QUESTION 1: Archery

http://www.flickr.com/photos/thecomicshop/223590417/

Sebastian and his friends decided to start of their super fantastic day by going out and playing archery together. His friends and him were feeling like the game should be more interesting. Who ever got to split his or her arrow first that's trying to send to the bulls eye will not have to pay for their dinner. What are the odds of Sebastian splintering his arrow that he is trying to send into the bullseye if he has arrows with 1 cm diameter shafts if the target is 30 cm in diameter, a bullseye of 3 cm in diameter and he cannot miss the target. (Breaking the lines of a target counts as a hit and moves to a higher point value).

SOLUTION 1: Archery

- Sebastian has to get the bullseye first. Find the probability of him hitting the target itself which is favorable outcomes over all possible outcomes.

- because we're dealing with circles, we'll be dealing with the area and where we want to hit and where all the possible places are to hit.

- Keep in mind that the favorable outcome is the bullseye but the person doesn't have to hit the bullseye with the whole arrow, only a portion of it has to hit. So, Sebastian's farvorable area to hit would be everything within the blue circle because the farthest the arrow could be would be just hitting the outer cicumference of the bullseye and that would count.

- a closer view of Sebastian's favorable area. The favorable area would be the bullseye area plus the whole area that the arrow can land around it while sitting on it.

- By using the radii of the two circles we have an area that is favored for the arrow to hit. We can than apply the same concepts to the outer edge of the target and we can continue our calculation.

- The radius of the first one would be 2 cm because the bullseye's diametre is 3 cm, the radius of the bullseye would be 1.5 cm. The diameter f the shaft of the arrow is 1 cm, the radius of the shaft must be 0.5 cm. Together the radii make the favorable arrow because that's where Sebastian can hit to get a bullseye. That's divided by all possible outcomes which is π(15.5)² because the diameter of the whole target is 30 cm, which must mean that the radius is 15.5 cm. The pi's reduce.




- Sebastian has to nick his own arrow and this uses the same logic. The radius of the arrow that his trying to hit is .05 cm and his own arrow is 0.5 cm, which gives us a radii of 1 cm together. Sebastian's favorable outcome is π(1)². All possible outcome is still the same because he can hit anywhere else on the target which is π(15.5)².


- Then to find the probability of this happening one arrow after the other, multiply the probabilities together.




- Sebastian's probability of splintering his own arrow that he's trying to send into the bullseye is:



- Which is really really small. However, that's the chance that everyone has as well. In the end no one got it so everyone decided to go out and eat.

QUESTION 2 : Dinner

http://www.flickr.com/photos/adc/402535628/

After archery and showing his friends whose really pro, Sebastian wanted to go out and eat with them. He wanted to finish in 45 minutes or less so they have more time to go back to his house and play poker. On the menu there were: 4 appetizers (1 of which have bread), 6 beverages (2 of which are smoothie-like), 10 meals (6 of which have steak), and 5 desserts (3 of which are brownie-like). It usually takes Sebastian 5 minutes to finish a regular appetizer, beverage, dessert each and 25 minutes to finish a regular meal. How many dinner choices does he have when it takes an extra 2 minutes to finish an appetizer with bread, 1 minute to finish a smooth-like beverage, 5 minutes to finish a meal with steak, and an extra 2 minutes to finish a brownie-like dessert in 45 minutes or less if he gets an appetizer, beverage, main course, and dessert?

SOLUTION 2: Dinner

| View | Upload your own

We'll start where it says How to Solve.
First make a table of what you know:

• 4 appetizers, 3 are REGULAR and take 5 minutes to finish and 1 has bread which takes 7 minutes to finish. A regular appetizer takes 5 minutes to finish but with bread it takes 2 MORE minutes to finish.

• 6 beverages, 4 are REGULAR and take 5 minutes to finish and 2 are smoothie like which takes 6 minutes to finish.

• 10 main course, 4 are REGULAR and take 25 minutes to finish and 6 have steak which takes 30 minutes to finish.

• 5 desserts, 2 are REGULAR and take 5 minutes to finish and 3 are brownie like which takes 7 minutes to finish.

Since Sebastian wants to have an appetizer, beverage, main course, dessert and finish in 45 minutes or less there's really only two cases. There are only two cases because:

• He can have everything that requires more time to eat or "extra" time except for the main course because that would elapse his time that he wanted to finish. It only takes him 40 minutes to complete dinner if everything was "regular" and since it takes a total of 5 minutes to finish an "extra" appetizer + "extra" beverage + "extra" dessert that works out perfectly to 45 minutes.

• He can also have everything remain regular and only change his main course because if he decides to have an "extra" as a main course it would take another 5 minutes to complete his dinner which works out to 45 minutes.

CASE 1

In the first case we would have four slots for four different dishes because each of those dishes have a number of choices to be chosen. There are 4 different dishes to choose for appetizers because he could either choose a regular appetizer or one that requires more time to finish. The second slot would be 6 because he could either choose a regular beverage or one that requires more time to finish. The third slot would be 4 because it HAS to be a regular dish. The reason why it has to be a regular dish is because if he choose the first one to be an extra and the second to be an extra he cannot have a main dish that requires more time or he'll go over 45 minutes. The fourth slot has to be 5 because he can either choose a regular or extra.

That number also includes if he chooses everything to be regular and if he would just have two things such as an appetizer and dessert to be the only ones that require a bit more time to eat.


CASE 2

In the second case the only thing that can be "extra" or can be changed is the main course because if it was regular it only takes 25 minutes and if the main course has steak it takes 5 minutes more to finish which then would work out to 45 minutes. So what we would have is all regular items and the main course would change.

He would only have 3 choices for an appetizer because they are regular and don't require any more time to finish. 4 choices for a beverage. For the main course 6 choices because they all have steak in them which would take 5 minutes more to complete. It's not 10 because in the previous case we already have main courses without steak with the rest being regular, so by making the main course 10 different choices, we would have dinners that were the same. 2 choices for dessert because there are 2 regular desserts.

To find the total number of dinner choices Sebastian can have we just simply add cases 1 and 2 and there's your answer.

QUESTION 3: Poker

http://www.flickr.com/photos/extraverage/1137829938/

It was a show down between Sebastian and his friend in the game of Texas Hold 'Em. They were the only two left playing and this last game was going to determine the winner. Since Sebastian went in blind he wanted to see if his chances are good and if he should go all in or not. His friend was nice enough to let him see his cards first and even the burn cards. What are the chances of Sebastian winning if whats on the table is a 7 and 5 of hearts, a king and queen of diamonds, and a jack of spade; the burn cards are a 3 of spade, 6 of club, and a 7 of club; his friends hand is a queen of hearts and a 2 of diamonds; and aces are high and 2's are wild cards?


Not sure how to play? click here to learn how to play

SOLUTION 3: Poker

| View | Upload your own

Since Sebastian knows what his opponent has and what cards are used, he knows that the only thing that would beat his friends hand is a straight or higher three of a kind. Theres no flush potential so everything else higher than a flush is eliminated. What Sebastian needs is a:

• king and a king, to make a higher three of a kind. There are three kings left and two of them can be his own cards. He doesn't know because he went in blind but it's possible but he has pocket kings.
• Two and a Two, to make a three of a kind kings, because two's are wild and can be any card. The highest face on the table is a king and the pocket two's can make a three of a kind.
• King and a Two, two's are wild and with the kings it can be a higher three of a kind.
• Ace and a Ten, that would make a straight because it'll be Ace, King, Queen, Jack, and Ten.
• Ace and a Two, the two can have a face value of a ten.
  
• Two and a Ten, the two can have a face of an Ace since it's a wild card.
• Ten and a Nine, that also makes a straight, king, queen, jack, ten, and nine.
• Ten and a Two, because the two can have a face value of a nine.

To get a king and a king or a Two and a Two:
Sebastian already knows that a king is used on the table so there are three kings left. Out of those 3 kings he only needs 2 and it doesn't matter the order in which he gets the kings because the outcomes still the same. So its only 3C2, because out of the three kings left he only needs two of them and suit doesn't matter. This also applies to pair of two's, same concept.

3C2 + 3C2 = 6

To get a King and a Two:
There are three kings left and he only needs one of them and since order doesn't matter it'll be 3C1. One of the Two's are used in the opponents hand so there are 3 Two's left and out of those 3 Two's he needs one of them which is also 3C1. Then they have to be multiplied.

(3C1)( 3C1)= 9

To get a Ace and a Ten:
There are four Aces left because they are not used and he only needs one of them so 4C1 and there are four Tens that are not used either and out of those four Tens he needs one of them so 4C1. This also applies to Nine and a Ten because the Nines are not used either so out of the four Nines he needs one.

(4C1)(4C1)(2)= 32

you multiply 16 by two because it's the same thing that happens to getting a Nine and a Ten.

To get an Ace and a Two:
There are four Aces left because they are not used and he only needs one of them so 4C1 out of the four Aces he needs one. For the Two there are three left because one of the Two's is in the opponents hand, so out of the 3 he only needs one, so 3C1. This also applies to getting a Ten and a Two, and a Nine and a Two.

(4C1)(3C1)(3)= 36



To find the favorable outcomes:
Add the numbers up because these are all the hands that he wants to get and that's what favorable outcomes are, the outcomes you want.

3C2 + 3C2 = 6

(3C1)( 3C1)= 9

(4C1)(4C1)(2)= 32

(4C1)(3C1)(3)= 36

total of 83


To find All possible outcomes:
Since ten cards were used (5 for the table, 3 for the burn cards, and 2 are the opponents cards) there are only 42 cards that remain. Out of those 42 cards he'll need two for his own hand, so it would be 42C2, which is 861.

To find the probability it would be favorable outcomes over all possible outcomes. And whatever that number is, in this case .0964, is the probability.

QUESTION 4: Homework

http://www.flickr.com/photos/jeremy512/1382345330/

After a nice day of just hanging out with friends, Sebastian had homework to finish. Homework was probably what he dreaded the whole day. He had done most of it before the weekend but was stuck on two questions. One of them involved logarithms and trigonometric identities.

Sebastian had to prove that, whatever that thing is. It's like a logarithmetric identity.