 • 0  1 / 15100% MAT 230 EXAM TWO
This document is proprietary to Southern New Hampshire University. It and the problems within
may not be posted on any non-SNHU website.
Jacqueline Amoah
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Directions: Type your solutions into this document and be sure to show all steps for arriving at
your solution. Just giving a final number may not receive full credit.
P
ROBLEM
1
This question has 2 parts.
Part 1:
Suppose that F and X are events from a common sample space with P (F ) 0 and P (X) 0.
(a)
Pro
v
e
that
P
(
X
)
=
P
(
X
F
)
P
(
F
)
+
P
(
X
F
¯
)
P
(
F
¯
).
Hin
t:
Explain
wh
y
P
(
X
F
)
P
(
F
)
=
P (X F ) is another way of writing the definition of conditional probability, and then use
that with the logic from the proof of Theorem 4.1.1.
P (X F )P (F ) = P (X F ) is
i
true
i
because
i
we
i
want
i
to
i
get
i
all
i
probabilities
i
of
i
both
i
events
i
occurring
i
but
i
want
i
to
i
avoid
i
counting
i
those
i
events
i
that
i
have
i
i
been
i
counted.
i
When
i
the
i
intersection
i
of
i
the
i
probabilities
i
is
i
subtracted
i
we
i
take
i
away
i
those
i
that
i
have
i
been
i
counted
i
twice.
Since
i
all
i
probabilities
i
must
i
i
up
i
to
i
1
i
and
i
no
i
more
i
we
i
can
i
say
i
that
i
the
i
i
of
i
the
i
original
i
probabilities
i
and
i
the
i
complements
i
of
i
said
i
probabilities
i
will
i
give
i
a
i
value
i
of
i
1.
(b)
Explain why P (F X) = P (X F )P (F )/P (X) is another way of stating Theorem 4.2.1 Bayes’
Theorem.
since
P(x)=
P
(
X
F
)
P
(
F
)
+
P
(
X
F
¯
)
P
(
F
¯
)
and bayes theorem is as follows:
p
(
X
|
F
)
p
(
F
)
¯
¯
p
(
X
|
F
)
p
(
F
)+
p
(
X
|
F
)
p
(
F
)
we
i
can
i
see
i
that
i
the
i
denominator
i
and
i
P(x)
i
are
i
equal
i
so
i
can
i
substitute.
Part 2:
A website reports that 70% of its users are from outside a certain country. Out of their users
from outside the country, 60% of them log on every day. Out of their users from inside the country,
80% of them log on every day.
(a)
What percent of all users log on every day? Hint: Use the equation from Part 1 (a).
O= OUTSIDE
I=INSIDE
E=EVERYDAY
p(O)=0.7
p(I)=0.8
p(E)=0.6
P(x)=
P
(
X
F
)
P
(
F
)
+
P
(
X
F
¯
)
P
(
F
¯
)
P(X)= (1-.8)(0.6)+(0.7)(0.8)=0.68
The probability is.68 or 68 percent.
(b)
Using Bayes Theorem, out of users who log on every day, what is the probability that they
are from inside the country? Baye’s Theorem:
p
(
X
|
F
)
p
(
F
)
¯
¯
p
(
X
|
F
)
p
(
F
)+
p
(
X
|
F
)
p
(
F
)
= (0.30)(0.8)/((0.3*.6)+(0.7*.4))=0.521
The probability is 0.521 or 52.1 percent. /
/
P
ROBLEM
2
This question has 2 parts.
Part 1:
The drawing below shows a Hasse diagram for a partial order on the set:
{
A, B, C, D, E, F, G, H, I,
J
}
Figure 1:
A Hasse diagram shows 10 vertices and 8 edges. The vertices, represented by dots, are as
follows: vertex J is upward of vertex H; vertex H is upward of vertex I; vertex B is inclined upward to
the left of vertex A; vertex C is upward of vertex B; vertex D is inclined upward to the right of vertex
C; vertex E is inclined upward to the left of vertex F; vertex G is inclined upward to the right of vertex
E. The edges, represented by line segments between the vertices are as follows: 3 vertical edges connect
the following vertices: B and C, H and I, and H and J; 5 inclined edges connect the following vertices:
A and B, C and D, D and E, E and F, and E and G.
Determine the properties of the Hasse diagram based on the following questions:
(a)
What are the minimal elements of the partial order?
Minimal elements hold
y=
x and y x.
On
i
the
i
diagram
i
it
i
would
i
be
i
the
i
points
i
with
i
no
i
connections
i
below
i
so,
i
I,A,F
i
are
i
the
i
minimal
i
elements.
(b)
What are the maximal elements of the partial order?
Maximal elements hold: y x an y x.
On
i
the
i
diagram
i
it
i
is
i
the
i
points
i
with
i
no
i
connections
i
above
i
them
i
so,
i
J,D,G
i
are
i
the
i
maximal
i
elements.
(c)
Which of the following pairs are comparable?
(A, D), (J, F ), (B, E), (G, F ), (D, B), (C, F ), (H, I), (C, E)
Comparable
i
ordered
i
pairs
i
are
i
pairs
i
that
i
can
i
be
i
reached
i
by
i
travelling
i
in
i
a
i
single
i
direction.
i
Out
i
of
i
the
i
list
i
the
i
following
i
are
i
comparable:
(A,D),(D,B),(G,F),(H,I) Part 2:
Consider the partial order with domain 3, 5, 6, 7, 10, 14, 20, 30, 60, 70 and with x y if
x evenly divides y. Select the correct Hasse diagram for the partial order.
(a)
Figure 2:
A Hasse diagram shows a set of elements 3; 5; 6; 7; 10; 14; 20; 30; 60, 70.
There are lines connecting 3 and 6, 6 and 30, 30 and 60, 5 and 10, 10 and 20, 20 and 60, 10
and 70, 7 and 14, 14 and 70.
This diagram is incorrect because the 10 is not connected to the 30 or the 60. Also the 5
is not connected to its multiple of 30.