where can I find solutions to probability problems?
147164
Ah, a variation on the Monty Hall problem:\n\nhttp://en.wikipedia.org/wiki/Monty_hall_problem\n\nIt is a probabilistic curiosity, this one. The highly unintuitive answer is that you actually **can** improve your chances by changing your mind.\n\nYour problem appears to only be different in that it starts with 5 boxes. The above link should give you sufficient information to solve the problem.
191190
check out www.quickmath.com\n\nAlmost everything is there.
460739
Here are the formulas:\nIf P(E) = a/b, then the odds for E are a to (b-a)\nTo find the odds against E just reverse the order.\nOdds against E are (b-a) to a\n\nIf the odds of E are a to b, then P(E) = a/(a+b)\n\nFor example,\nif P(R) = 1/10, then the odds for R are\n1 to (10-1) or\n1 to 9.  \nThus, the odds against R are 9 to 1.\n\nAnother example,  say the odds of R are 5 to 3.  \nThe probability of R is 5/(5+3) = 5/8.\nThe probability of not R would then be \n1 - 5/8\n= 3/8
180283
Not exactly sure what you are looking for but try this:\n\nhttp://en.wikipedia.org/wiki/Problem_solving
625162
Try dictionary.com, m-w.com, or one of the thousand other good dictionary websites on the web, as for problems, try search on http://mathforum.org/
540053
In the experiment of tossing a coin ten times.\nTotal number of possible choices = 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 \n\n      = 1024 \n => n = 1024\n\nLet A be the event of getting number of heads is double the number of tails.\n\nFor event A\n------------\nNumber of favourable choices  =  Number of chances of getting heads is double the number of tailes.\n\n                  = No chances for occurance of event A\n                  = 0\n           =>  m  = 0\n\nProbablity of occurance of an event A =  Number of favourable choices / total number of possible choices\n     => P(A) =  m/n\n     => P(A) =  0/1024\n     => P(A) =  0\nTherefore the probability is zero.\n\n\n\nFor all about probabilty:\n\nhttp://www.futureaccountant.com
154434
The problem cannot be attended to without knowing the total number of possibilities.\n\nhow many sections are there on the spinnner..\n\n\nhttp://www.futureaccountant.com/probability/\nhttp://www.schoolingkids.com/
911284
You would use monte Carlo simulation when you need to find the distribution of a variable, and this depends on other variables which are random, especially when the interact in a complicated way.\n\nRather than trying to work out an answer, you just run the model using lots of different random numbers as inputs.  The outputs should approximate your answer if you have enough inputs.\n\nThese can be checked by using solvable cases.  I.e. consider the average value when you through a dice =(1+2+3+4+5+6)/6=3.5\n\nYou could also attempt to answer this by Monte Carlo methods by throwing a dice lots of times and taking the average of your results.  If you make enough throws the answer should be close enough for all practical purposes.
412043
I can't believe that this question is being asked in the sixth grade.  Maybe half of my college students could answer this.\n\nThe best way to answer this question is to make a table.  Make the rows the results of the first die and the columns the results of the second die.  Then put the product of the two roll inside of the table.\n            1   2   3   4   5   6\n          ------------------\n        1| 1   2   3   4   5   6\n        2| 2   4   6   8  10  12\n        3| 3   6   9  12  15  18\n        4| 4   8  12  16  20  24\n        5| 5  10  15  20  25  30\n        6| 6  12  18  24  30  36\n\n(Sorry I can't get the columns to even out!)\n\nNotice that there are 36 possible rolls of the two dice.  To get the two probabilities, just count the number of times you get the event in the table and divide by 36.\n\nProbability of an even product = 27/36\nProbability that the product is less than or equal to six = 14/36\n\nI see a lot of incorrect answers up there.  I teach statistics at the college level.  Trust me, this is right.
167958
There are several solutions other than eBay, including the site below:
375277
Mathworld.com.  Or, if you type the subject your interested in followed by "Forum" in google, you'll find some good help that way.  Fore xample, "calculus forum".  Also, you can ask me whatever you want.
815085
Not every problem should be solved with an elephant gun.\nJust ask yourself a different question. What is the probability that x2 is closer to x1 than to x3. since the answer to that question is the same as the question you ask and the only 2 possibilities are certain\nand mutually exclusive, the answer should be the same, 1/2.\n\nWhat about if x2 is exactly in the middle you ask?\n\nIt's like asking what is the probability that a random number in an interval is the midpoint. 1/infinity=0
887044
to calculate just time the total of first choice and the total of second choice or until the total of n choice.\nor you can try this: if you toss n coins total probability is 2^n\nif you toss n dice total probability is 6^n\n\nor if you wanna know the probability just arrange it and try this\nex: the probability if you take two alfabets from A - I are\nAA,AB,AC,ETC\nBA,BB,BC,ETC\nETC\n\n: )
102166
Yeah, try searching google for math problems
902378
http://thesaurus.maths.org/mmkb/entry.html?action=entryById&id=3206\n\nhttp://mathforum.org/library/drmath/view/51899.html\n\nhttp://www-staff.mcs.uts.edu.au/~don/pubs/solving.html\n\nhttp://thesaurus.maths.org/mmkb/entry.html?action=entryById&id=3206
911469
I'll do some for you to show how they are done. In all cases probability P = #number of good throws/ # of all throws.\nAlso 1-P = #number of bad throws/ # of all throws\nOften you can arrive at a solution by considering the dices one by one.\n1) First dice can be everyting, second dice can be one of the five numbers not yet thrown, third dice can be one of the four numbers not yet thrown, etc. This leads to P = (6/6) * (5/6) * ... *(1/6) =720/46656 = 0.0154\n2) Divide the dice in three pairs, named A,B and C. The probability to throw a pair of two's with pair A is 1/36. Idem for B and C. But we have 6 posibilities to arrange the pairs ABC, ACB, ... And we have 6x5x4x3x2x1 possibilities to arrange the dice into the three pairs. Therefore P = 6 x 720 x (1/36)^3 = 0.0926\n3) First dice 1, second dice 1, third dice 1 gives a probability of (1/6)^3 = 1/216. But there are 20 possibilities to arrange the dice.\nP=20/216 = 0.0926
1186148
There used to be a lot of them, but with computers these days they are being solved by brute force. Examples include Fermat's Last Theorem and the Four Color problem in topology. You can look those up to see what they are, and you will probably find sites listing other unsolved mathematical puzzles.