(
X=4
)
=0.2900
p
(
5
)
=P
(
X=5
)
=0.1589
pf3
pf4

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Homework (1)

A construction site is needed to be excavated and prepared for the next stage of a project.

That work requires haul trucks to transport the excavated materials.

X is the number of haul trucks working at the construction site.

Y is the number of the days the haul units can finish the job.

Marginal probability mass function of X and Y:

p ( x )= P ( X = x )=

y j = 5

10

p ( x , y

j

p ( y )= P ( Y = y ) =

x

i

= 1

6

p ( x

i

, y )

Marginal PMF of X:

p ( 1 )= P ( X = 1 )=

y j = 5

10

p

1 , y

j

p ( 2 )= P ( X = 2 )=

y

j = 5

10

p

2 , y

j

p ( 2 )= P ( X = 3 )=0.

p ( 4 )= P ( X = 4 )=0.

p ( 5 )= P ( X = 5 )=0.

p ( 6 )= P ( X = 6 ) =0.

Marginal PMF of Y:

p ( 5 )= P ( Y = 5 )=

x

i

= 1

6

p

x

i

p ( 6 )= P ( Y = 6 )=0.

p ( 7 )= P ( Y = 7 )=0.

p ( 8 )= P ( Y = 8 )=0.

p ( 9 )= P ( Y = 9 )=0.

p ( 10 )= P ( Y = 10 )=0.

The conditional probability mass function of X when Y=

p

x

i

y = 5

= P

X = x

i

∨ Y = 5

P ( X = x

i

,Y = 5 )

P ( Y = 5 )

p ( x

i

p ( 5 )

p

= P

X = 1 ∨ Y = 5

p

XY

p

Y

p ( 2 ∨ 5 ) = P ( X = 2 ∨ Y = 5 )=

p

XY

p

Y

p

= P

X = 3 ∨ Y = 5

p

XY

p

Y

p

= P

X = 4 ∨ Y = 5

p

XY

p

Y

p

= P

X = 5 ∨ Y = 5

p

XY

p

Y

p

= P

X = 6 ∨ Y = 5

p

XY

p

Y