a number cube has side labeled 4,5,6,7,8,and 9.  Which of the follwing events is equally likely to occur.?
90357
F- even and odds.  There are 3 even sides and 3 odd sides. So the likelihood is the same for both.
832564
The answer is one chance in 2^5  or 1 in 32.\n\nNotice that the chances are the same as flipping five heads in a row.
553417
if there are only 10 numbers to chose from, say 0-9 then the odds are 10*10*10, if only nine number say 1-9 then it is 9*9*9. that help ya?
1113295
(1, 2, 3,4) or (-1, -2, -3, -4) or (1, 2, -3, -4) or (1, -2, -3, 4)\nYou also have\n(-1, -2, 3, 4), (-1, 2, -3, 4), (-1, 2, 3, -4), and (1, -2, 3, -4)\nSo there are 8 possibilities.\nEach having a probability of 1/8 or occurring.\nFor (1,2,3,4) and (-1,2,-3,4), the prob. of an even natural number is 2/4 = 1/2\nMultiply this by 1/8 to get the conditional prob.\n(1/2)(1/8) = 1/16\nFor each of the following: (1,2,-3,-4), (1,-2,-3,4), (-1, -2, 3, 4), and (-1, 2, 3, -4), the prob. of an even natural number is 1/4\n(1/4)(1/8) = 1/32\nFor (-1,-2,-3,-4) and (1,-2,3,-4), the prob. of an even natural number is 0.\n0(1/8) = 0\nAdd all these together.\n2(1/16) + 4(1/32) + 2(0)\n= 1/8 + 1/8\n= 2/8\n= 1/4\nThis doesn't match your solution, but you can see if this makes any sense to you.
893743
8\n9\n7\n4\n4\n5\n7\n7\n9\n8\n5\n4\n8\n8
452386
7*6*5*4 = 840 if order matters.\n\n7*6*5*4 / 4! = 35 if order does not matter
600596
There are 6 possible outcomes for the roll. There is only one outcome when the number is 4. Therfore the chances of getting a 4 are 1/6.
1009251
Ahh!  True, 6 consecutive flips of a coin in which each lands heads has a (1/2)^6 probability of happening.  But what are the odds that the seventh flip will be heads, too?
247655
0,1,1,5,7,11,13,13,21,23,27.......
305696
5\nIt is helpful if you know the first 5 or 6 cubes.\n1^3 = 1\n2^3 = 8\n3^3 = 27\n4^3 = 64\n5^3 = 125\n6^3 = 216
893652
with 1q, 2 ways (0,1 n)\nwith 3d, 1 ways (0 n)\nwith 2d, 3 ways (0,1,2 n)\nwith 1d, 5 ways (0,1,2,3,4 n)\nwith 0d, 7 ways (0,1,2,3,4,5,6 n)\n\nTotal of 18 ways
686219
1,24 24,1 2,12 12,2 3,8 8,3 4,6 6,4 -1,-24 -24,-1 -2,-12   \n-12,-2 -3,-8 -8,-3 -4,-6 -6,-4 and if x and y don't have to be intergers, there are basicly an infaninte number of possibilitys, such as 1/2,48 1/3,72 1/4,96..... so on, and the negatives. I hope x and y must be intergers, or you're stuck. easiest way is to graph it, or put like y=24/x or x=24/y. Not much help, but it's right.
842276
1,2,3,4,6,8,12,16,24,48 \n\nlolz hope this helps :)
783821
5 out of the 8 items in Box A are non-defective.\n3 out of the 5 items in Box B are non-defective.\n\nThe probability that an item from Box A and an item from Box B are drawn and are both non-defective is:\n\n5/8 * 3/5 = 3/8\n\nThe probability that an item from Box A and an item drawn from Box B and both defective is:\n\n3/8 * 2/5 = 3/20.\n\nThis means that the probability that one of the two items drawn is defective is:\n\n1 - p(both are not defective) - p(both are defective)\n=1 - 3/8 - 3/20\n= 19/40\n\nWe can see this by showing all possible outcomes. The items in Box A are shown on the top edge, and the items in Box B are shown on the right edge\n\nN  N  N  N  N  D  D D\nA  A  A   A  A  B  B  B  N\nA  A  A   A  A  B  B  B  N\nA  A  A   A  A  B  B  B  N\nB  B  B  B  B  C  C  C D\nB  B  B  B  B  C  C  C D\n\nThe events marked "A" represent 2 non-defective items. The events marked "B" represent 1 defective and 1 non-defective item. The events marked "C" represent 2 defective items.\n\nOut of 19 events where there is 1 defective and 1 non-defective item, the defective item came from box A 9 times.\n\nSo if 1 item is defective and 1 is not, the probability is 9/19 that it came from box A
938348
There are many combinations. One of them is 1,2,3,4,5,16. Is there some specific property you wanted these six numbers to have?
1070909
Five and seven\n\nBy giving one, it's four and eight (twice)\n\nBy givng one, it's six and six
133517
the solutions is 7,8,3,2 or 7,6,5,2. The problem works with both sets of numbers.