332:321
Probability and Stochastic Processes
Examination 1
October 14, 2004
SOLUTION (version 1)
This exam was given in two versions. The questions were mostly the same but with different numeric
values. Solutions are included for both versions.
You have 80 minutes to answer the following
four
questions in the notebooks provided. You are permit-
ted 1 double-sided sheet of notes. Make sure that you have included your name, Rutgers netid and signature
in each book used. (5 points) Read each question carefully. All statements must be justified. Computations
should be simplified as much as possible.
Make sure to read both sides of this exam paper.
1.
30 points
SHORT ANSWER You must explain your answer.
(a) Events
A
and
B
are independent,
P
[
A
]
=
P
[
B
] and
P
[
A
∪
B
]
=
P
[
A
]
+
P
[
B
]. What is
P
[
A
]?
For any events A and B,
P
[
A
∪
B
]
=
P
[
A
]
+
P
[
B
]
−
P
[
AB
]
.
It follows that P
[
]
=
0
. Since A and B are independent and P
[
A
]
=
P
[
B
]
,
P
[
A
]
2
=
P
[
A
]
P
[
B
]
=
P
[
]
=
0
.
Thus P
[
A
]
=
0
.
(b)
K
is a Poisson
(α
=
2
)
random variable. What is
P
[
K
<
E
[
K
]]?
Since K is a Poisson
(α
=
2
)
random variable, K as PMF
P
K
(
k
)
=
(
2
k
e
−
2
/
k
!
k
=
0
,
1
,
2
,...
0
otherwise
Since E
[
K
]
=
2
,
P
[
K
<
E
[
K
]]
=
P
[
K
<
2]
=
P
K
(
0
)
+
P
K
(
1
)
=
2
0
e
−
2
/
0!
+
2
1
e
−
2
/
1!
=
3
e
−
2
(c) On each turn of the knob, a gumball machine is equally likely to dispense a red, yellow, green
or blue gumball, independent from turn to turn. After eight turns, what is the probability
P
[
R
2
Y
2
G
2
B
2
] of receiving 2 red, 2 yellow, 2 green and 2 blue gumballs?
In this problem, we are drawing gumball samples from the machine without replacement. The
requested probability is given by the multinomial probability
P
[
R
2
Y
2
G
2
B
2
]
=
8!
2! 2! 2! 2!
±
1
4
²
2
±
1
4
²
2
±
1
4
²
2
±
1
4
²
2
=
8!
4
10
≈
0
.
0385
.
2.
50 points
A particular birth defect of the heart is very rare; a random newborn infant will have the
defect with probability
P
[
D
]
=
10
−
4
. In the regular exam of a newborn, a particular heart arrhythmia
A
occurs with probability 0
.
99 in infants with the defect
D
. However, the arrhythmia
A
also appears
with probability 0
.
1 in infants without the defect
D
. When the arrhythmia is present, a lab test is
performed. The result of the lab test is either positive (event
T
+
) or negative (event
T
−
). The lab does
the test correctly with probability
p
=
0
.
999, independent from test to test. When the test is done
correctly, the test result correctly indicates whether the infant has the defect. When the lab does the
test incorrectly, the test result is always wrong. If the arrhythmia is present and the test is positive,
then the heart surgery (event
H
) is performed to repair the defect.
1