Download Homework Week 3 Solutions 1. In the daily Play 4, players select a

Homework Week 3 Solutions
1. In the daily Play 4, players select a number between 0000 and 9999. The lottery machine
contains 4 bins, each with 10 ping-pong balls. Each bin has its ping-pong balls labeled
0, 1, 2, . . . , 9. The state then “randomly” selects a ball from each bin and forms a number
between 0000 and 9999.
(a) What is the probability you select the exact “4-digit” number?
1
1
=
4
10
10000
(b) You can choose to play a “BOX”, where if your numbers match in any order, then you
win.
i. If you select to play “BOX” and your numbers have the form 1112, what is the
probability that you win?
There are four choices for where the 2 goes (1112, 1121, 1211, 2111), so the proba4
1
bility of winning is 10000
= 2500
ii. If you select to play “BOX” and your numbers have the form 1122, what is the
probability that you win?
·µ ¶¸
4
There are six choices
for where the 2’s go (1122, 1212, 1221, 2112, 2121,
2
6
3
2211), so the probability of winning is 10000
= 5000
iii. If you select to play “BOX” and your numbers have the form 1123, what is the
probability that you win?
There are four choices for where the 2 goes and three choices for where the 3 goes,
12
3
so the probability of winning is 10000
= 2500
iv. If you select to play “BOX” and your numbers have the form 1234, what is the
probability that you win?
There are 4! ways to arrange the four numbers, so the probability of winning is
24
3
10000 = 1250
2. In Powerball, you select 5 (different) numbers between 1 and 53. You then select the red
Powerball, which is between 1 and 42. The Powerball Gods then select 5 different numbers
between 1 and 53 and the Powerball. You win the Jackpot if you match all the Powerball
Gods’ numbers, including the red Powerball.
(a) What is the probability that you match the (red) Powerball?
1
42
(b) What is the probability that you win the Jackpot?
1
¡53¢¡42¢ =
5
1
1
120, 526, 770
(c) What is the probability that you match exactly 4 numbers and the Powerball?
You have to choose which four balls you have right and then an incorrect ball from the
remaining 48 balls (that were not selected by the Powerball Gods).
¡5¢¡48¢
4
¡53
¢¡142¢ =
5
1
240
1
≈
120, 526, 770
502195
(d) If the Jackpot is over $150, 000, 000, then how come Bill Gates doesn’t just buy all
possible tickets and win the Jackpot? (Ignore the tax issue and the morality issue.)
If somebody else wins, they split the pot and Mr. Gates is out of A LOT of money.
HW pp. 65 − 68 :
µ ¶
20
#39a.
because you are just choosing 6 of 20 sticks.
6
#39b. You can think of this as having an integer equation x1 + x2 + · · · + x7 with x1 ≥ 0, x7 ≥ 0,
and the other variables xi > 0 for i = 2, 3, . . . , 6. This is because if you lay out the stick and circle
the first (or last) stick, nothing is wrong. If you circle two consecutive inner sticks, then that would
mean xi = 0 for some i = 2, 3, . . . , 6. Because you are selecting 6 sticks, that is why you have 7
variables.
Adjust the variable to make them all nonnegative (letting y1 = x1 , y7 = x7 and yi = xi − 1 for
iµ= 2, 3, . . ¶
. 6, soµ all¶yi ’s nonnegative), we get y1 + y2 + · · · + y7 = 14 − 5 = 9. So the answer is
9+7−1
15
=
.
7−1
6
#39c. You can think of this as having an integer equation x1 + x2 + · · · + x7 with x1 ≥ 0, x7 ≥ 0,
and the other variables xi > 1 for i = 2, 3, . . . , 6. As in part b, adjust the variable to make them
all nonnegative (letting y1 = x1 , y7 = x7 and yi = xi − 2 for
all yi ’s nonnegative),
µ i = 2, 3,¶. . . 6,µso ¶
4+7−1
10
we get y1 + y2 + · · · + y7 = 14 − 10 = 4. So the answer is
=
.
7−1
6
#42 Case 1: Everyone gets at least one orange drink
Choose who gets the lemon (4µchoices)
¶ andµwho
¶ gets the lime (3 choices): (x1 + x2 + x3 + x4 = 10,
10 − 1
9
with xi ≥ 0 for i = 1, 2, 3, 4.)
=
· 4 · 3 = 1008
4−1
3
Case 2: Exactly one person does not get an orange drink and that person gets either the lemon or
the lime drink.
There are 4 ways to choose which person does not get a drink, 2 choices for the drink that person
does get, 3 choices for who gets the other drink (of the lemon and lime). The rest is stars and bars
(with x1 + x2 + x3 = 10 and xi ≥ 0 for i = 1, 2, 3.)
µ
¶
10 − 1
4·2·3·
= 864
3−1
Case 3: Exactly two people do not get an orange drink.
·µ ¶
¸
4
Choose which two people did not receive drinks
= 6 ways .
2
Choose which of the two people get the lime (the other gets the lemon):µ 2 choices.
¶
10 − 1
The rest is stars and bars (with x1 + x2 = 10 and xi ≥ 0 for i = 1, 2.):
=9
2−1
The total for Case 3 is: 6 · 2 · 9 = 108
The final answer is: 1008 + 864 + 108 = 1980.
µ
¶ µ ¶
20 + 5 − 1
24
#45a: 5 bars and 20 stars:
=
= 10626.
5−1
4
#45b: Each book has 5 choices for the shelf: 520 .
#45c: Determine the number of books on each of the 5 shelves and arrange all the books in an
ordered pile. Then place the books on the shelves. For example, if you had 6 books on the 1st
shelf, 3 on 2nd shelf, 10 on the 3rd shelf, 0 on the 4th shelf, and 1 on the last shelf, you would put
the 1st 6 books from
¡ the
¢ pile on the 1st shelf, the next 3 on the 2nd shelf.... The total number of
ways to do this is: 24
4 · 20!
HW pp. 83 − 85 :
#5: Make n sets out of the 3n elements: {1, 2, 3}, {4, 5, 6}, . . . , {3n − 2, 3n − 1, 3n}. If you choose 2
from the same set, then the difference among the two elements is at most 2. By the PHP, you much
choose two from the same set, since you only have n sets and you are choosing n + 1 elements.
#15: Consider the following (n) differences: an+1 − a1 , an+1 − a2 , . . . , an+1 − an .
If any of these differences are divisible by n, then we are done. Therefore, we can assume that none
of them are divisible by n, and therefore the remainders when each difference is divided by n, is
between 1 and n − 1, i.e., there are n − 1 choices for the remainders. Since we have n differences
and n − 1 remainders, two of the differences must have the same remainder (when dividing by n).
Let’s say an+1 − aj and an+1 − ak have the same remainder when dividing by n. Thus we can write
an+1 − aj = pn + r and an+1 − ak = qn + r.
If you take the difference of these differences, i.e., (an+1 − aj ) − (an+1 − ak ) = ak − aj and (an+1 −
aj ) − (an+1 − ak ) = (pn + r) − (qn + r) = n(p − q). Therefore, ak − aj is divisible by n.
OR Jered’s Solution: You have n + 1 numbers. If you divide each number by n, there are n
choices for the remainder. This means two of the numbers (say ai and aj ) have the same remainder
r when divided by n. This means ai = pn + r and aj = qn + r for some integers p and q. Taking
the difference ai − aj , you get n(p − q), which is divisible by n.
#16: The number of possible acquaintances of each person is between 0 and n − 1. If two people
know the same number of people, then we are done. If not, then that means:
there is exactly one person who knows 0 people,
there is exactly one person who knows 1 person,
there is exactly one person who knows 2 people,
...
there is exactly one person who knows n − 1 people.
If this were the case, then there would be a person who know nobody and a person who knew
everybody. THAT CAN’T BE! Thus, two people must know the same number of people.
#17: The number of possible acquaintances of each person is between 0 and 98. If three people
know the same number of people, then we are done. If not, then that means:
there is exactly two people who know 0 people,
there is exactly two people who know 1 person,
there is exactly two people who know 2 people,
...
there is exactly two people who know 98 people.
If this were the case, then there would be two people who know nobody and a person who knows
98 people. THAT CAN’T BE since the person who knows 98 people cannot know herself or the
two people who know nobody. This would leave only 97 people left to know. Thus, three people
must know the same number of people.
#18: Divide the square up into four 1” by 1” squares. Because you are choosing 5 points, two of
the points must come from the same 1” by 1” square. The maximum distance between two points
in
√ a 1” by 1”
√ square is when the two points are at the corners of the square. This distance is
2
2
1 + 1 = 2.