Arthur

1
Arthurs 8th Birthday Party!
2
It was during the summer of the year 2007, and
Arthur was finally going to turn eight years old
on August 15th. He was extremely excited for his
birthday, because this year he was planning to
hold a party at his house. Arthur knew that it
would take a lot of time and preparation to
organize his party, but he was thrilled that this
year he was finally old enough to make all the
decisions for his birthday bash.
Hmmtheres so much planning to do for my party.
But first I should decide who to invite to my
birthday. Since I can only invite six people to
my party, not all of my friends can come.
3

• Arthurs Friends
• Francine
• Sue Ellen
• Buster
• Fern
• George
• Binky
• Muffy
• Prunella
• Brain

How many different combinations of six friends
can Arthur choose from to invite to his party?
4
Since there are nine friends to choose from and
only six can attend, the combination formula
would be n - total of friends to choose from
r - the of people that can attend the
party nCr __n!__ (n-r)!r! 9C6
___9!___ (9-6)!6! 9C6 ___9!___
(3!)(6!) 9C6 84 Since there is a
total of nine friends to choose from, and only
six are invited to the party. There are a total
of 84 different combinations of friends that
Arthur can choose to invite.
5
If Arthur randomly picks six of his friends
randomly, what is the probability that he will
choose more girls than boys to his party? Let A
be the event that Arthur will select four girls
and two boys. Let B be the event that Arthur
will select five girls and a boy. We need to
separate these events into two different cases.
This is a mutually exclusive event since there is
no overlap in the events. Case 1 Four girls
and two boys 5C4 4C2 30 different
combinations Case 2 5 Girls and one boy 5C5
4C1 4 different combinations Probability of
Arthur choosing more girls than guys to his
birthday party will be P(A or B)
P(A) P(B)________ Total number of
combinations (304) / 9C6
34/84 17/42 Therefore, Arthur has a
40.48 chance of choosing more girls than boys to
his 8th birthday party.
6
Arthur finally decides who to invite to his
birthday bash, which is the most important part
of his decision making. Now, all is left is the
little bits and pieces of the planning that needs
to be done for his party. Since all his friends
are going over to his house, Arthur will need to
prepare snacks for his friends. His favorite junk
food is cookies, so he decides to make a list of
possible types of cookies he could serve at his
birthday bash. The list of cookies included
chocolate chip cookies, double chocolate cookies,
oatmeal cookies, white chocolate cookies and
raisin cookies. On your birthday, I will
prepare other snacks that would be more
nutritious, so you can buy a dozen cookies from
the bakery, Arthurs mom said.
7
In the store, they sell 5 chocolate chip, 5
double chocolate, 3 oatmeal, 3 white chocolate
and 4 raisin cookies that was just baked fresh
from the oven (20 cookies in total). Since
chocolate chip cookies are an original flavor,
Arthur wants to buy at least two chocolate chip
cookies. How many different combinations of a
dozen cookies can be bought? Find the number of
combinations by using the indirect method. This
method will be a lot quicker because there will
only be two separate cases. Instead of finding
the combinations for purchasing 2, 3, 4 or 5
chocolate chip cookies, we find the number of
ways that there will be no chocolate cookies, or
just one. After adding the combinations for no
chocolate chip or one chocolate chip, subtract it
from the total number of combinations to solve
it.
Case 1 No chocolate chip cookies 15C12
455 Case 2 One chocolate chip cookie 5C1
15C11 6825 Total number of Combinations 20C12
125970 Total number of combinations no
chocolate chip 1 chocolate chip 125970 - 455
- 6825 118690
As a result, Arthur has 118690 different ways he
can purchase a dozen cookies, given that he will
have at least two chocolate chip cookies.
8
The storeowner of the bakery shop wants to know
which type of cookies are the most popular and
the least popular. Over the years, she has
noticed that the customers usually pick their
cookies in order of preference, meaning that
their first choice would be their favorite. The
owner thinks that the white chocolate chip
cookies are the most popular and oatmeal is the
second top favorite. What is the probability that
Arthur will buy 1 white chocolate cookie and 1
oatmeal cookie in that particular order?
Therefore, Arthurs choices would be supporting
the owners hypothesis of the top two favorite
types of cookies. Note (5 chocolate chip, 5
double chocolate, 3 oatmeal, 3 white chocolate
and 4 raisin cookies)
9
Let A be the event that Arthur chooses white
chocolate chip as his first cookie Let B be the
event that Arthur chooses oatmeal as his second
cookie In this situation, the probable outcome
of an event, B, depends directly on the outcome
of another event, A. When this happens, the
events are said to be dependent. P (white
chocolate, oatmeal) 3/203/19
9/380
Given that there are 3 white chocolate cookies in
the store, there will be a 3/20 chance of Arthur
picking it first. The probability of him choosing
an oatmeal cookie second would be 3/19 given that
a white chocolate cookie was chosen already and
there are only 19 cookies left. This is a
conditional probability because the probability
that B occurring is given that A has already
occurred. As a result, Arthur has a chance of
9/380 or 2.37 of choosing one white chocolate
cookie and one oatmeal cookie in that order.
10
Finally the big day has arrived and it was
officially Arthurs birthday. It was 8oclock in
the morning and Arthurs alarm went off. The day
he has been waiting for has finally come and he
was very excited for his birthday party. After
all his planning, he has decided to invite
Buster, Binky, Francine, Muffy, Brain and Sue
Ellen to his birthday bash. Today is my
important day, and everyone will be coming to
celebrate with me. I must pick the perfect outfit
to wear. Hmmmwhat should I wear? Arthur
wondered. Arthur went through his closet and
found four different colored sweaters he has a
red, blue, yellow and a green sweater. In his
drawer, he finds a pair of jeans and a pair of
shorts. How many different outfits can Arthur
choose from?
11
Tree Diagram
Blue Sweater
Green Sweater
Red Sweater
Yellow Sweater
Jeans
Jeans
Jeans
Jeans
Shorts
Shorts
Shorts
Shorts
The sample space for Arthur's choice of outfits
is Red sweater and jeans, red sweater and
shorts, blue sweater and jeans, blue sweater and
shorts, yellow sweater and jeans, yellow sweater
and shorts, green sweater and jeans, green
sweater and shorts. There are a total of 8
different possible outfits that Arthur can wear
on his birthday. This process is made up of
stages with separate clothing apparel choices and
the totally number of choices is nm. Where m is
the number of choices of sweaters, and m is the
number of choices of pants (42 8). This
process is also known as the fundamental counting
principle.
12
What is the probability that Arthur will wear his
yellow sweater and his blue pair of jeans? P
(yellow sweater blue jeans) P (yellow
sweater) P (blue jeans) 1/4 1/2
1/8 Since the probability of choosing a sweater
and a pair of jeans are not affected by one
another, they are considered to be independent
events. In the end, Arthur decides to wear his
yellow sweater because it is his favorite color
and a pair of blue jeans because they are
comfortable. He knows that this was the perfect
outfit because it is fitting and casual.
13
It was 2pm in the afternoon, and Arthurs friends
arrives one after another. They were all having a
blast and time flew by really quickly. After
dinner it was finally time to eat the birthday
cake. Arthur has been waiting forever to blow out
his candles and make his special birthday wish.
By that time, it was already 9pm and parents
started arriving to pick up their kids. When all
the guest left the house, Arthur knew it was time
to open all his birthday presents! Wahooothis
is my first year receiving so many birthday
presents! Arthur said. In total, Arthur
received 10 different presents for his 8th
birthday. He decides to align the presents in a
row where he could open one gift at a time. How
many different ways can the 10 presents be
arranged in a line?
Method 1 10! 3628800 Method 2 n total
of gifts r size of arrangement nPr
__n!__ (n-r)! 10P20 __10!__
(10-10)! 10P10 3628800
There are a total of 3628800 different
arrangements of presents, or also known as
permutations if they are placed side by side in a
row.
14
What is the probability that Arthurs moms gift
is first in line and his grandmas present is
placed second in line? There is only one way for
Arthurs moms gift to be the first and grandmas
to be second. P __
1____________ Total number of arrangements
P __1____ 3628800 This is also known
as theoretical probability, where a particular
event is deduced from analysis of the possible
outcomes. Therefore the probability that Arthurs
moms gift is first, and grandmas gift is second
in line is 1/3628800.
15
By the end of the day, Arthur was exhausted from
his birthday party. He had an amazing time with
his friends and family and couldnt have asked
for anything better. Arthur couldnt wait until
his next birthday where he could organize his
party again and have a blast like he did this
year!