Homework #2 — Math 173 — Spring ’09
Solutions
(1) 5.1; 16)
(a) The dice are distinguishable. So the first pair of rolls yields 6·6 possibilities. We
get 6 ·6 possibilities for the second pair as well. Since they are all independent,
by the MR we have 64possibilities total.
(b) No restrictions are placed on the first pair of rolls. But we have only one choice
for each die in the second pair of rolls (i.e., whatever it showed the first time).
So this event contains 6 ·6·1·1 outcomes. By the probability principle, the
answer is 62/64= 1/36.
(c) It is tempting to have our sample space consist of the twelve possible sums (2
through 12). However, each of these outcomes is not equally likely. For example,
out of the 36 possible outcomes that can arise from rolling a red and a white
die, there is only one way to get a sum of 2 (red 1 and white 1), but two ways
to get a sum of 3 (red 1, white 2 or red 2, white 1).
Computing directly, we find the following values:
sum 2 3 4 5 6 7 8 9 10 11 12
# of ways 1 2 3 4 5 6 5 4 3 2 1
So, the probability of getting the same sum both times is
(12+ 22+ 32+ 42+ 52+ 62+ 52+ 42+ 32+ 22+ 12)/362
which equals 146/1296.
(d) By the previous part and the DR, the probability of getting unequal sums is
1−146/1296. Getting the higher sum on the first pair of rolls is equally likely
as getting the higher sum on the second pair of rolls. So getting the higher
pair on the second roll must happen half the time. Hence, the answer is (1 −
146/1296)/2 = 575/1296.
(2) 5.1; 24) The restriction means that the first, second, eleventh and twelfth digits must
be 3’s. There are no restrictions on the remaining 8 digits. So the answer is 108.
(3) 5.1; 32) There are ten letters in the word RECURRENCE and eight in RELATION.
So there are 10 ·8 = 80 possible choices. The only letters in common are R, E and
N. There are 3 ·1 ways to choose an R from each word, 3 ·1 ways to choose an E
from each word, and 1·1 ways to choose an N from each word. So by the probability
principle, the answer is (3 + 3 + 1)/80 = 7/80.
(4) 5.1; 46) The important fact to note is that I can specify rows in which the rectangle
lies without placing any restrictions on the columns it lies in. Of course, it is not
enough to specify how many rows, I must also specify which rows. Now, if the
rectangle has height one (i.e., lies in only one row), there are eight possibilities for
which row. If the rectangle has height two there are only seven possibilities (the top
row of the rectangle can’t be the bottom row of the chessboard). If the rectangle
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