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Corribus
Hero of Order
The Abyss Staring Back at You
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posted June 14, 2010 02:32 AM |
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I'd still like the problem to be explained more clearly.
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ohforfsake
Promising
Legendary Hero
Initiate
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posted June 14, 2010 10:22 AM |
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Hey FOG, thanks for posting it here. [Not sure I understand the relevance of using 'at least' in the question, you're most welcome to explain that part.  ]
I wrote the question wrong, I see.
I think the information was that the kid you know was a boy, was also born on a tuesday, but I've decided to simply copy the question into this thread:
"Suppose I have two children, and one of them is a boy
born on a Tuesday. What is the probability that both
my children are boys?" taken from: http://sci.tech-archive.net/Archive/sci.math/2010-05/msg01621.html
Now, what I wonder is the following:
Isn't both kids not always born a given weekday no matter what, and isn't there a symmetri in that all weekdays affects the outcome equally?
So, the boy is born on a tuesday, but if this wasn't told, the boy would still be born on one of the days, and since all days are symmetrical in the way the affect the outcome, you can just as well assume it to be any day, e.g. tuesday, without this should changing the probabiliy, I'd imagine. Though that's not how it goes from going through the problem.
The first part: "Two kids, one is a boy" gives the following scenario:
GG, BG and GB. But it doesn't tell anything about how often GG, BG and GB should each be counted. From the information that it's equally likely to get a boy and a girl and that it's not specified which of the two kids is a boy, it seems correct that all outcomes, GG, BG and GB are of same likelyness, therefore making it more likely the other child is a girl [though in reality it's a 0 or 1, because it's already determined].
But we know that both of the kids are born on some weekday, I mean they have to, and using this information we can from the start write the possibilities:
BG give 49 possibilities.
GB give 49 possibilities
BB give 49 possibilities
But using that it has to be one certain weekday and that all weekdays, except when it's the same, generates the same outcome, can one not just say:
G born on a given, unkonwn, weekday [X] and B born on a given, unkonwn, weekday [Y], gives the possibilities:
XY and YX
B born on a given, unkonwn, weekday [Z] and B born on a given, unkonwn, weekday [Y], gives the possibilities:
ZY and YZ
This is equal amount of possibliities, making it a 50/50. I'm pretty certain I'm doing something wrong, but I can't quite figure out. As I see it, the weekday information, is something always true, but not always considered. When taking account of all possible weekdays, it seems to me, like one is actually answering the question about what is most likely to be born, not what is the best guess of who got born.
Ah well, I know I'm not very good at wording this kind of thing, if I get a better idea of wording it, I'll do so, my main concern is that a given person is born on a given weekday is always true, so it does not need to be an information in the question, yet it still applies and should be taken account for, like a kind of hidden assumption.
____________
Living time backwards
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friendofgunnar
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able to speed up time
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posted June 15, 2010 08:02 AM |
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The year is 2186. Mankind has been at peace for 50 years thanks to the introduction of the Yezno robots. These machines have one purpose, to execute anybody that says something illogical.
It is a highway rest stop somewhere in the (greatly depopulated) United States. A man is traveling to meet his family and strikes up a conversation with a Yezno. He decides to test out how clever the robot is by telling it a riddle.
Man: "Suppose I have two children, and one of them is a boy
born on a Tuesday. What is the probability that both
my children are boys?"
Yezno: The answer is clearly 6/13
The man is surprised. He's told this riddle to many people before and they were all stumped. After he revealed the answer they all agreed that was correct and also that it was a clever riddle. So now he thinks he's identified a flaw in the robots program, does the Congress of Reason know about this? This could undermine everything modern society is built on. He confidently tells the robot the correct answer.
Man: Well no, the answer is 13/27.
Yezno: The answer is clearly 6/13. Under article 3 of the Constitution of Reason you are now have an opportunity to retract your illogical statement. If you refrain from retracting your illogical statement you will be executed immediately.
Suddenly the man is struck with terror. This flaw in the robot could cost him his life. He sits paralyzed with fear until he realizes he might know a way out.
Man: Under article 7 you must explain to me what the illogical statement is before you can execute me.
Yezno: One by one, describe every possibility that holds true for your original statement.
So the man starts describing each possibility. When he gets to the possibility that his first son is was born on a tuesday and also that his second son was born on a tuesday the robot stops him.
Yezno: This possibility, where two of them are boys born on a tuesday, clearly contradicts your original statement where you said: "one of them is a boy born on a Tuesday."
The man realizes the robot is right and he is now going to die.
Man: But that's not what I meant. I meant that I had one son in my mind and also that he was born on a tuesday.
Yezno: The limitations of the English language is not my responsibility. I am giving you one last chance to retract your illogical statement.
Man: Okay, yes, I retract my statement "the answer is 13/27."
Yezno: Very well. You are now on probation until further notice. So how about them Mets....
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ohforfsake
Promising
Legendary Hero
Initiate
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posted June 15, 2010 09:01 AM |
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Funny post FOG 
So, have I understood correct, that if someone states that one is born on a tuesday, you read it as, "at most one", or "only one" is born on a tuesday?
Anyway, my main concern is that, as I see it, both kids are born on a certain day of the week no matter what and it doesn't matter if the days are any combination of the possible weekdays. So the possible amount of configurations should be the same with or without the information, that is, the amount of success divided with the total should be the same (there are just less to count) when knowing the day.
At least that's how my intuition goes, but when actually counting the possibilities and the success, it's not the same and that I find weird.
I'd imagine, if you say, at least one boy is born on a tuesday, you'd get the same set of possible configurations as if it was any other weekday, only with another labelling for the weekdays. However I'd then imagine, if one were to sum over all these configurations one should end up with the same end result as if we did not know what day the given person was born. If not, where did those possibilities dissapear to? Not sure if I make that much sense, I probably need to sit down and think some more about the problem, because I've no problem in understanding the math, and I believe I've no problem in understanding the reasoning, I just don't think the result is in coherance with real life.
____________
Living time backwards
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friendofgunnar
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able to speed up time
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posted June 15, 2010 09:34 PM |
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Quote:
Q21. Friend or No Friend?
A friend of yours visits you and asks if you would like to play a card game. The card game works as follows. He deals four cards from a standard deck of cards onto a table. If the four cards are the same suit, he pays you one dollar. If the four cards are one of each suit, you pay him one dollar. If the four cards are any other arrangement of suits, you both keep your money.
Should you play?
The odds for four of the same suit are 1 * 12/51 * 11/50 * 10/49
The odds for four of a different suit are 1 * 39/51 * 26/50 * 13/49, which is approximately 10 times more likely, which means he's no friend of yours!
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friendofgunnar
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able to speed up time
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posted June 15, 2010 09:37 PM |
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Q23. Ruller and Compass
You have a ruler and compass and are given a circle. Draw a square that is the same area as the circle.
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ihor
Supreme Hero
Accidental Hero
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posted June 18, 2010 06:16 PM |
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Seems interesting. I will think when have time, now I have to prepare to Tuesday state exam - thats crazy, there may be tasks from all math subjects we studied from the beginning (integrals, series, differential equations, groups, rings, operators, metric and topological spaces... OMG). I afraid I forgot a lot.
BTW
I want to present one problem like the last one.
Q23-A Ruler and Calipers
You have ruler and calipers and pencil.
Draw a line from a point which is perpendicular to another given line.
Note 1. This is not so easy, I recall I spent enough much time 5 years ago when found this problem. But that was bloody interesting .
Note 2. Calipers is a compass which has 2 pins. You can measure same distances (on lines !), but you can't draw any circles.
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ihor
Supreme Hero
Accidental Hero
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posted June 26, 2010 02:16 PM |
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Edited by ihor at 14:18, 26 Jun 2010.
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Finally me exams are behind. Fuhh.
Since it is related to Math I'd like to publish my tasks from the tuesday state exam. Maybe it will be interesting for somebody of you.
1) Find a power series represantation for function
f(x)=x/sqrt(1-2x).
2) Using the definition of Riemann integral prove that the following function is non-integrable on [-1;5]
f(x)=(x^2)/(1-x)
3) Prove that eigen values of self-adjoint operator in Euclid space are real.
4) Find a distance from a point M0(5;5;5) to line l: x=2y=z/3
5) Solve a differential equation:
(y^2-2xy)dx + (x^2)dy = 0
6) Let x be uniformly distributed on [-Pi/2;Pi/2] random variable and y=sin(x). Find density, mean value and variance of y.
7) Banach fixed point theorem.
8) Consider the following transformation w=z^2 + 3i. Find the region in w-plane, which is image of the following region in z-plane.
D={z : |z|<2, |arg(z)|<Pi/4}
I thought some time about Q23 but have no idea for now. Little hint would be appreciated. 
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friendofgunnar
Honorable
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able to speed up time
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posted June 27, 2010 06:31 AM |
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oops, I solved the caliper problem for a parrallel line not a perpendicular one:
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Binabik
Responsible
Legendary Hero
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posted June 27, 2010 08:55 AM |
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1) Draw a line AB
2) Set the dividers (that's the correct word, not caliper) to the length of line AB
3) Use the dividers starting at point A to create an isosceles triangle ABC
4) Create a random point along line AB
5) Use dividers to create another isosceles triangle ADE
6) Lines BC and DE are parallel
7) Lines BD and CE are equal length
8) Set dividers to length BD
9) Use dividers to create 2 more isosceles triangles BDG and CEF
10) Draw a line from point A through the intersection of the triangles BDG and CEF
11) Lines AH and BC are perpendicular
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Binabik
Responsible
Legendary Hero
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posted June 27, 2010 09:05 AM |
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I just realized that you don't need the points F and G, all you have to do is draw lines BE and DC. I was too lazy to edit and upload the graphics though.
Actually I was going to ask if they still do these tings in school now that everyone uses CAD for everything. I guess the answer is yes. It's still a good learning tool even if people don't draft by hand any more.
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ihor
Supreme Hero
Accidental Hero
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posted June 27, 2010 02:14 PM |
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Edited by ihor at 14:23, 27 Jun 2010.
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@FoG
Explain how could you draw that blue line.
@Binabik
Quote:
3) Use the dividers starting at point A to create an isosceles triangle ABC
Sorry you can't do that. You can measure equal distances if lines have been already drawn. But at the beginning you have only point A and line BC. (See my explanation in the original post)
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friendofgunnar
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posted June 27, 2010 07:22 PM |
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In frame 7 (the frame before the blue line) you can see three blue dots. You get those dots by:
1. putting one end of the caliper on the top blue dot
2. one on the left blue dot
3. swing the bottom "pincer" around to the right until it hits the original line again. It's basically a mirror image of the green line.
Now that you can draw a parallel line through a dot if you're given a line and a dot, you can also solve the problem. In frame 7 you can see that each set of one blue dot and one green dot can make a perpendicular line to the original line. So now you have a perpendicular line plus the original dot. If you make a parallel line through that dot (using the steps I described in the picture) you will have a line going through the dot which is perpendicular to the original line.
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ihor
Supreme Hero
Accidental Hero
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posted June 28, 2010 08:58 AM |
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Again the same problem. It is not legal by rules of using calipers.
I guess I should explain it better - my fault.
See the image.
1) Left side. If you have point A and two lines as at the image you can find using calipers a point B so those distances are equal
2) Right side. If you have points A, B and two lines (AB and bottom line), then you can't that easy find a point C, so AB=AC.
Sorry such a rules. Basically you can use calipers(or dividers - doesn't matter) to measure equal distances on lines.
I believe now you get it.
Finally FoG, try to draw a parallel line within these rules. Parallel line is step 1 in the build  .
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Binabik
Responsible
Legendary Hero
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posted June 28, 2010 09:34 AM |
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Quote:
Draw a line from a point which is perpendicular to another given line.
Are you saying that the line is given and the point is given? For example line AB is already given, and point C is already given. And you have to draw a line through point C that is perpendicular to line AB?
Also, does the new line have to actually intersect line AB?
(I probably won't have time to look at it tonight, but just asking for future reference)
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Ecoris
Promising
Supreme Hero
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posted June 28, 2010 11:38 AM |
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Quote:
Are you saying that the line is given and the point is given? For example line AB is already given, and point C is already given. And you have to draw a line through point C that is perpendicular to line AB?
Yes.
Quote:
Also, does the new line have to actually intersect line AB?
Lines have no endpoints; line segments do. Non-parallel lines always intersect.
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ihor
Supreme Hero
Accidental Hero
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posted June 29, 2010 05:43 PM |
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Edited by ihor at 17:43, 29 Jun 2010.
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If someone waits for my confirmation then yes, Ecoris's answers are correct.
It would be very nice if you'll try to solve it by steps.
Step 1. How to draw a parallel line to the given line through a point?
Remember you have ruler and calipers.
I guess there are several methods to do it, FoG was pretty close to one of them.
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OmegaDestroyer
Hero of Order
Fox or Chicken?
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posted June 29, 2010 05:47 PM |
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You guys are all nerds.
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The giant has awakened
You drink my blood and drown
Wrath and raving I will not stop
You'll never take me down
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ihor
Supreme Hero
Accidental Hero
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posted June 29, 2010 05:56 PM |
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We are .
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friendofgunnar
Honorable
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posted June 29, 2010 06:53 PM |
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Can you lift the calipers off the page and plunk them down somewhere else?
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This thread is pages long: 
