FALL 2006 FINAL EXAM SOLUTIONS
MATH 218, DECEMBER 13, 2006
This is a solution set for the Math 218 final common exam held from
2:00 to 4:00 PM on Wednesday, December 13, 2006.
There were ten
problems, weighted equally (for a total of
200
points).
These solu-
tions are rather more full and complete than anything we could have
expected from any student.
Problem 1
(20 pts)
.
30
marbles are sitting on a table:
6
red,
9
blue, and
15
green. All of the red marbles,
3
of the blue marbles,
and
10
of the green marbles are covered in chocolate, because a
child played with them while eating a piece of pie. The remaining
marbles were out of the child’s reach and remained clean.
(a)
Draw a probability tree describing the situation. Be sure to
include the labels of events, probabilities, and conditional
probabilities.
(b)
If a randomly selected marble is covered in chocolate, what
is the probability that it is red?
(c)
If a randomly selected marble is red, what is the probability
that it is covered in chocolate?
(d)
Are the events “red” and “covered in chocolate” indepen-
dent?
Solution.
(a) The marbles are partitioned in two ways: according to color,
and according to whether they’re covered in chocolate or are
clean. There are, accordingly, two ways you can draw the tree:
in the first, the first level of the tree is based on color, and the
second level based on chocolate/clean; in the second, the first
level of the tree is based on chocolate/clean, and the second is
based on color. The two ways are about equally easy.
Our first tree uses the color-first approach. Of course, you
were instructed to reduce fractions to lowest terms, so instead
of “
10/15
”, for example, as a conditional probability in the
tree, you should have written “
2/3
”. We have left the fractions
unreduced to make it more obvious where they came from.
Also, the calculation of the
∩
probabilities at the right edge
was not, strictly speaking, asked for; but they’re too useful for
the rest of the problem, and since they’re trivial to calculate,
you should probably always write them down.
Here’s the first tree:
R
B
G
6/30
9/30
15/30
Choc
Choc
Choc
Clean
Clean
Clean
6/6
0/6
3/9
6/9
10/15
5/15
P
(
Choc
∩
R
) =
1
5
P
(
Clean
∩
R
) =
0
P
(
Choc
∩
B
) =
1
10
P
(
Clean
∩
B
) =
1
5
P
(
Choc
∩
G
) =
1
3
P
(
Clean
∩
G
) =
1
6
In the second possible tree description, we have to count the
number of chocolate-covered marbles and the number of clean
marbles. Obviously there are
19
chocolate-covered marbles (
6
red,
3
blue, and
10
green), leaving
11
clean marbles (
0
red,
6
blue, and
5
green).
This tree has the advantage that all the
probabilities are already in lowest terms: