Vinh Nguyen
Stat 120A (Fall 2007): HW 2 Solutions
October 10, 2007
Text:
Probabilty and Statistics
, by Degroot and Schervish (3
rd
ed)
Problems:
1.7.5, 1.7.8, 1.8.6, 1.8.18, 1.9.8, 2.1.9, 2.1.10, 2.2.8, 2.2.16, 2.3.2, 2.3.8
1.7.5 Four dice are rolled and would like to compute the probability that each of the four numbers that appear
will be different. We would like the following to occur:
Dice
Result
Comment
Possibilities
1
D1
Any of 6 side appears
6
2
D2
Any side that is not D1
5
3
D3
Any side that is not D1 or D2
4
4
D4
Any side that is not D1, D2, or D3
3
Thus, the probability is
6
6
×
5
6
×
4
6
×
3
6
=
5
18
1.7.8 Each of the five elevator passenger will choose one of seven floors to get off. The probability for getting off
at a floor is
1
7
for each consumer. We want the probability that the five passengers will choose a different
floor to get off of (call this event E). This is exactly like Example 1.7.4 on p.25. The number of possible
outcomes for event E is
P
7
,
5
and the total number of possible outcomes is 7
5
(each person could get off at
any of the 7 floors, and they are all independent). Thus we have
p
=
P
7
,
5
7
5
=
7!
(7

5)!7
5
.
1.8.6 We have
n
people to be seated at random in a theater row of
n
seats. We want two particular people A
and B to be seated next to each other. Since they are to be seated next to each other, we could view the
two people as one when making the seating arrangements.
That is, treat the arrangement as arranging
n

1 people. The number of possible arrangements is therefore (
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 Spring '09
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 Statistics, Probability, Probability theory, possible ways, DeGroot, ﬁrst roll

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