MATH 218 FINAL EXAM SOLUTIONS
MAY 3, 2005
Solutions prepared by Ronald Bruck, May 2005.
N
ote
: Answers have been
marked in red.
The instructions specifically noted that answers were to be given either as
exact fractions reduced to lowest terms (when that was possible), or (if given
numerically) to four
exact
digits after the decimal point. Also, answers were
to be placed in answer boxes, where provided.
It is a very fortunate thing for some students that the examiners didn’t insist
on these conditions. They will not always be so lucky.
A remark on calculation: to compute to “four exact digits” requires comput-
ing to
five
digits. The extra digit is to prevent roundoff error, and is called a
“guard” digit. What
does not
work is to round off 4
/
3 to 1
.
3, use 1
.
3 through-
out a calculation, and then try to get the answer to four decimals! Once you’re
rounded off, you’ve lost that extra precision forever. Rounding 4
/
3 to 1
.
3 re-
sults in two significant figures of precision, at most.
Problem 1
(20 points)
.
Your company receives widgets from 3 suppliers, A,
B and C. Sixty percent of the widgets come from A, thirty percent come from
B, and the rest come from C. Company A is excellent, and only 5 percent of
their widgets are defective. Company B is not so good, and 20 percent of their
widgets are defective. Company C is pretty bad, and half of their widgets are
defective.
You take a random widget arriving at your company and test it. Write D for
the event that it is defective.
1a)
Draw a tree diagram to represent this situation. Include all events
and probabilities involved—individual, conditional, and joint (inter-
section) probabilities. [
REMINDERS:
branches from the root are
labeled with individual probabilities, further branches are labeled
with conditional probabilities, and terminal nodes represent inter-
sections and should be labeled with the (joint() probability of that
intersection].
1b)
What is the chance that the tested widget is defective?
1c)
Suppose the tested widget is defective.
What is the chance that it
came from A?
1d)
Suppose the tested widget is defective.
What is the chance that it
came from B?
1e)
Suppose the tested widget is defective.
What is the chance that it
came from C?
1f)
Suppose the tested widget
is NOT
defective. What is the chance that
it came from A?
1g)
[Yes or no.] Should the probabilities for 1c), 1d), and 1e) add up to
1?
1h)
[Yes or no.] Should the probabilities for 1c), 1f) add up to 1?
Solution.
1a) The tree is in Figure (1), on the next page.
A
D
D
0.05
0.95
P(A&D)=0.6x0.05=0.03
P(A&D)=0.6x0.95=0.57
B
D
D
0.20
0.80
P(B&D)=0.3x0.2=0.06
P(B&D)=0.3x0.8=0.24
C
0.6
0.3
0.1
D
D
0.50
0.50
P(C&D)=0.1x0.5=0.05
P(C&D)=0.1x0.5=0.05
F
igure
1. Tree diagram for widget suppliers (A, B, C) vs.
defective (D)
1b)
D
=
(
D
&
A
)
∪
(
D
&
B
)
∪
(
D
&
C
)
, and these are mutually exclusive
and have probabilities 0
.
03, 0
.
06, and 0
.
05, thus
P
(
D
)
=
0
.
03
+
0
.
06
+
0
.
05
=
0
.
14
.