Arthur’s 8th Birthday Party!

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.

“Hmm…there’s 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.”

Arthur’s 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?

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.

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 boys5C4 * 4C2 = 30 different combinations

Case 2: 5 Girls and one boy5C5 * 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

= (30+4) / 9C6 = 34/84 = 17/42

Therefore, Arthur has a 40.48% chance of choosing more girls than boys to his 8th birthday party.

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,” Arthur’s mom said.

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 cookies15C12 = 455

Case 2: One chocolate chip cookie5C1 * 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.

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, Arthur’s choices would be supporting the owner’s 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)

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) = 3/20*3/19 =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.

Finally the big day has arrived and it was officially Arthur’s birthday. It was 8’oclock in the morning and Arthur’s 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. Hmmm…what 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?

Red Sweater

Blue Sweater

Yellow Sweater

Green Sweater

Jeans Jeans Jeans JeansShorts Shorts Shorts Shorts

Tree Diagram:

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 n*m. Where m is the number of choices of sweaters, and m is the number of choices of pants (4*2 = 8). This process is also known as the fundamental counting principle.

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.

It was 2pm in the afternoon, and Arthur’s 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!

“Wahooo…this 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.

What is the probability that Arthur’s mom’s gift is first in line and his grandma’s present is placed second in line? There is only one way for Arthur’s mom’s gift to be the first and grandma’s 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 Arthur’s mom’s gift is first, and grandma’s gift is second in line is 1/3628800.

By the end of the day, Arthur was exhausted from his birthday party. He had an amazing time with his friends and family and couldn’t have asked for anything better. Arthur couldn’t wait until his next birthday where he could organize his party again and have a blast like he did this year!