HW1 Solutions
Dilip Raghavan
September 6, 2008
1
Section 5.1
Problem 16.
We model the outcomes as 4–tuples (
R
1
,
W
1
,
R
2
,
W
2), where
R
1 represents the first roll of the
red die and
R
2 the second, and similarly for
W
1 and
W
2.
a. There are 6 possible outcomes for each of
R
1,
R
2,
W
1 and
W
2. So there are a total of 6
4
outcomes.
b. We need R1
=
R2 and W1
=
W2. So if R1 and W1 are choosen, R2 and W2 are determined. There are
6 possible outcomes for each of R1 and W1. So there are 6
2
favorable outcomes. So the probability is
no. of favorable outcomes
no. of total outcomes
=
6
2
6
4
.
c. The sum can be anything from 2 to 12. It is best to break up the problem into cases depending on
what the sum is. Notice that given a fixed value for the sum, if the value of
R
1 is choosen, then the
value of
W
1 is automatically determined. So for any given sum, it is enough to figure out how many
outcomes there are for
R
1 and for
R
2.
(a) The sum is 2.
R
1 can only be 1. Same for
R
2. So 1
2
outcomes.
(b) The sum is 3.
R
1 can be either 1 or 2. Same for
R
2. So 2
2
outcomes.
(c) The sum is 4.
R
1 can be anything from 1 to 3. Same for
R
2. So 3
2
outcomes.
(d) The sum is 5.
R
1 can be anything from 1 to 4. Same for
R
2. So 4
2
outcomes.
(e) The sum is 6.
R
1 can be anything from 1 to 5. Same for
R
2. So 5
2
outcomes.
(f) The sum is 7.
R
1 can be anything from 1 to 6. Same for
R
2. So 6
2
outcomes.
(g) The sum is 8.
R
1 can be anything from 2 to 6. Same for
R
2. So 5
2
outcomes.
(h) The sum is 9.
R
1 can be anything from 3 to 6. Same for
R
2. So 4
2
outcomes.
(i) The sum is 10.
R
1 can be anything from 4 to 6. Same for
R
2. So 3
2
outcomes.
(j) The sum is 11.
R
1 can be anything from 5 to 6. Same for
R
2. So 2
2
outcomes.
(k) The sum is 12.
R
1 has to be 6. Same for
R
2. So 1
2
outcomes.
So the number of favorable outcomes is:
1
2
+
2
2
+
3
2
+
4
2
+
5
2
+
6
2
+
5
2
+
4
2
+
3
2
+
2
2
+
1
2
.
1
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d. The number of outcome where
R
1
+
W
1
R
2
+
W
2 can be gotten by subtracting the answer to part
c. from the total number of outcomes. Since the situation is symmetrical, the number of outcomes
where
R
1
+
W
1
<
R
2
+
W
2 is half of the number of outcomes where
R
1
+
W
1
R
2
+
W
2. So the
number of favorable outcomes is
1
2
6
4

1
2
+
2
2
+
3
2
+
4
2
+
5
2
+
6
2
+
5
2
+
4
2
+
3
2
+
2
2
+
1
2
.
Problem 22.
The first slot can have any of 0, 1 or 2. So there are 3 possibilities for the first slot. Every slot
after the first one can have anything
except
what is in the slot right before it. So there are 2 possibilities for
every slot after the first one. So there are a total of 3
×
2
9
outcomes.
Problem 23.
In all three problems it is best to first count the possibilities between 1,000 and 9,999 and then
to deal with 10,000 separately.
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 Natural number, Universal quantification

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