An example would be,I roll the dice and get 6.This is my point I have to obtain.If I roll 7 or 11 after I rolled my point I would lose.Therefore the question means I can roll 7 or 11 and this would take me to my next roll whereupon I must roll 7 or 11 or roll any other point other than 2,3 or 12.This number is my point.Examples are 4,5,6,8,9,10.If I rolled any of these numbers,coming out or starting off then this is my point.I must roll that number instead of 7 or 11 because I would not obtain my point before losing.If I roll 7 or 11 coming out that means I move onto my next roll.I hope this helps in getting a correct answer.
136914
You are dealing with probabilities here and your odds are going to be exponential. But here is what I have for you.\n\nOdds in rolling a:\n4 = 1/12\n5 = 1/9\n6 = 5/36\n7 or 11 = 2/9\n8 = 5/36\n9 = 1/9\n10 = 1/12\n\nNow for each roll after that you would take the odds and times it to the "n" power, where "n" is the number of the roll. So to roll any of these numbers 12 times in a row would mean taking the fraction to the 12th power. Below are those answers. (Hope you understand scientific notation.)\n\n(1/12)^12 = 1.13876838470415E+55\n(1/9)^12  = 1.13771030934265E+55\n(5/36)^12 = 5.1524055910299E-11\n(2/9)^12  = 1.14868734219891E+55\n\nIn other words you would be more likely to win the lottery several times before this would happen. Hope this helps.
482780
As the previous poster said, you don't have enough information.  If you have n-sided polyhedra, are the sides numbered 1 to n?  Is the expected value in reference to a throw of the dice?  If so, are we talking about two dice?\n\nFor two dice with the sides numbered 1 to n, the answer is n + 1.
257931
8. One for each corner of the cube. Think about how many corners are on a die or a box. Each of these is a point.
10512
Sometimes, it's easier to find the probability of the opposite case you are looking for.  This is one of those times.  Try to find the probability that you don't get the right one.\n\nAfter taking you two chances, one of the three objects is left.  What are the chances that this one that is left is the right one? The one that is left is one of three, so there is a 33.3% chance that it is left behind- which means there is a 33.3% chance that you DID NOT get the right one.  Therefore, there is a 66.7% chance that you WILL get the right one. \n\nIf you are still not convinced, it is easy to show that your reasoning that got you to 83% is wrong.  What if you used the same reasoning with 4 objects and 3 chances to get the "right" one.  Your logic would say there is a 25% chance the first time, a 33% chance the second time, and a 50% chance the third time.  If you just added them (as you did to get 83%), you would find that 25% + 33% + 50% = 108%.  Clearly you can't have a percentage above 100%, which illustrates the problem with that reasoning.\n\nHere's why that reasoning dosn't work; 33% of the times you play this game, your second try has a 0% probability of getting the right one because you already got it on the first try.  The other 66% of the times you play the game, your second try has a 50% probability.  So the AVERAGE probaility of the second try is (33.3% x 0%) + (66.7% x 50%), or 33.6%.  Add this 33.3% to the 33.3% probability of the first try, and you get the correct answer of 66.7%.\n\nIf you still aren't convinced, come over some time for a "friendly" game of dice.
121896
Ummm... take the number of points you have, divide by 5 and there you have it.\n\n979 in my case.
639254
This is a binomial distribution with a 5/6 chance of losing a dollar and a 1/6 chance of winning $5... winning six while paying the one to play.  The mean expected value of wins after "n" rolls of the die is (n/6), and the standard deviation is [(5n)^(1/2)]/6.\n\nThe thing is, you're equally likely to lose money are you are to win money at this game.\n\nWhen n gets really large, the expected value of your net winnings will move farther away from (n/6), but the variance and standard deviation will become smaller.  For examples, it would take 8 rolls of the die to get a standard deviation of possibilities net wins of 1, but 29 rolls to get to a standard deviation of 2, 65 rolls to get to 3, 116 rolls to get to 4, and 180 rolls to get to 5.\n\nIn other words, the more you play, if you're on a winning streak and you play a long time, your winnings are more likely to increase, though at a much slower rate.  But if you're on a losing streak, the longer you play makes it equally as likely to never catch up to break even.\n\nThe bottom line is the answer to maximize your profit depends on whether you're ahead or behind after your first few dozen throws or so.  If you're losing, quit.  If you're ahead, play 'til you're bored.
333435
a)\nthe two actions are independent, so it doesnt make any difference, so\n\nthe answer is 3/10\n\nb) (3/10)x(2/9) + (7/10)x(3/9)\n\n(if it was red on the first draw), + (if it was white on the first draw)
148610
Ok, I got it.  I recommend writing the numbers 1-12 and crossing them out as you use them.\n\nYou've got your 5,6,1,and 8 on the inside.  So on the upper row of four, 5+6=11, so you need 15 more to make 26.  The only numbers that are left where 2 add up to 15 are 12&3 or 11&4.  I tried 11&4 first and will just tell you it didn't work, so it's 12&3.  Now to decide which goes where.  If you put the 3 on the left, it lines up with the 1 making 4, and there's no other numbers that will add up to the required 22 - so 12 must go on the left and 3 on the right.\n\nNow you can see the line  from the bottom point up to the 3.  8+3=11, so you need 15 again, and the only choices left are the 11&4.  Put the 11 at the bottom point and the 4 between the 8&3.\n\nNow looking at the line that goes from 12 down to the 11 you have 12+1+11=24, so you must use the 2 here.  Now you have 7,9 & 10 left.\n\nLooking at the line with the 5&2, that makes 7, so you need 19.  The only way to get that is with the 10 & 9, but you don't know which goes where.  You DO know, though, that the only other point that is left gets the 7.  That's the bottom right point.\n\nThis made a line with 7,6,4 and the very top point.  7+6+4=17, so you need 9 at the top to make 26, which leaves the space for the 10 at the bottom left.\n\nIn order, from top to bottom/left to right you have\n\n9\n12,5,6,3\n2,4\n10,1,8,7\n11
93286
will any answerer plz tell me what exactly is the SIX AND OUT rule is??
466495
The next number is 6 (and every subsequent number is 6 as well).\n\nIt has to do with converting bases.  If you have a_n=6, then n has to be greater than or equal to 2.  Therefore, \n\n6 in base 2 = 110\n6 in base 3 = 20\n6 in base 4 = 12\n6 in base 5 = 11\n6 in base 6 = 10\n6 in base 7 = 6\n\nand so on....\n\nHope that helps!
591113
It means a number between 5 and 200.  You might have to round to the nearest number if it doesn't take decimals.
535461
I understand what you are saying when you mention the permutations for 10 digits, but I think you're missing what it really means.  First, this isn't right to think of it this way to think about this.  Non-repeating doesn't mean that I can't find a string in the decimal expansion that's the same as an earlier string (for example, finding a piece 14159 later in pi isn't impossible).  What it means is that there is no point where the digits in pi *completely* repeat (i.e. it becomes something like 141591415914159.... on and on forever).  So you're right, it can include some kind of partial repeat, but non-repeating means it can't have a total repeat.\n\nYour comment about the permutations (not combinations) on the numbers {0,...,9} is right but it's used incorrectly also.  It's wrong because of what I mentioned above.  Basically it's assuming that once you pick a number you can't repeat it again in a 10 digit sequence, but that's clearly wrong (again, 14159 appears in pi, where 1 occurs twice).  So you have several right ideas here but you're combining them seomwhat incorrectly.\n\nHope this clears things up!!
103962
In the experiment of choosing the 3 digits\n\nTotal no. of possible choices \n= (no.of ways in which the first digit can be chosen) * (no.of ways in which the second digit can be chosen) * (no.of ways in which the third digit can be chosen)\n==> n = 9 * 10 * 10\n==> n = 900\n\n[Since the first digit cannot be zero, there would be 9 possibilities for choosing it, 10 for second and third]\n\nLet "A" be the event that all the 3 digits are different and "B' the event that the 3rd digit is different from the first 2 digits.\n\nFor Event "A"\n\nNo.of favorable choices\n= (no.of ways in which the first digit can be chosen) * (no.of ways in which the second digit can be chosen) * (no.of ways in which the third digit can be chosen)\n= 9 * 9 * 8\n==> m(a) = 648\n\n[Since the three digits to be chosen should be different, there would be 9 possibilities for choosing the first digit (since 0 is excluded), 9 possibilities for the second (excluding the one chosen for the first and including 0) and 8 for the third (excluding the ones chosen for the first and the second and including 0)\n\nProbability that all the digits are different\n==> Probabilty of occurance of event "A" = m(a)/n\n==> P(A) = 648/900\n==> P(A) = 18/25\n\nFor Event "B"\n\nNo.of favorable choices\n= (no.of ways in which the first digit can be chosen) * (no.of ways in which the second digit can be chosen) * (no.of ways in which the third digit can be chosen)\n= 9 * 10 * 8\n==> m(a) = 720\n\n[Since the first two digits can be any of the ten and the last should be different from the first two, there would be 9 possibilities for choosing the first digit (since 0 is excluded), 10 possibilities for the second (any digit can be chosen including 0) and 8 for the third (excluding the ones chosen for the first and the second)\n\nProbability that the 3rd digit is different from the first two\n==> Probabilty of occurance of event "B" = m(b)/n\n==> P(A) = 720/900\n==> P(A) = 4/5\n\n\nhttp://www.futureaccountant.com/probability/study-notes/\n\nhttp://www.schoolingkids.com/
498549
If John had a 1 then Sheila's number would need to be 2\n\nIf John had a 5 then Sheila's number would need to be 4\n\nSo when John says "I don't know your number", he has just effectively told Sheila he does not have 1 or 5\n\nThe same logic dictates Sheila cannot have 1 or 5 and also lets you say:-\n\nIf Shiela had a 2, she would know John had a 3 (Since she knows he does not have a 1)\n\nIf Shiela had a 4, she would know John had a 3 (Since she knows he does not have a 5)\n\nSo when Sheila says "I don't know your number" she is effectively saying, "I don't have 2 or 4"\n\nSo Sheila's number must be 3
184310
7-10-1? three pins? that's much easier than just 7-10. take a 8 lb ball, go from left and aim at the right of the 1 pin. let the 1 bounce and hit the 7 while the ball will also bounce to hit the 10! Got it?
141781
A fair dice is one in which the probability of getting one side is 1/6.\n\nSum of 6 or 11 can be obtained in the following ways: \n(1,5) (2,4) (3,3) (5,1) (4,2) (5,6) (6,5)\n\nThere are a total of seven elements here.\nTotal possible elements = (6 X 6) = 36\n\nHence probability = 7/36
54139
That's an extremely good question. \n\nThe answer is simple. Before that, i will explain what your question means to those who don't get. Well, it means that we must get 24 using 1,3,4 and 6 any number of times however involving operations +,-,* and / only once. \n\n  Then, the answer is very simple :\n\n  {[(6*6)/(3-1)]+[6]}  =  24\n\n---------------------------------------------------------\nWOKRING :\n\n {[(6*6)/(3-1)]+[6]}   =  {[(36)/(2)]+[6]}\n                       =  18 + 6\n                       =  24\n\n----------------------------------------------------------\n\nI hope you post more questions like these. They evoke a lot of thought. I hope the solution serves th purpose.