St. Louis University
Spring 2009
Solutions: Final Exam
Math 403–01
Prof. G. Marks
1.
(15 pts.)
A box contains
100
dice.
Of these,
99
dice are cubes, and the remaining die is a tetrahedron (a
pyramid with a triangular base, having four faces). Each cubical die is labeled with the integers
1
through
6
,
and
these numbers occur with equal probability when the die is rolled. The tetrahedral die is labeled with the integers
1
through
4
,
and these numbers occur with equal probability when the die is rolled.
Someone randomly selects two dice from the box and rolls them. The sum of the two numbers appearing on the
selected dice is
9
.
What is the probability that one of the two dice selected was tetrahedral?
Solution.
Let
A
be the event that two cubes are selected from the box; let
B
be the event that one cube and one
tetrahedron are selected. Let
N
be the event that 9 is the total on the two selected dice. Then
Prob(
A
) =
parenleftbigg
99
2
parenrightbigg
parenleftbigg
100
2
parenrightbigg
=
49
50
,
Prob(
B
) = 1

Prob(
A
) =
1
50
,
Prob(
N

A
) =
1
9
,
Prob(
N

B
) =
1
12
(since there are four ways to roll 9 on two cubical dice out of 36 ways to roll anything, and there are two ways
to roll 9 on a cube and tetrahedron out of 24 ways to roll anything).
Given that 9 was rolled, the conditional
probability that one of the selected dice was tetrahedral is
Prob(
B

N
)
=
Prob(
B
∩
N
)
Prob(
N
)
=
Prob(
N

B
)
·
Prob(
B
)
Prob(
N
)
=
Prob(
N

B
)
·
Prob(
B
)
Prob(
N

A
)
·
Prob(
A
) + Prob(
N

B
)
·
Prob(
B
)
(by Bayes’s Theorem)
=
1
12
·
1
50
1
9
·
49
50
+
1
12
·
1
50
=
3
199
.
2.
(16 pts.)
The card game bridge is played by dealing a standard deck of
52
cards
*
to four players, known as
North, South, East, and West, so that each player receives
13
cards. If South has been dealt precisely five spades,
what is the probability that North has been dealt three or more spades? (Assume the deck was shuffled before the
deal so that every arrangement of the
52
cards is equally likely.)
Solution.
There are
parenleftbigg
39
13
parenrightbigg
different hands that North can be dealt, since this is the number of ways of selecting 13 cards from among the 39
cards not in South’s hand. The number of North hands that contain precisely
k
spades (for
k
∈ {
0
,
1
,
2
, . . .,
8
}
) is
parenleftbigg
8
k
parenrightbigg
·
parenleftbigg
31
13

k
parenrightbigg
,
since there are
(
8
k
)
ways to select
k
spades for North’s hand from among the eight spades not in South’s hand,
and
(
31
13

k
)
ways to select 13

k
nonspades for North’s hand from among the 31 nonspades not in South’s hand.
*
A standard deck consists of four suits, clubs, diamonds, hearts, and spades; each suit contains cards of 13 denominations.
1
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Thus, given that South has five spades, the conditional probability that North has
k
spades is
parenleftbigg
8
k
parenrightbigg
·
parenleftbigg
31
13

k
parenrightbigg
parenleftbigg
39
13
parenrightbigg
for
k
∈ {
0
,
1
,
2
, . . .,
8
}
.
This is the end of the preview.
Sign up
to
access the rest of the document.
 Spring '10
 NTD
 Probability theory, ........., probability density function, Willie Mays

Click to edit the document details