CS237. Practice Problems Set 4: Basic Probability.
Graded Problems due Thurs Feb 17 11:59PM.
February 20, 2011
Reading.
Schaum’s Chapter 4. L&L Chapter 1920.
Optional Extra Practice.
Any of the problems in Schaums’s Chapter 4. MitzenmacherUpfal
problems 1.11.16.
1
Practice problems, ungraded
Hint.
Use tree diagrams where ever possible!
Exercise 1.
There are three caves in Boston. A wolf is looking for a home; he moves into cave 1
with probability
5
9
, cave 2 with probability
1
9
and cave 3 with probability
1
9
, and with the remaining
probability he moves to San Diego. A goat moves into one of the unoccupied caves, choosing his
cave with equal probability. These caves are in Boston, so it snows with probability
9
10
. The goat
always leaves tracks in the snow, but the wolf does not (he’s clever and has learned how to cover
his tracks). What is the probability that the wolf lives in cave 2 given that the there are no tracks
in front of cave 1?
Solution: Let
W
1
,
W
2
,
W
3
and
W
S
be the event that the wolf moves into caves 1, 2, 3 and San Diego respectively.
Similarly, let
G
1
, . . . ,
G
3
be the events that the goat moves into the caves
1
, 2 and 3 respectively. Let
S
be the event of
snow, and
¯
S
be the event of no snow. If there are no tracks in cave 1 (call this event
A
), either the goat is not in cave 1,
or it didn’t snow; so we can make a tree diagram to solve this (first step  wolf choose cave, second step  goat chooses
cave, third step  it snows or not). Draw the tree, you’ll find the following:
Pr[
A
] = Pr[
¬
G
1
∪
¯
S
] = Pr[
W
1
] + Pr[
W
2
G
3
] + Pr[
W
2
G
1
¯
S
] + Pr[
W
3
G
2
] + Pr[
W
3
G
1
¯
S
] + (Pr[
W
4
]

Pr[
W
4
G
1
S
])
= 5
/
9 + (1
/
9)(1
/
2) + (1
/
9)(1
/
2)(1
/
10) + (1
/
9)(1
/
2) + (1
/
9)(1
/
2)(1
/
10) + (2
/
9

(2
/
9)(1
/
3)(9
/
10))
Now, we want the probability that the wolf is in cave 2 given no tracks in front of cave 1. Let event
B
be the event that
the wolf is in cave 2. We want
Pr[
B

A
] =
Pr[
A
∩
B
]
Pr[
A
]
We already worked out
Pr[
A
]
.
Referring again to our tree, we see that event
A
∩
B
contains only the outcomes
{
W
2
G
1
¯
S, W
2
G
3
¯
S, W
2
G
3
S
}
, so it occurs with probability
Pr[
A
∩
B
] = (1
/
9)(1
/
2)(1
/
10) + (1
/
9)(1
/
2)(9
/
10) + (1
/
9)(1
/
2)(1
/
10)
We get the answer after plugging the numbers into our calculators.
Exercise 2.
Dirty Harry puts 2 bullets in a 6cell cylinder of his revolver. He gives a cylinder a
random spin and says “Feeling lucky”? He pulls the trigger.
1. What’s the probability that his victim gets shot?
1
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2. Suppose his victim didn’t get shot, but now Harry pulls the trigger again (without giving the
gun another random spin). What’s the probability that his victim gets shot the second time
Harry pulls the trigger?
3. Now suppose you noticed that DH put the bullets one next to the other in the cylinder. How
does this change the answer to the previous two questions?
Solution: Let
S
1
be probability of being shot the first time, and
S
2
the probability of being shot the second time.
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 Spring '11
 Goldberg
 Probability, Probability theory, Harry, San Diego

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