Mathematics Department, UCLA
T. Richthammer
winter 09, sheet 5
Jan 30, 2009
Homework assignments: Math 170A Probability, Sec. 1
051. An apartment complex is equipped with an alarm system that is supposed to directly give
alarm at a police station if there is a burglary. If a burglary occurs it gives alarm with a
probability of 99%. If there is no burglary it gives alarm with a probability of 0
.
1%. It
is known that a burglary at this apartment complex occurs at about 1 in 1000 days. One
day the police station is getting an alarm from the system. What is the probability that
there really is a burglary? (First describe a probabilistic model!)
Answer:
Model:
S
=
{
b, n
} × {
a, n
}
(burglary/no burglary and alarm/no alarm.) We have
p
b
=
0
.
001,
p
n
= 0
.
999,
p
a

b
= 0
.
99,
p
n

b
= 0
.
01,
p
a

n
= 0
.
001 and
p
n

n
= 0
.
999. By the lecture
this characterizes the model completely. For
A
=
{
b, n
} × {
a
}
and
B
=
{
b
} × {
a, n
}
we
have to calculate
P
(
B

A
). Using the formula of Bayes or the tree diagram we obtain
P
(
B

A
) =
P
(
A

B
)
P
(
B
)
P
(
A

B
)
P
(
B
) +
P
(
A

B
c
)
P
(
B
c
)
=
p
a

b
p
b
p
a

b
p
b
+
p
a

n
p
n
≈
49
.
8%
.
052.
Variants of the famous Monty Hall Problem.
Suppose you’re on a game show, and you’re
given the choice of three doors: Behind one door is a car; behind the others, goats. You
pick a door (at random), and the host opens another door. There turns out to be a goat.
He then says to you, ”Do you want to stick with your door, or do you want to switch?”
Is it to your advantage to switch your choice? Answer the question based on a suitable
probabilistic model,
(a) assuming that the moderator picks one of the two remaining doors at random (so it
is possible that he ﬁnds a car behind the door he opened).
(b) assuming that the moderator always picks a door with a goat behind it.
Answer:
Let
S
=
{
c, g
1
, g
2
}
2
, the projections
X
1
, X
2
describe what is behind your door and what is
behind the host’s door, where
c
denotes the car and
g
1
,
g
2
are the two goats. For (a) and (b)
we get
p
c
=
p
g
1
=
p
g
2
= 1
/
3, but we have diﬀerent conditional probabilities for the second
step. In both cases we have to calculate
P
(
X
1
=
c

X
2
±
=
c
), i.e. the probability that you
win if you don’t switch, given the host found a goat behind his door. Using Bayes formula:
P
(
X
1
=
c

X
2
±
=
c
) =
P
(
X
1
=
c
)
P
(
X
2
±
=
c

X
1
=
c
)
P
(
X
1
=
c
)
P
(
X
2
±
=
c

X
1
=
c
)+
P
(
X
1
=
g
1
)
P
(
X
2
±
=
c

X
1
=
g
1
)+
P
(
X
1
=
g
1
)
P
(
X
2
±
=
c

X
1
=
g
1
)
=
p
c
(1

p
c

c
)
p
c
(1

p
c

c
)+
p
g
1
(1

p
c

g
1
)+
p
g
2
(1

p
c

g
2
)
(a) Here
p
g
1

c
=
p
g
2

c
= 1
/
2,
p
c

g
1
=
p
g
2

g
1
= 1
/
2, and so on. So
P
(
X
1
=
c

X
2
±
=
c
) =
1
/
3
·
1
1
/
3
·
1+1
/
3
·
1
/
2+1
/
3
·
1
/
2
=
1
1+1
/
2+1
/
2
=
1
2
. Thus it doesn’t matter if you switch or not.
(b) Here