What is the probability of rolling a perfect square with one roll of number cube.?
402919
ok i'm not doing your homeowkr for you, but i'll help.  you're unclear about some things:  number cube?  you mean a six-sided die?  in that case, one roll can give you: 1,2,3,4,5, or 6.  which ones are perfect squares?  let's see: 1^2=1, 2^2=4, 3^3=9>6 therefore just 1 and 4 are perfect squares.  so what's the probability or rolling a 1 or a 4 out of the possibilities 1,2,3,4,5, or 6?\n\nprobability=what you want/what you CAN get.
106376
Also, if you agree that the formula for the number of combinations of n objects taken r at a time is\n\nn!/(r!(n-r)!)\n\nand that there is only one way to get a combination of n objects taking all n at a time, you'd get\n\nn!/(n!(n-n)!)  or n!/(n!0!), or canceling, 1/0! which must equal 1.\n\nSo 0! must equal 1.  More of a demonstration than a proof but it's easy to understand.
284048
In the experiment of rolling two dice\n\nTotal number of possible choices = 36 ==> n=36\n\nLet "A" be the event of rolling the dice such that the sum on the dice is 5\n\nFor Event "A"\n\nNumber of favorable choices = 4  ==> m = 4\n[(1,4), (2,3), (3,2), (4,1)]\n\nProbability of getting a sum of 5 on rolling the dice \n==> Probabilty of occurance of event "A" = m/n\n==> P(A) = 4/36\n==> P(A) = 1/9\n\nhttp://www.futureaccountant.com/\nhttp://www.schoolingkids.com/probability/
296539
This is a good question, and is the basis behind the theory of statistical thermodynamics. \n\nWhen you roll one die (i'm assuming a standard, fair,  six sided die) the probability of getting any particular number is 1/6. That is pretty obvious, each number is equally likely.\n\nInteresting things start to happen when you roll multiple dice however. Lets take all the possible outcomes for rolling two standard dice.\n\nDie 1   -   Die2   -   Total\n  1          1           2\n  1          2           3\n  1          3           4\n  1          4           5\n  1          5           6\n  1          6           7\n  2          1           3\n  2          2           4\n  2          3           5\n  2          4           6\n  2          5           7\n  2          6           8\n  3          1           4\n  3          2           5\n  3          3           6\n  3          4           7\n  3          5           8\n  3          6           9\n  4          1           5\n  4          2           6\n  4          3           7\n  4          4           8\n  4          5           9\n  4          6           10\n  5          1           6\n  5          2           7\n  5          3           8\n  5          4           9\n  5          5           10\n  5          6           11\n  6          1           7\n  6          2           8\n  6          3           9\n  6          4           10\n  6          5           11\n  6          6           12\n\nNow each die is independent of the other one. This means that getting 1-6 is a different result to getting 6-1. Thus it's easy to see that there are 36 possibile outcomes as shown above (6*6). Take a look at the totals of the outcomes though.\n\nThe first total i've written down is '2'. This comes from getting a 1 and another 1. You'll notice that this total does not appear anywhere else at all. The only possible way to get a total of 2 is if you get a 1 on both dice.\n\nAlso look at the last total, '12'. This is from getting a 6 and another 6, and is also the only place you'll see this total. There is only 1 possible way to get 12 as a total.\n\nHowever if you look at the total '7', you'll notice this appears 6 times. There are 6 different ways to get a 7 (1-6, 2-5, 3-4, 4-3, 5-2, 6-1). Therefore you are 6 times more likely to get a total of '7' than you are to get a total of either '1' or '12'! If you were to plot the average score received after rolling the two dice and plot them on a probability distribution curve (just graph the average scores along the bottom with the probability of getting that score up the side) you will see that it starts of with a low probability for getting an average of 1, rises to a peak at getting an average of 3.5, and then drops back to a low probability got getting an average of 6.\n\nNow with two dice there is still a reasonably good chance of scoring an average of 1, and a reasonably good chance of scoring an average of 6 (you'll see it happen every now and then if you roll the dice for a few minutes), but what happens if we go to 3 dice? Well the number of possibile outcomes is multipled by 6 (don't worry, I won't list them all!) But you'll notice there is STILL only 1 possible way of getting an average score of '1' (1-1-1). And there is STILL only one possible way of getting an average score of '6' (6-6-6). Therfore, the probability of you getting these numbers as an average of rolling 3 dice is much less than with two. Also the probability of getting an average of 3.5 is much higher (i'll leave that to the reader to confirm). If you plot the results for 3 dice onto the probability curve you'll notice that it is a bit steeper. It still has it's peak at 3.5, but it is much sharper.\n\nThe really interesting things happen when you go to large numbers of dice. Imagine that you have one die for every atom in a sample of some substance. Ther
731770
You have 5 dice, with 6 numbers each, for a total of 6^5 possible rolls on your first roll.\n\nThere are 6 ways you can get a Yahtzee:\n1-1-1-1-1\n2-2-2-2-2\n3-3-3-3-3\n4-4-4-4-4\n5-5-5-5-5\n6-6-6-6-6\n\nTherefore your chances of getting a Yahtzee on your first roll is 6/6^5 = 6/7776 = 1/1296 ≈ 0.00077\n\n~ ~ ~\n\nIf you already have a 5-5-5, you only roll two dice -- with two dice you have 6^2 = 36 possible outcomes.\n\nP(yahtzee on next roll) = P(5-5 on next roll) = 1/36\n\nP(full house on next roll) = P(1-1 or 2-2 or 3-3 or 4-4 or 6-6 on next roll) = 5/36\n\n{ notice I left out 5-5, but technically 5-5-5-5-5 is a full house too, in addition to a yahtzee, so depending on how your teacher interprets "full house" the answer to this might be 6/36 rather than 5/36 }\n\nP(4 of kind) = P(5-1 or 5-2 or 5-3 or 5-4 or 5-6) = 5/36\n\n~ ~ ~\n\nYou have 2,3,4,5 and are rolling one die -- you have 6 possible outcomes for rolling that one die\n\nP(large straight) = P(1 or 6) = 2/6 = 1/3\nP(small straight) = 100% since you already have a small straight with your 2-3-4-5\n\n~ ~ ~\n\nYou have 3-4-5 and are rolling 2 dice -- you have 36 possible rolls with those 2 dice\n\nP(full house) = 0, since a full house is 3 of one number and 2 of another, and you already have three different numbers showing!\n\nP(3 of a kind) = P(3-3 or 4-4 or 5-5) = 3/36 = 1/12\n\nP(large straight) = P(1-2 or 2-6) = 2/36 = 1/18\n\n~ ~ ~ \n\nI hope you understood how I approached and solved these, so that you can do them on your own when it's time to take your test on this material.  Good luck!\n\n~ ♥ ~
467797
The chance of not rolling a four is 5 out of 6. So five times:\n\n5/6 ^ 5 = 0.4019\n\nor about 2 in 5.
967346
This probability problem can be used using Binomial Distribution.  \n\nGiven a number of 'x' success, the probability in a binomial experiment is:\n\nP(x) = [n choose x]*p^x*q^(n-x)\n\nwhere:\n\nn choose x is a combination problem: n!/[(n-x)!x!]\np = probability of success\nq = probability of failure\nn = total selection\nx = successes\n\nIn this problem, we are looking for at least one case has a success (which is one of the highest 9 numbers).  In essence, we are solving for:\n\nP(x=1) + P(x=2) + P(x=3) + P(x=4) + P(X=5).  \n\nHowever it would be easier to find the probability of the failures and subtract from the total probability because we only have to calculate 1 probability.  This would be:\n\n1 - P(x=0) is what we are looking for.  Now all we need is to calculate the probabilty.\n\nFor P(x=0), we have the following:\n\nn = 5\nx = 0\np = (9/26) = probability of success\nq = (17/26) = probability of failure\n\nPlugging into P(x), we get:\n\nP(x=0) = (5!/5!)(9/26)^0(17/26)^5\n\nSolving, you get P(x=0) = 0.12\n\nTherefore, \n\n1 - P(x=0) is ------>\n\n1- 0.12 = 0.88\n\nBelow is a website devoted to Binomial Distribution exercises\n---------\n\nHope this helps
749852
1.Probability of getting a number from 1-4 on throwing the dice\n         = 4/6\n\n2.Probability of a correct throw = 1/4\n\n3.Total number of questions = 160\n\nYour Answer = (4/6 * 1/4) raised to the power 160
955894
x1*p(x1) + x2*p(x2)\n\nx1 = 1 for  event dots> 3,\nx2 = 4 for event  dots<=3.  \nThese are equally probable because there are 6 dots on the die.  \n\n1*(1/2) + 4*(1/2)
878193
EDIT: This answer has been corrected.  See note below.\n\n\nIn the case of the dice, you have discrete values in the probability distribution.  It is not a "normal" (bell-curve) distribution.  But you can calculate the std dev. as follows:\n\nFor every possible value v of a throw i, compute P(vi).  In a large number of throws N, the expected number of times a value vi occurs, nvi = N*P(vi).  The mean value would then be m = (1/N)∑nv*v = (1/N)∑N*v*P(v) = ∑v*P(v).  The standard deviation would then be\n\nd^2 = [1/(N-1)]∑(xi - m)^2 =  [1/(N-1)]*{∑[N*vi*P(vi)]^2 - 2*N*m*∑v*P(vi) +N*m^2}\n\nd = √ [1/(N-1)]*{∑[N*vi*P(vi)]^2 - 2*N*m*∑vi*P(vi) +N*m^2}\n\n\nYou can't eliminate N (the total no of triails) but you can compute the result for a large N without actually rolling the dice.\n\nEDIT:  There is an error in the formula: the terms N and P(vi) should not be squared in the first term of the sum. The formula should be \n\nd = √ [1/(N-1)]*{∑N*P(vi)*vi^2 - 2*N*m*∑vi*P(vi) +N*m^2}\n\nThen, for large N, N-1≈1 and the N's cancel out to give\n\nd = √ [∑P(vi)*vi^2 - 2*m*∑vi*P(vi) +m^2]\n\nSo for all possible values of the dice (vi), compute the probabiity of that value P(vi), the mean value m = ∑vi*P(vi),  and sum over all value according to the above formula.\n\nPlugging the mean value into the formula gives the simplification:\n\nd = √ [∑P(vi)*vi^2 - 2*m^2 +m^2] = √ [∑P(vi)*vi^2 - m^2]\n\nFINAL:  Made final correction.  Above formula is now correct; the m^2 is not under the summation sign.\n\nCalculated for an ordinary pair of dice: m = 7, d = 2.415\nYou can see the calculation here http://img452.imageshack.us/img452/1734/dicestddevlk8.png
330379
Count the total number of EQUAL sections of a spinner. THese must all have the same size. Or find the angle of each section formed.\n\nIf it could be divided up into equal sections:\n\nDivide the total number of times the spinner could land on the event you want. If you were looking for theoretical probability of finding purple, and there were 2 purple sections and the spinner was divided up into 5 equal sections, the probability would be 2/5. Just divide the number of times that YOUR event could occur(ie your color), divided by the number of equally likely events (ie spaces).\n\nIf you found the angle of the section(s) your looking for:\n\nAdd up the angle measures. If you found three purple spaces, one measuring 48, another measuring 80, and another measuring 16 degrees, you would add all of these up to get 144 degrees. Because there is a total of 360 degrees in the circle, and it is equally likely to get a point on each degree, you can divide 144 (the number of degrees that your purple spaces have) by 360 (the total number of degrees to get 144/360. This is reducable and we get 2/5 as our theoretical probability.
864738
The odds of getting all six are 13,983,816 to 1\n\nNote: the above answer forgets to divide by through the possible combinations of the six numbers. \n\nIf we multiply the numbers 49 x 48 x 47 x 46 x 45 x 44 we get 10,068,347,520. However, each possible group of six numbers (combination) can be drawn in different ways depending on which number in the group was drawn first, which was drawn second, and so on. There are 6 choices for the first, 5 for the second, 4 for the third, 3 for the fourth, 2 for the fifth, and 1 for the sixth. Multiply these numbers out to arrive at 6 x 5 x 4 x 3 x 2 x 1 = 720. We then need to divide 10,068,347,520 by 720 to arrive at the figure 13,983,816 as the number of different groups of six numbers.\n\nEdit: Yes, well the above answer has now changed their response. Well done and all that.
876444
Okay, I'm gonna assume that you're playing with all six-sided dice.\n\nIf you roll the one die, you want a 2, and nothing else helps.  The probability of getting a two in two rolls is\n\nP(2) = (1/6) + (5/6)(1/6) = 11/36,\n\nas if you get a two on the first roll you won't roll again, and if you don't get it on the first roll you have to roll again, to try to get the two.  \n\nNow let's consider if you roll the two dice, keeping the 3,4, and 5.  You want, basically, some sort of straight.  Since you are keeping the 3, 4, and 5, it's obvious that you need a 2 for any straight, whether it ends in a 5 or a 6.  Then you need either a one or a six.    Let's try this first with just one roll.  Say you roll two dice, one after the other (I say that just because it helps me think about it, even though you're rolling them at the same time).  If the first is a 1 or 6, you need the next to be a 2, and if the first is a 2, you need the next to be a 1 or 6.  Thus, with only one roll, the probability is:\n\nP = P(2) * (P(6) + P(1)) + (P(6) + P(1)) * P(2)\n\n   = 2 * P(2) * (P(6) + P(1))\n\n   = 2(1/6)(1/6 + 1/6)\n\n   = 1/9.\n\nThis is less than 11/36, but again, that's with only one roll.  Now we generalize it to two rolls, taking into account that your decision for the second roll will change depending on what you get in the first (but assuming that you'll always go for a straight).\n\nWe know the odds of you getting a straight in one roll are 1/9, so we don't need to calculate that.  What remains to be determined is what happens when you get a two but not a 1 or 6, a 1 or 6 but not a 2, or neither.  We'll try the last one first.\n\nThe odds of you not getting a 2, a 6, or a 1 in one roll of two dice is:\n\nP = (3/6)*(3/6) = 1/4,\n\nsince there is a 3/6 chance on each die of getting something besides those three.  If that happens, you roll both again, and the probability of getting a straight then, we've established, is 1/9.  (Now, you can see the just rolling the two dice twice in a row gives you a probability of 1/9 + 1/9 = 2/9 of getting that straight with either or both rolls.  This is by itself only 8/36, though, so we gotta calculate further.)\n\nSo the probability of either getting a straight your first roll, or getting nothing helpful your first roll and getting a straight on the second, is\n\nP = 1/9 + (1/4)*(1/9) = 5/36.\n\nNow for if you get a 2 on your first roll.\n\nThe probability of getting a 2 but not a 1 or 6 is\n\nP = P(2) * P(3, 4, or 5) * 2 + P(2) * P(2)\n\n   = (1/6) * (3/6) * 2 + (1/6)*(1/6)\n\n   = 7/36,\n\nand the following probability of you getting a 1 or 6 on your next roll is\n\nP = 2/6 = 1/3, \n\nso the total probability of you getting a 2 on your first and a 1 or 6 on your next is\n\nP = (7/36) * (1/3) = 7/108.\n\nNow for the probability of you getting a 1 or 6 on your first roll but not a 2, and getting a 2 on your next roll.\n\nThe probability of getting a 1 or 6 but no 2 is\n\nP = P(1) * P(not 1 or 2) * 2 + P(6) * P(not 2 or 6) * 2 + P(6) * P(6) + P(1) * P(1)\n\n   = (1/6) * (4/6) * 2 + (1/6) * (4/6) * 2 + (1/6) * (1/6) + (1/6) * (1/6)\n\n   = 8/36 + 8/36 + 1/36 + 1/36\n\n   = 1/2.\n\nThe probability of then getting a 2 with one die is 1/6, so the probability of doing both is\n\nP = (1/2) * (1/6) = 1/12.\n\nAdding up all these probabilities, you have that the total probability of getting a straight if you roll two dice on your second roll instead of just one, is\n\nP = 5/36 + 7/108 + 1/12\n\n   = 15/108 + 7/108 + 9/108\n\n   = 31/108,\n\nwhich is less than 11/36, which equals 33/108.\n\nSo your odds are a bit less going for a straight with two dice than they would be if you went with the one die.  Then again, if you roll two you have an additional small possibility of getting two sixes or something, which may also help a bit.  But if you gotta have a straight, go with the one die..\n\nHope this helps with something (or is at least mildly entertaining).\n\n
657146
Time for Pascal's Triangle.\n.......1\n......1 1\n.....1 2 1\n....1 3 3 1\n...1 4 6 4 1\n..1 5 10 10 5 1\n1 6 15 20 15 6 1\n\n1 chance in 64 of 6 Heads and 0 Tails\n6 chances in 64 of 5 Heads and 1 Tail\n15 chances in 64 of 4 Heads and 2 Tails\n20 chances in 64 of 3 Heads and 3 Tails\n15 chances in 64 of 2 Heads and 4 Tails\n6 chances in 64 of 1 Heads and 5 Tails\n1 chance in 64 of 0 Heads and 6 Tails\n\nChance of exactly 3 tails: 20 in 64\nChance of 3 or more tails: 42 in 64\nChance of more than 3 tails (4 or more) : 22 in 64\n\nAnyone who think the chance of 3 tails in 6 tosses is 50% is generalising wildly.\n\nThe chance of 1 tail in 1 toss is 50%\nThe chance of 1 head & 1 tail in 2 tosses is 2 chances in 4\nThe chance of 2 heads & 2 tails in 4 tosses is 6 chances in 16. (37.5%)\nThe chance of 3 heads & 3 tails in 6 tosses is 20 chances in 64. (31.25%)\nThe chance of 4 heads & 4 tails in 8 tosses is 70 chances in 256. (27.34%)\nThe chance of 5 heads & 5 tails in 10 tosses is 252 chances in 1024. (24.61%)\n\n(read the middle number in each row of Pascal's Triangle, and extend it downwards)\nThe bigger the number of tosses, the probability of tossing exactly half of the tosses as tails keeps on falling. This should not surprise us. There are so many more possible outcomes.
791422
Each individual toss is 1 / 2, a 50% probability.\n\nThe possibilities of all outcomes is 1 / (2 x 2 x 2), a 12.5% probability.\n\nOnly 2 possible outcomes do not meet the desired outcome (all heads, all tails).\n\n2 x 12.5% is 25% (but that's the opposite of what you want)\n\n100% - 25% = 75%
947747
Pick one knight anywhere.\nIf you pick either of the knights directly adjacent to him, you are done.  This happens 2/24 of the time.\n....OOXXOO...\n\nIf you pick a knight that is separated by exactly one seat (2/24 of the time), then there are 3 out of 23 seats that are adjacent.\n...OXOXO....\nIf you pick a knight that is more than one seat away (20/24), then there are 4 out of 23 seats that are adjacent.\n...OXO...OXO...\n\nSo the probability is:\n1/12 + 1/12 * 3/23 + 10/12 * 4/23\n\nAdding it up:\n23/276 + 3/276 + 40/276, or 66/276 = 33/138\n\nThis is approximately 23.91%
551111
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.