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Counting Story Performance Task Detective Conan - A Meaningful Day -. MDM4U1-02 Hall Lo. Conan today goes to school as usual. Today the teacher wants to form a committee in class which is responsible for collecting homework. The committee should consist of 5 students.
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Counting Story Performance TaskDetective Conan- A Meaningful Day - MDM4U1-02Hall Lo
Conan today goes to school as usual. Today the teacher wants to form a committee in class which is responsible for collecting homework. The committee should consist of 5 students. To be fair, the teacher decided to choose the group randomly: as there are 30 students in the class, so the teacher prepared 30 small sheets of paper having one student’s name on each of them without repetition . Then she will choose 5 of the 30 paper out without replacement.
Here are Conan’s 4 friends: Mitch Amy Ai George What is the probability that Conan and his 4 friends will be chosen to be the committee at the same time?
Since there are 30 students in the class, the total number of choosing 5 students out can be calculated by using the Combination formula. nCr = The reason why we use combination is because the choosing order of the students isn’t important.So, the total number of ways to choose 5 students from the 30 students is 30C5 = 142506. By using the same formula, we can also calculate the number of choosing Conan and his four friends. As there Total number of items can be chosen n!r! (n – r)! Number of items has to be chosen are 5 of them and 5 students have to be chosen from them, the number of ways is 5C5 = 1.
From the data from last slide, we can calculate the probability by applying the Probability formula. P(A) = As stated above, the sample space includes all possible outcomes. An event is a group of outcomes, and an outcome is a possible result that the event can happen. So, the probability of Conan and his 4 friends are chosen into the 5-student committee 5C5 1 30C5 142506 Oh… they all are selected as the committee!They are so lucky today! Number of outcome in which Event A can occur n(A) n(S) Total number of possible outcomes (Sample space) = = = 0.0007017…%
After getting Conan and his 4 friends as the committee, the teacher wants to choose their positions. They only have 5 subjects which have homework. The 5 subjects are English, Mathematics, History, Music and Science. Conan wants to be the Mathematics homework collector, while his 4 friends all want to be the English homework collector. So the teacher decides to choose the positions for them. She uses 5 pieces of paper this time, on each of them there is one of the 5 students’ name. The teacher will choose one piece of paper for the Mathematics homework collector first, and then choose other pieces of paper for other subjects’ homework collectors, without replacement.
What is the probability that Conan will be chosen as the Mathematics homework collector while his 4 friends can be any of the other subjects’ homework collector? I want to be the Mathematics homework collector…
As Conan must be the Mathematics homework collector, that means there is only 1 choice for the Mathematics homework collector. After that, we can apply the Permutation formula to calculate number of ways of arranging the remaining 4 friends. nPr = As there are 4 friends that have to be arranged for the remaining 4 subjects’ homework collectors, so we use 4P4(4 x 3 x 2 x 1) for the number of arrangements for them. So the total number of arrangements is 1 x 4P4. We use multiplication because these two events has to happen at the same time. This is also called multiplicative counting principle. Total number of items n! (n – r)! Number of items has to be taken at one time 1 x4x3x2x1no. of arrangements for number of arrangements for the rest toone person to be chosen be chosen for the remaining 4 subjects’ for the math homework homework collectorscollector
SAMPLE SPACE Mathematics English Science Music History 4 3 2 1 1 Number of arrangements = 1 x 4 x 3 x 2 x 1 = 24
For the total number of arrangements, we can apply the Permutation formula again. As there are 5 students (Conan and his friends) have to be arranged for the different positions, total number of ways will be 5P5, which will be 5! = 120 From the previous slide, we know that there are 24 arrangements for Conan being the Mathematics homework collector. So, we can calculate the probability by using the Probability formula again. 24 120 1 5 Probability = ——— = —— Conan really is selected as the Mathematics homework collector! He is so lucky! Unbelievable….. I am so lucky today!Am I Dreaming?
After few lessons, it is now lunch time. The students have 3 choices for their lunch: Set A, Set B or Set C. Set A: Salad + chocolate milk Set B: Bread + egg + sausages + chocolate milk Set C: Sandwich + grilled Fish + soup The lunch is randomly given to the students.
Conan doesn’t want to have Set A as his lunch as he thinks that it is strange to have salad and chocolate milk at the same time… But he likes Set B or Set C. What is the probability that he will not get Set A as his lunch?
We can get the answer by getting the probability of getting Set A as lunch first. After that, as the probability of NOT getting Set A as lunch is a complement of that of getting Set A as lunch, we can simply use 1 to minus the probability of getting Set A as lunch. This is known as a formula also. Probability of getting Set A as lunch = = 1/3So, probability of NOT getting Set A as lunch = 1 – 1/3 = 2/3 Outcomes make up A P(A) + P(A’) = 1 Outcomes make up A’ (event A doesn’t happen)
Oh, Conan is still lucky… as he gets Set C for his lunch. He is happy at first, but then he discovers that the grilled fish doesn’t taste good. So he finishes his lunch unhappily. Tastes bad… :(
Time flies…..And they enjoy the visit of the Museum. They are so hardworking that they decided to go to the library and do the project together.
There are 4 parts in the project: Introduction part, Information part, Q & A part, Conclusion part. They decide that every person is responsible for one part, and after that they will discuss and combine the parts into a project. Amy is very good in designing, so she just had to design the layout of the project and didn’t need to do any of the 4 parts.
To be fair, they numbered the parts and prepared 4 papers having numbers 1 – 4 on each of them. Then each of the 4 people will choose 1 of them. The numbers refer to the following parts: 1 Introduction2 Information3 Q & A4 Conclusion They will get the papers in this order: Conan, Ai, George, Mitch. What is the probability that Conan will do the Conclusion part and Ai will do the Introduction part?
1 2 3 4 1 2 4 3 1 3 2 4 1 3 4 21 4 2 31 4 3 22 1 3 42 1 4 32 3 1 42 3 4 12 4 1 32 4 3 13 1 2 43 1 4 23 2 1 43 2 4 13 4 1 2 3 4 2 14 1 2 34 1 3 24 2 1 34 2 3 14 3 1 24 3 2 1 SAMPLE SPACE These are dependent events, as the papers Ai can get will be affected by what Conan has taken. Probability that Conan will do the Conclusion part and Ai will do the Introduction part = 2/24= 1/12
We can check the answer we just got by using the Conditional Probability formula. Conditional probabilityis the probability that B occurs, given that A has already occurred. Let event A be Conan does the Conclusion part.Let event B be Ai does the Introduction part. By applying the formula, we can get the P(A and B), which represents Conan will do the Conclusion part while Ai will do the Introduction part.P(A) = 1/4 (There are 4 parts left and Conan only want 1 of them)P(B|A) = 1/3 (There are 3 parts left and Ai only want 1 of them)So, P(A and B) = 1/4 x 1/3 = 1/12We can see that we got the correct answer! :) P(B|A) x P(A)= P(A and B) Conditional Probability
So, what if the question becomes this: What is the probability that Conan will do the introduction part OR Ai will do the introduction part? For this question, we have to get the probability that Conan will do the introduction part and the probability that Ai will do the introduction part respectively first. P(Conan do the introduction part) = 1/4P(Ai do the introduction part) = 1/4 So, the probability that Conan will do the introduction part OR Ai will do the introduction part will be 1/4 + 1/4 = 1/2. We use additive counting principle here because the 2 tasks are mutually exclusive. They cannot do the introduction part at the same time as each of them should be responsible to each different part. We cannot do the introduction part at the same time…
After doing the project for 2 hours, they feel hungry. They decided to go to a nearby restaurant to have dinner together. Yeah!Dinner Time!
They look at the menu. They find that some of the dinner set contains vegetables, some contains meat, and some contains egg as following: - 6 of the dinner set contain meat - 8 of the dinner set contain vegetable - 7 of the dinner set contain egg - 3 of the dinner set contain meat and vegetable - 2 of the dinner set contain meat and egg - 2 of the dinner set contain vegetable and egg - 1 of the dinner set contain meat, vegetable, egg How many dinner sets are there altogether?
S Meat Egg We will use a Venn diagram to demonstrate this question. 2 1 4 Subset 1 2 1 6 Vegetable We find that these are non-mutually exclusive events. There are some overlapping in the Venn diagram. Total number of sets = 2+1+1+2+4+1+6 = 17
After a while, Conan receives his coffee. There are 3 bottles of powder: 1 containing sugar, 1 containing salt, and 1 containing milk powder. Conan is so lazy that he will just choose any 2 of them with replacement and add them into the coffee and drink it! He will choose 1 and add it to the coffee and then put the bottle of powder back and choose another 1 and add it to the coffee again. What is the probability that he will get sugar and milk powder? Sugar Salt Milk Powder
Let event A be getting a sugar bottle.Let event B be getting a milk powder bottle. These are independent events as there is replacement. The probability of event A will be 1/3, as it is to choose 1 of the 3 bottles out. The probability of event B is also 1/3, because after event A the bottle is put back, and there still have 3 bottles for Conan to choose. So the probability of event A and event B is 1/3 x 1/3 = 1/9. Wish me luck!
Actually, Ai knows about the bottles. She knows which of them contains which kind of powder. However, she doesn’t want to tell Conan directly about the correct bottles. She just let him choose one of the bottles for 9 times with replacement, and then she will tell him how many times he gets the bottles correctly. So, Conan get one of the bottles with replacement for 9 times. That means he carries out 9 trials. A : the bottle containing sugar.B : the bottle containing salt.C : the bottle containing milk powder. ABACCBACA OUTCOMES Which one should I choose?
From the results of last slide, we can calculate the experimental probability. The experimental probability that Conan get the bottle containing sugar (Bottle A) is 4/9, the bottle containing salt (Bottle B) is 2/9, and the bottle containing milk powder is 3/9 = 1/3. We discover that the experimental probability is different from the theoretical probability. This is because of chance variations. The theoretical probability will always the same for an event, while the experimental probability can be different every time for an event. With the help of Ai, Conan get the 2 correct bottles. He enjoys the coffee. Thanks Ai…
After they finished their dinner, they decide to go home. It is already evening. They really had a meaningful day! ~~~~~~~~~Thanks for watching! ~~~~~~~~~
Credits • YTV Japanhttp://www.ytv.co.jp/conan/(for the screenshots of Detective Conan from the TV episodes) • Xenical http://www.xenical.com.ph/eatingright.html(for the pictures of food) • Cloverleaf International http://www.cloverleaf.com.hk/fadeProducts2.php(for the picture of sugar bottle) • APTX forumhttp://bbs.aptx.cn/ (sources of Detective Conan TV episodes) • Gosho Aoyama(for he creates Detective Conan) • Shogakukan(Publisher for Detective Conan)