- Anyone?
Contents
Counting
What I've found most confusing is knowing when to subtract repetitions of things, like in problem 44 on the homework, where we're asked to find how many strings there are with five consecutive 0s or five consecutive 1s. What's a good method for knowing or deciding how to count something like that?
Suggestion
The goal is to count each thing exactly once. Counting everything a certain number of times is almost as good, as you can just divide by that number.
If you know how many times individual elements were counted, that is the next best thing. You count how many elements of that type exist, and divide by that number. For example, you may have counted everything with 7 ones like 1111111000, 0111111100 and 1011111110 (and so on) 3 times (for 3 different sets of '11111'), and thus just count these separately and divide by 3.
What may be less work than this is just to devise a method that actually only counts exactly once instead of a different number of times. It may take a little more thought, but I think it's worth it in the end.
I quite disagree with the last statement. Although in some problems it may be worth it to just count each one only once and that's it, but with many problems, this will take MUCH longer than if you were to do it in a Boolean logic fashion. Count the total number of possibilities with no restrictions and then subtract the overcount, then add back the overcount of the overcount, then subtract...etc...etc. This may sound confusing, but problems like the binary question makes things much easier.
The way that I did the problem was that I only looked at a single string of 5 numbers that had to be there (I assumed 0, but it could have been one) and then in the end I multiplied by 2 then subtracted the over count. Ex: the first line would be 00000XXXXX, where the X's symbolizes either 0 or 1 (thus multiply by 2). The next would be 100000XXXX, which MUST be different from the first, and so on until you reach 1111100000. I Then multiplied by two, but then subtracted the overcount of 1111100000 = 0000011111 and such, when switching the 0s and 1s. Adjust to your method, and complete.
--Norlow 14:43, 10 September 2008 (UTC)
Counting
I also have the most trouble with counting. I know what I need to do, but I never know how to set the problem up. For example, problem 38 in section 5.1 asks: How many subsets of a set with 100 elements have more than one element? I know I need to take the total number of possible subsets of a set of 100 elements and subtract 100 (the total number of subsets that contain 1 element). The problem is finding the total number of possible subsets. Any suggestions?
I thought of #38 as a sum of combinations. I have $ \sum_{r=2}^{100} \left(\!\!\! \begin{array}{c} 100 \\ r \end{array} \!\!\!\right) $. This should count all possible subsets and exclude those containing only 1 element. I don't know how to simplify this, and I'm thinking there's a better way to do it. But it's a start.
Suggestion
The problem is finding the total number of possible subsets.
The total number of subsets is $ 2^n $. I don't remember the way Uli explained it in class (sorry about that), but the way I learned it before 375 was for each element, you can either put it in the set, or take it out. (2 choices) You can make this choice for every element, so 2*2*2*2... = $ 2^n $. (You multiply, since the choices for one element don't affect any other element)
Here's an example. If the elements are A, B, C, then you can either put A in, or leave A out. You can do the same with B and C. If you decide to leave A out, then you would have () (B) (C) (BC), depending on if you put B and C in or left them out. If you decided to put A in, you would have (A) (AB) (AC) (ABC by similar reasoning.
(Also don't forget to subtract out the 1 set of 0 elements at the end. The total sum goes from 0 to 100.)
--Norlow 23:38, 11 September 2008 (UTC)
Overcount
I find that the hardest thing for me is recognizing when I will have overcount and then how to correct it (subtract or divide)
I too have trouble with overcounting. I can never tell when to compensate.
Suggestion
They way I remember if I need to adjust is by asking myself if order matters? If it does there is no over count. If order does not matter then I have to adjust for over count.
- This is a problem that I have as well. But this is a really good suggestion.
- I find it very helpful to make a venn diagram and draw the picture out. Once you do that you can visulatize and see what needs to be added and subtracted.
- Yeah Venn Diagrams are pretty sweet. Although I don't feel that just having a diagram is very helpful.
RHEA
Really, the only notion that has frustrated me is this silly website. I finally learned today what was going on with this thing. How does Uli even know who's writing what for the weekly credit? I guess I will just sign my name at the bottom of this to make sure I get credit. Jacob Ahlborn
- Man i could not agree more. What is with this? But we have to do it so whatev.
- Guys, every thing you do on this site is recorded with your name and a time stamp. In fact, the whole world can see what you write, as well as the history of every page, including the names of all the contributors and the time stamp for each contribution. Just click on the "history" tab above or the "Recent changes" link on the left and you will see. Welcome to the 21st century. --mimi
I have a very hard time remembering to contribute by Sunday. I wish our due date was during the week, maybe on Tuesday night, so we could be reminded to contribute in class. I've had Sunday homework before and usually my mind is on what is due Monday, not what is due Sunday night.
I totally agree with the above statement. My computer broke two weeks ago and because of that, I have forgotten to post on Rhea in the past 2 weeks since I'm not reminded to do so by my computer every day I use it. I keep telling myself to post when I'm doing homework in a lab, but I always forget after I get into my other larger assignments... By: Yu Suo
- Just check it like you would check blackboard, your e-mail, facebook, or any sites you look at. The goal is to help other people in a web manner. Hopefully that's what we are all doing with contributions (like even this post.) If a weekday due date is easier, you can make the contribution on Tuesday/Wednesday/whenever before it's due.
If you look above this section, there is a (user) name and timestamp. If you want to sign whatever, just press the second button from the right (to the left of the line) -- it says "your signature with timestamp". You can add that on... it looks like this -- Norlow 22:05, 17 September 2008 (UTC)
Another problem I have is I don't know what to say on here. I'm not use to having to contribute to a website for a math class. None of my classes in the past have had anything like this. It seems like it just doesn't fit into the class.
Combination
I am having trouble with the idea of combination. What i mean is when we say "choose", for example nine choose three. What does that even mean. I can plug it into my calculator and get an answer but that does not help me understand it.
My main problem with Combination is the idea of permutation against the idea of combination. I think I learned it backwards in high school so I always get confused when I look at it now.
Suggestion
Maybe think of "combination" as combining things in a soup or stir fry? Then the order does not matter, as the stuff will all get in the soup regardless. Think of it as choosing 3 stir fry ingredients out of 9 which are possible. It doesn't matter how you choose them, as long as you choose them. 9 "choose" 3 means you have 9 items in total, and you choose 3 to do something with. This is written as C(9, 3).
You have 9 items to choose from, but once you choose that item, you only have 8 items left for you next selection. Then once you choose that, you only have 7 left. So it would be 9*8*7.
But, you could choose "A, then B, then C", and "B, then A, then C", or "A, then C, then B", or... with this method, and you only want to count "one A, one B, one C" once.
You would count the same thing (and every other group of 3 items) 3*2*1 times, so you divide by that. (This is what overcount refers to) It could be written $ \frac{9*8*7}{3*2*1} $ and that would be correct, but a pain to write, especially if you had 30 choose 10. So instead, it's written as $ \frac{9*8*7*6*5*4*3*2*1}{(6*5*4*3*2*1)(3*2*1)} $ which seems longer, but using factorials makes it simpler. This way it's just $ \frac{9!}{(6!)(3!)} $
Overcount
I agree with a lot of the other people that overcount is a difficult concept. I get hung up a lot of the time when I think about whether objects are unique or not, and, if so, whether or not I should factor in overcount. Sometimes the wording of a question can make it more obvious but other times the question is too vague to tell.
Does that mean rather than writing everything in factorial, we can just use the C( , ) and get the same number? And will that also count the "overcount"? I'm totally confused but if this idea is true, then I'm a bit relieve. ::Mohamad- --hussaini 20:46, 21 September 2008 (UTC)
THE GAME SHOW HOST PROBLEM!
Has anyone seen the movie 21? If you have, you know what I'm talking about... If you haven't, I am having a little trouble understanding the game show host problem! The teacher asks Ben:
Ben, suppose you are on a game show. You are given the chance to choose between 3 doors. Behind one of the doors is a new car... behind the other 2 are goats
Ben chooses door #1.
The game show host opens door #3 to reveal one of the goats. He then proceeds to ask Ben, "Ben, do you want to stay with door #1 or go with door #2"?
Here is the punchline. Is it in Ben's interest to change his answer?
Answer
Yes, when Ben was first asked to choose a door, he had a 33.3% chance of picking the car. When the professor re-offers the question, Ben has a 66.7% chance of finding the car behind door #2. My question is how, when anyone in the world knows that there is a goat and a car behind 2 doors, there should be a 50% chance of choosing right? The answer is solved by accounting for "variable change", but unfortunately, this guy has know idea what that means and how it could be applied in a problem that makes more sense to me?
Conditional Probability
This problem can be solved using conditional probability. Once the door is opened, you have some extra information.
Related Links
This problem is also studied in ECE302. See the Monty Hall video here and look at the corresponding homework problems that the students had to solve in this engineering course. --Mboutin 16:45, 24 September 2008 (UTC)
