Download HW3 Solution - UCSD VLSI CAD Lab

Yuncong Chen
Assignment HW3 due 04/20/2014 at 11:59pm PDT
CSE21 Spring2014
allowed at most 1. Let’s pass give each girl one red jellybean so
that we only have 6 left to distribute. Now, we must break the
problem into the different possibilities:
1. (3 pts)
REMEMBER that you can use C(n,k) or P(n,k) as functions
in your answer.
1. The 6 jellybeans are distributed to all 4 girls (and none go
to the boys)
2. 5 jellybeans go to the 4 girls, and 1 goes to one boy
3. 4 jellybeans go to the 4 girls, and 2 go to two boys (the
boys can have at most one)
4. 3 jellybeans go to the 4 girls, and 3 go to the three boys
How many ways are there to distribute 10 red jellybeans and 8
blue jellybeans to 4 girls and 3 boys if:
◦ There are no restrictions?
We do not have anymore cases because we cannot give 4 beans
to 3 boys, since they are allowed at most one jellybean. Now
let’s find the number of distributions for each case:
◦ Everyone gets at least one red jellybean?
1. This isvery similar to the previous problems. There are
6+4−1
ways to distribute 6 jellybeans among 4 girls.
4−1
2. Now we have 5 jellybeans for the girls. There are 5+4−1
4−1
ways to distribute them. However, we still need to give one
of theboys a jellybean. Since there are three boys,
3 there
are 31 ways to do this. Thus, there are 5+4−1
4−1
1 possibilities for this case.
3. This case can be calculated very similarly to case 2. This
time, there are only 4 beans for the 4 girls, and we need to
choose two
3boys
to give the last two beans to. This gives
us 4+4−1
4−1
2 ways to distribute 4 beans among the girls
and 2 beans among the boys.
4. This last
case is also
similar to case 2. There are
3+4−1 3
3+4−1
=
ways
to distribute 3 jellybeans to 4
4−1
3
4−1
girls, and 3 boys to three boys such that each boy can only
receive one bean.
◦ Every girl gets exactly two blue jellybeans and at least one
red jellybean, and every boy gets at most one red jellybean
and at most one blue jellybean.
SOLUTION:
(a) The number of ways to distribute all the jellybeans is the
number of ways to distribute the red jellybeans times the number
of ways to distribute the blue jellybeans. Once the problem is
separated this way, we can think stars and bars. To distribute the
red jellybeans, consider the 10 jellybeans to be 10 starsthat must
be distributed among 7 people (6 bars). This is 10+7−1
. We can
7−1
do the same thing for blue jellybeans: 8+7−1
.
So,
the
number
7−1
10+7−1 8+7−1
of ways to distribute all the jellybeans is 7−1 · 7−1 .
Since these cases are disjoint, we can add them together to get
the total number of ways to distribute the jellybeans with the
given
constraints:
6+4−1
5+4−1 3
4+4−1 3
3+4−1
+
+
+
4−1
4−1
1
4−1
2
4−1
(b) This can be solved the same way as above, though we need
to modify the number of ways to distribute the red jellybeans.
Each girl and boy must get at least one, so give each of them one
red jellybean. This leaves us with 3 red jellybeans that need to
be distributed to the 7 people. This means that there are 3+7−1
7−1
ways to distribute the red jellybeans, making the totalnumber
8+7−1
of ways to distribute all the jellybeans 3+7−1
7−1 · 7−1 .
Correct Answers:
• C(10+7-1,7-1)*C(8+7-1,7-1)
• C(3+7-1,7-1) * C(8+7-1, 7-1)
• C(6+4-1, 4-1) + C(5+4-1,4-1)C(3,1)
+
C(4+4-1,4-1)C(3,2) +
2. (1 pt)
Let A and B be events with P(A) = 17 , P(B) = 12 , and P((A ∪
B)c ) = 37 . What is P(A ∩ B)?
(c) The first thing to recognize is that each of the four girls must
have exactly 2 blue jellybeans. This means that there are no
blue jellybeans to distribute among the boys, and there is only
one way to give each girl exactly 2 blue jellybeans. So, we will
ignore the blue jellybeans for the rest of this problem. For the
red jellybeans, each girl must receive at least 1, and each boy is
1
SOLUTION:
4. (2 pts)
(a) What is the coefficient of x2 y3 in the expansion of (2x−3y)5 ?
P(A ∪ B) = P(A) + P(B) − P(A ∩ B)
This can be rewritten as:
(b) What is the coefficient of xy2 z3 in the expansion of (x + 2y +
3z)6 ?
P(A ∩ B) = P(A) + P(B) − P(A ∪ B)
Recall that P(A ∪ B) = 1 − P((A ∪ B)C ) so:
P(A ∩ B) = P(A) + P(B) − (1 − P((A ∪ B)C ))
So plug in the given probabilities:
P(A ∩ B) =
SOLUTION:
1 1
3
+ − (1 − )
7 2
7
(a) Theorem 8 in your book, the Binomial Theorem, states:
n
Correct Answers:
(x + y)n =
• [1]/7 + 1/2 - (1-[3]/[7])
3. (2 pts)
n
∑ k xn−k yk
k=0
We can use this equation to find the coefficient we need. n = 5,
since the expression is (2x − 3y)5 , and k = 3 since we are
looking
for x2 y3 . So, the term when n = 5 and k = 3 is
5
5!
5−3
· 22 x2 · (−3)3 y3 = (10 · 4 · −27)x2 y3 =
(−3y)3 = 3!2!
3 (2x)
−1080 · x2 y3 .
5 light bulbs are chosen at random from 16 bulbs of which 6 are
defective.
(a) What is the probability that exactly 2 are defective?
(b) For (x + 2y + 3z)6 we first need to derive a formula. We can
do this by using the Binomial Theorem twice:
(b) What is the probability that at most 1 is defective?
(x + y + z)n = (x + w)n where w = y + z
n
= ∑
k=0
SOLUTION:
n
= ∑
(a) The probability that exactly 2 bulbs are defective is the number of ways to count exactly 2 defective bulbs divided by the
number of ways to count 5 bulbs. When we count bulbs, we do
not care about the order.
This means that the number of ways
to count 5 bulbs is 16
5 . The number of ways to count exactly
6 10
2 defective bulbs is 2 3 because there are 6 defective bulbs,
so there are 62 ways to count them, and there are 10 working
bulbs, and we need to count 3 of them. So the probability of
(6)(10)
finding exactly 2 defective bulbs is 2 16 3 .
(5)
k=0
n
= ∑[
k=0
n
n n−k k
w
k x
n n−k
(y + z)k
k x
n n−k
k x
k
= ∑ ∑
k=0 j=0
n
k
= ∑ ∑
k=0 j=0
(b) The probability to count at most 1 defective bulb is the probability to count exactly 1 defective bulb plus the probability to
count 0 defective bulbs. Each probability can be solved like in
(a). So, the final answer is
(61)(104) (60)(105)
+ 16
(165)
(5)
n
k
= ∑ ∑
k=0 j=0
n
k
= ∑ ∑
k=0 j=0
k
∑
j=0
k k− j j
z
j y
n k n−k k− j j
y z
k j x
k!
n!
n−k yk− j z j
(n−k)!k! (k− j)! j! x
n!
n−k yk− j z j
(n−k)!(k− j)! j! x
n−k k− j j
n
y z
(n−k),(k− j), j x
Correct Answers:
Now that we have a formula, we can find the coefficient of
• (C([6],2) C([16-6],[5-2]))/C([16], [5])
• C([6],1)C([16-6],[5-1])/C([16],[5]) + C([6], 0) C([16-6],
xy2 z3 in[5])/C([16],[5])
(x + 2y + 3z)6 . n = 6, n − k = 1, k − j = 2, and j = 3.
2
6 So, the term we are looking for is 1,2,3
(1x)1 (2y)2 (3z)3 =
6!
2
2
3
3
2
3
2 3
1!2!3! x2 y 3 z = (60 · 4 · 27)xy z = 6480 · xy z .
(a) There is exactly one arrangement - (1,2,3,4,5,6,7,8,9)
(b) We do this by counting those arrangements that have ai ≤
ai+1 except, perhaps, for i = 5. Then we subtract off those that
also have a5 < a6 . In set terms:
◦ S is the set of rearrangements for which a1 < a2 < a3 <
a4 < a5 and a6 < a7 < a8 < a9 ,
◦ T is the set of rearrangements for which a1 < a2 < a3 <
a4 < a5 < a6 < a7 < a8 < a9 , and
◦ we want |S \ T | = |S| − |T |.
Correct Answers:
• C(5,3)2ˆ(5-3)(-3)ˆ3
• 6!/(1!2!3!)*2ˆ2*3ˆ3
5. (1 pt)
A rook on a chessboard attacks its own square, and every square
in the same column or row on the board. Given an 8 x 8 chessboard with all 64 squares distinguished (columns a-h, rows 1-8),
if 8 rooks are placed onto 8 random squares, what is the probability that they are mutually non-attacking (that is, no rook attacks another)?
An arrangement in S is completely determined by specifying the
set {a1 , · · · , a5 }, of which there are C(9,5)=126. In (a), we saw
that |T|=1. Thus the answer is 126-1=125.
Correct Answers:
• 1
• 125
SOLUTION:
7. (6 pts)
Since rooks are not distinguished, the total number of configurations of 8 rooks is C(64,8).
You draw 9 cards at random, one at a time WITH replacement,
from an ordinary 52-card deck.
Now we want to count the number of choices of squares for
rooks that are non-attacking.
In any configuration of non-attacking rooks, there can be at most
one rook per row (rank). In addition, since rooks can attack
along columns (files), the column of rooks in each row must be
different.
Thus we can imagine enumerating every configuration of nonattacking rooks by originally placing rooks along the diagonal
of the board, then permuting the rows of the board.
There are 8! ways that rows of a chessboard can be permuted,
thus 8! non-attacking rook configurations.
(a) What is the probability that no Ace appears in any of the
draws?
(b) What is the probability that at least one King appears?
(c) What is the probability that at least 2 Queens appear?
So the probability is 8!/C(64,8).
Correct Answers:
• 8!/C(64,8)
6. (2 pts)
Now, answer the same three questions assuming that the 9 cards
are drawn at random WITHOUT replacement.
Consider all possible 9-lists (a1 , a2 , . . . , a9 ) without repetition
that can be made from the 9-set {1,2,3,4,5,6,7,8,9}. For example, (9,2,4,5,6,1,3,7,8) is one such list.
(a) How many have the property that ai < ai+1 for all i ≤ 8?
(d) What is the probability that no Ace appears in any of the
draws?
(e) What is the probability that at least one King appears?
(b) How many have the property that ai < ai+1 for all i ≤ 8 except i = 5?
(f) What is the probability that at least 2 Queens appear?
SOLUTION:
3
SOLUTION:
(a) Suppose that, among the initial 5 cards, exactly three cards
are of the same suit. The player decides to keep these three
cards, and replaces the other two cards.
(a) To compute probability of no Ace, we need to find the total
number of 9-card hands and the number of 9-card hands without
an Ace. The number of ways to draw ten cards with replacement
is 529 and the number of ways to draw 9 cards with replacement
9
and not pull in Ace is 489 . Thus P(no Ace) = 48
.
529
What is the probability of her ending up with a flush?
(b) The number of ways to draw 9 cards with replacement and
draw at least one king is the same as using the complement and
subtracting the number of hands without kings from the number
of possible hands. So, the probability to draw at least one king
9
9
9
= 1 − 48
.
is 52 52−48
9
529
(b) Suppose that, among the initial 5 cards, there is exactly one
pair (e.g., 8♥, 8♣) and one “kicker” (e.g., A♠). The player decides to keep the pair and the kicker, and replaces the other two
cards.
(c) For this problem we do the same as part b, but we also
must subtract the number of ways to draw exactly 1 Queen.
The number of ways to draw exactly one Queen is 4 · 488 · 9,
since there are 4 Queens to draw from, and 10 places to draw
as the Queen. So, the probability to draw at least 2 Queens is
9
8
529 −489 −4·9·488
= 1 − 48
− 4·9·48
.
529
529
529
What is the probability of her ending up with:
◦ 3 of a kind? (note: this means exactly 3-of-a-kind, not including those that can also fall in a higher category. The
same goes for the following two questions)
◦ 2 pair?
If without replacement, the total number of 9-card hands is
C(52, 9). All answers below are counts for the specific event.
To get the probability of the event, simply divide the count by
the total number, C(52, 9).
◦ full house?
(d) There are C(48, 9) 9-card hands that do not contain Ace.
(e) This is the total number of 9-card hands minus those that do
not contain King. The count is C(52, 9) −C(48, 9).
SOLUTION:
(f) Similar to (e), except that we also need to remove those
hands that have exactly one Queen. There are C(4, 1) ·C(48, 8)
hands of this kind. So the count is C(52, 9) −C(48, 9) −C(4, 1) ·
C(48, 8).
(a) Suppose the suit of the three cards is spade. The total number of ways to choose 2 cards from the deck of remaining cards
is C(52 − 5, 2). Out of these, there are C(13 − 3, 2) cases where
both cards are spade. So the probability of ending up with a
flush is C(13 − 3, 2)/C(52 − 5, 2).
Correct Answers:
•
•
•
•
•
•
48ˆ[9]/52ˆ[9]
(52ˆ[9] - 48ˆ[9])/(52ˆ[9])
(52ˆ[9] - 48ˆ[9] - 4*[9]*48ˆ[9-1])/(52ˆ[9])
C(48,[9])/C(52,[9])
(C(52,[9])-C(48,[9]))/C(52,[9])
(C(52,[9])-C(48,[9])-C(4,1)*C(48,[9-1]))/C(52,[9])
8. (4 pts)
In 5-card draw poker, a player receives an initial hand of 5 cards,
and is then allowed to replace up to three of her cards with the
remaining cards in the deck.
(b) Suppose the rank of the pair is 8, and the kicker is Ace.
To end up with a three-of-a-kind, the two newly-drawn must
contain exactly one 8 and no Ace. The probability of this is
2 · (52 − 5 − 2 − 3)/C(52 − 5, 2).
(c) To end up with a two pair, either she draws exactly one Ace
and no 8, or she draws exactly a pair that are neither Ace or 8. Of
the former, there are 3·(52−5−2−3) cases. Of the latter, there
are (13 − 2) ·C(4, 2) cases. Finally, you must remove the 6 pairs
that used the two cards you threw away. The probability is therefore, (3 · (52 − 5 − 2 − 3) + (13 − 2) ·C(4, 2) − 6)/C(52 − 5, 2)
(d) To end up with a full house, either she draws (8,A), or
she draws (A,A). There is 2 · 3 = 6 cases for the former, and
4
C(3, 2) cases for the latter. The probability is therefore, (2 · 3 +
C(3, 2))/C(52 − 5, 2)
10. (1 pt)
Correct Answers:
•
•
•
•
A bin contains 8 Red marbles and 6 Blue marbles. 4 marbles
are randomly drawn without replacement. After that, one more
marble is drawn.
C(13-3,2)/C(52-5,2)
2 * (52-5-2-3) / C(52-5,2)
(3 * (52-5-2-3)+(13-2)*C(4,2) - 6 )/C(52-5,2)
(2 * 3 + C(3,2))/C(52-5,2)
What is the probability that this last marble drawn is Red?
9. (1 pt)
In seven-card games (7-card stud, Texas Hold-em) you make the
best five-card hand possible out of seven available cards.
SOLUTION:
How many combinations of seven cards contain a full house as
the best five-card hand?
Simple solution:
We are not given the results of the 4 randomly drawn marbles,
so we can answer this question as though the 4 marbles were
drawn with replacement. Since there are 8 red marbles, and a
total of 8 + 6 = 14 marbles, the probability that the last marble
8
drawn is red is 14
.
SOLUTION:
With 7 cards, a full house may be constructed in 1 of 3 ways:
Alternate solution:
◦ 1 triple, 1 pair and 2 kickers
If the above solution is unsatisfactory, we can calculate this using the many cases:
◦ Draw 4 red marbles + 0 blue marbles, then draw a
red marble: For this problem, we do not care about
the order of the first 4 marbles, we only care how
many of each color were drawn, and how many ways
we can choose the number of marbles of each color.
The
number of ways to choose 4 of 8 red marbles is
8
and
the number of ways to choose 0 of 6 blue mar4
bles is 60 . So, the number of ways to choose 4 red
marbles and 0 blue marbles is 84 60 . Lastly, we need
to account for the number of ways to draw the fifth
marble as red. Since there were 8 marbles, and we
already pulled out 4, there are 4 marbles left, or 41
ways to choose the last marble as red. This makes the
total number of ways for this case:
The triple may be 1 of 13 ranks, and by definition 3 of the
4 of that rank are chosen. The pair may be 1 of the remaining 12 ranks, and (again, by definition) 2 of the 4 of
that rank are chosen. The ranks of the 2 kickers are chosen
from the remaining 11 ranks, and 1 of the 4 of each rank
are chosen. Thus, the total number of full houses in this
form is: C(13, 1)C(4, 3)C(12, 1)C(4, 2)C(11, 2)C(4, 1)2 =
3, 294, 720
◦ 1 triple and 2 pairs
The triple is chosen the same way as before, the ranks
of the two pairs are chosen from the remaining 12 ranks,
and the 2 of the 4 of each rank are chosen as usual.
Thus, the total number of full houses in this form is:
C(13, 1)C(4, 3)C(12, 2)C(4, 2)2 = 123, 552
8 6 4
4 0 1
◦ Draw 3 red marbles + 1 blue marbles, then draw a
red marble:
8 6 5
3 1 1
◦ Draw 2 red marbles + 2 blue marbles, then draw a
red marble:
8 6 6
2 2 1
◦ Draw 1 red marbles + 3 blue marbles, then draw a
red marble:
◦ 2 triples and 1 kicker
The ranks of both triples are chosen from the 13, then
the rank of the kicker is chosen from the remaining 11
ranks. Thus, the total number of full houses in this form
is: C(13, 2)C(4, 3)2C(11, 1)C(4, 1) = 54, 912
Thus, the total number of full houses is 3, 473, 184:
Correct Answers:
• 3473184
5
8 6 7
1 3 1
◦ Draw 0 red marbles + 4 blue marbles, then draw a
red marble:
8 6 8
0 4 1
So the total number of ways to choose 4 marbles, and then
choose a red marble is:
8 6 4
8 6 5
+
4 0 1
3 1 1
probability of drawing a white marble in the second draw. There
are 4 white marbles, but now the total number of marbles in the
4
jar is 8 + 4 + 2 = 14, making this second probability 14
. So, the
probability of drawing a red marble in the first draw and then
4
9
× 14
.
drawing a white marble in the second draw is 15
(b)) The process for this question is very similar to part a. The
4
probability of drawing white on the first draw is 15
and the probability of drawing white on the second draw, assuming that it
3
. Thus, the probability of drawwas white in the first draw is 14
ing a white marble in the first draw and then a white marble in
4
3
the second draw is 15
× 14
.
8 6 6
8 6 7
8 6 8
+
+
+
2 2 1
1 3 1
0 4 1
(c)) For yellow not to be drawn at all, then either red or white
need to be drawn on the first draw, and then we need to draw either red or white on the second draw. Since there are 9 + 4 = 13
red or white marbles, and there are a total of 9 + 4 + 2 = 15 marbles in the jar, the probability of not drawing a yellow marble in
the first draw is 13
15 . Assuming that we drew red or white on the
first draw, then there are 13 − 1 = 12 red or white marbles left
in the jar, and a total of 14 marbles left in the jar. So, the probability of not drawing yellow in the second draw is 12
14 , making
13
12
the probability of not drawing yellow at all to be 15 × 14
.
= 280 + 1680 + 2520 + 1120 + 120 = 5720
Lastly, we need to divide this total by the number of ways to
randomly choose 4 marbles and then draw a fifth one. The number of ways to draw 4 of 14 marbles is 14
. Then, there are 10
4 marbles left to choose the last marble, or 10
1 ways. So, the total
14 10
number of ways is 4 1 = 10010.
Correct Answers:
So, the probability to choose a red marble on the fifth draw is:
8
5720
10010 = 14
• (9/15) * (4/14)
• (4/15)*(3/14)
• (13/15)*(12/14)
Correct Answers:
• 8/14
12. (2 pts)
11. (3 pts)
(a) How many one-to-one (i.e., injective) maps are there from
the set {1, 2, 3, 4} to {a, b, c, d, e, f}?
Two marbles are drawn randomly one after the other, WITHOUT replacement, from a jar that contains 9 red marbles, 4
white marbles, and 2 yellow marbles. Find the probability of
the following events:
(b) How many onto (i.e., surjective) maps are there from {a, b,
c, d, e, f} to {1, 2, 3, 4}?
(a) A red marble is drawn rst followed by a white marble.
(b) A white marble is drawn rst followed by a white marble.
SOLUTION:
(a) A map f : A → B is one-to-one if f(x) = f(y) implies that
x = y. In other words, every element in A must be mapped to
a unique element in B. If A = {1, 2, 3, 4} and B = {a, b, c,
d, e, f}, then we want to find the number of ways we can map
the four elements in A to 4 of the 6 elements in B. There are
6 ways ways to map the first element, then 5 for the second, 4
for the third, and lastly 3 for the fourth. This gives us a total of
6 · 5 · 4 · 3 = 360 mappings.
(c) A yellow marble is not drawn at all.
SOLUTION:
(a) First we need to find the probability of drawing a red marble
in the first draw. There are 9 red marbles and 9 + 4 + 2 = 15
9
marbles total in the jar, giving us a probability of 15
. Assuming
we drew a red marble in the first draw, now we need to find the
(b) A map f : A → B is onto if for b ∈ B, there exists a ∈ A
such that f(a) = b. In other words, every element in B must be
mapped to at least once. So, if A = {a, b, c, d, e, f} and B =
6
{1, 2, 3, 4}, then we need to find how many ways we can map
the 6 elements in A to the 4 elements in B. One way to think
of this is to count all the ways we can break up the 6 elements
into 4 groups, and then assign the groups to an element of B. As
can be read about in your book, the number of ways to break up
6 elements into 4 groups is S(6, 4). Since this only gives us 4
sets, we also need to calculate the number of ways we can assign the elements of B to the 4 sets: 4!. Thus, the total number
of mappings is S(6, 4) · 4! = 65 · 4!.
Correct Answers:
• 360
• 65*4!
c
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