I'll post a probability problem, first person to give me a correct answer (show your work!) wins 20 forum cash and +rep. Let's say I have a number line that goes from 0 to 1. I generate three random numbers (between 0 and 1) and plot their positions on that number line. So it'll look something like this: Code (Text): 0---A-----B---C--------1 As you can see, the three random numbers separate the number line into 4 segments. What is the probability that the sum of the lengths of the two medium length segments exceeds 1/2? Example: A = 0.3 B = 0.5 C = 0.9 The lengths of the segments will then be: 0.3, 0.2, 0.4, and 0.1 The two medium length segments would have lengths 0.2 and 0.3 And the sum would be 0.5
you drew your lengths wrong...cause currently there's 1.7 between 0 and 1 :lol: oh and is there required to be space between any two letters?
Who said each "-" corresponded to 0.1 unit lengths? I just drew them arbitrarily Uh, it only makes an infintesimal difference in the probability. But if it makes you feel better, then no, 0 lengths are acceptable.
omg... How many times do I have to say this? DO NOT GUESS Next guess on any of my competitions will result in -20 cash and -rep :nope:
This problem is WAYYYYYY too difficult for 5 forum cash. I refuse to solve it The average sum of the middle two pieces is 5/12, in case that helps anyone out.
I agree. I posted it thinking it wasn't too difficult but it turns out I can't solve it either Anyone who solves it gets 20 forum cash and +rep.
Let P = Probability . 0-A = 0.3 , A-B = 0.5 , B-C = 0.3 , C-1 = 0.8 ( As Shown On No. Line ) Actually all can be solved and is > 1/2 . Etc : 0-A and A-B = 0.3 + 0.5 = 0.8 If 1/2 is 0.5 , thus its greater than 1/2. I worked out all the solutions ... i got the answer 3/19/100. Correct?
Uhhhhhhhhhhhh no It looks like you tried to do something with the example that I gave you which is not what I asked.
oh i get it .. so its any three numbers between 0-1 . so its ( 0.1 + 0.4 ) , ( 0.2 + 0.3 ), (0.3 + 0.2) (0.4 + 0.1 ) , izzit smth like that?
Spoiler For the sake of making this easier on my brain, I'm going to call that number line 0 through 10 and upgrade .5 to 5 and convert 'em back at the end. Same difference, anyways. I'm also assuming you simplified your example, and the numbers .999 (essentially 1) or .000001 (essentially 0) are possible random numbers. I'll try to do another solution assuming you mean to cap it at .9, if I can work this one out. I do hope this won't be considered brute forcing it, but here goes. I've never been so elegant with math, or proofs. AB + BC must = 5. Because B is simply the dividing line between the two segments, and we're looking at the sum of the segments, the answer is immediately obvious; throw out B. B is useless and sidetracking. Simply always assume that B exists between the two random numbers A and C, and you're fine. Now we come up with a better formula; A subtract C must = at least 5. This is a much simpler solution. There are 66 total combinations of two numbers chosen at random between 0 and 10 (because 0 through 10 is 11 numbers to pick from, assuming A can be so low as to be virtually 0). Because C - A must = 5 or greater, and C has a hard cap of 10, any value for which A exceeds must be thrown away. Giving C the max value possible, 10, there are 5 values for A which return an answer of at least 5. Reducing C to 9 drops the value count to 4. Put shortly, 5+4+3+2+1 = 15 possible combinations for which C - A = 5 or more. Put more elegantly, the number of combinations for C - A being over 5 is equal to the total sum of (C - 5), where C = 10, 9, 8, 7,6. 15/66 = .2272727 continuing probability that the sum of the two line segments will = 5 or more. Assuming you meant me to work with 1 through 9, it just takes some tweaking. The number of possible combinations reduces to 45. The number of combinations for C - A being over 5 is equal to the total sum of C - 5, where C = 9,8,7,6. A total of ten combinations. 10/45 = .222 continuing probability that the sum of the two line segments will = 5 or more. Edit; fixed a mistake where I left out part of the second answer "combinations for C- is equal to" changed to "combinations for C - A being over 5 is equal to" The above is me failing hardcore at understanding simple instructions. Hugs to all.
I didn't really read your whole explanation, so I apologize if my critique is incorrect, but I think you have misunderstood the question. It seems to me that you are assuming that the two medium segments are the segments AB and BC...these are the two MIDDLE segments, but not the 2 medium sized segments... Thus, the rest of your proof is moot, and I did not really look at it. TRUST ME PEOPLE, THIS PROBLEM IS ABSOLUTELY RIDICULOUS. A BRUTE FORCE PROOF WOULD BE PAGES LONG AND WOULD INCLUDE VARIOUS DIAGRAMS (I think 3d space is required to visualize the relationships between the line segments). PLEASE DO NOT ATTEMPT THIS CHALLENGE, IT IS A WASTE OF YOUR TIME.
Well I'll be. You're right, I totally misunderstood x_X 30 hours with no sleep makes Syn incredibly stupid.
Yeah, this problem is way too difficult. I'm locking to stop people from posting incorrect answers. The answer is 1/4. If you can PM me a proper proof of it, I'll give you the prize.
Well...let's just say I was really bored on the airplane today (back in ATX <3) so I ended up giving this problem some thought (sudoku only takes so long...). >___< Rephrasing the question.... I spent like 5 freakin' minutes trying to figure out how the hell the answer is 1/4 before I realized that my understanding of the problem was just wrong. The way I first understood it, the points a, b, and c had to stay in that order, left to right. But if that were true, an answer of 1/4 would make no sense. The probability that point a is in a location where point c has the possibility of being .5 units away is 1/2. That would mean that in order for the overall probability to be 1/4, for every possible location of point a from which point c can be located .5 units away, there must be a 1/2 possibility that point c is indeed in such a location. But that's impossible.... So yeah, I had a completely wrong understanding of the question. And I think a few other people may have not understood it either. Basically what it's asking is "What is the probability that two arbitrary points between the units 0 and 1 inclusive are a distance of greater that .5 units apart?" So point b isn't really worth mentioning and a---c is not the only possible orientation; c---a is also valid. Solution... I'm sure there are better and more "elegant" ways to solving this (a visual solution would be pretty cool =P) but I <3 calculus so that's what I'm using. Plus, calculus is like the brute force method of mathematics (pretty hard to mess it up), and it makes for a pretty easy solution. XD So basically...I added all the probabilities of point c being .5 units away given the position of point a. Consider the reference: Lets say we divide the length of 0 to 1 into 100 parts (as in like 0.01, 0.02, 0.03....0.97, 0.98, 0.99) and that the points a and c have to fall on one of these marks. At this point, we will spit the problem into two cases: Case 1: If point a is at a point less that point c (this orientation: a----c)..... So then given that point a is at 0.00, the probability that point c is at least .5 units away is 1/2 (point c can be anywhere from 0.51 to 1.00 inclusive). So then the probability of this two point system is (1/100)*(1/2) or .005. Then you calculate the probability of the two points given that a is at 0.01, then at 0.02 and so forth until point a reaches 0.50, at which point it is no longer possible for point c to be .5 units away from point a. A summation can be used to easily express and evaluate the total probability that the two points are at least .5 units away given that point c is at a greater point than point a: probability: 1/8 Case 2: If point c is at a point less that point a (this orientation: c----a)... Same as above. All we've done is switch the two points. probability: 1/8 Solution to this example: Just add the two parts together (1/8 + 1/8) and you get 1/4 as the answer And now imagine: What if we had divided the distance between 0 and 1 into 1,000 segments, or 1,000,000 segments, or... Calculus <3 So...since I don't like infinite sums (or rather...my calculator doesn't like them =P), we're going to take a limit. Of course, that 1/8 has to be doubled since there are two cases (a----c and c----a) to give a final answer of 1/4. long explanation I know...but I had a lot of time on the plane =P inelegant brute force solutions ftw gah...so when I was on the plane, I thought there were 4 points (a,b,c,d)...not 3 points (a,b,c) while that changes the solution very little, there may be parts where I wrongly name a point (if I say point b...I actually mean point a)