U are tossing 2 fair number cubes number 1 to 6 and recording product. What's probablity product will be even?
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.
515436
plz complete the question. i like to solve the questions on probability, but yur question is incomplete.
131088
try this site\n\nhttp://www.futureaccountant.com/probability/
152796
try this site...\n\nhttp://www.futureaccountant.com/probability/
627413
OK.  Here are the possibilities.  It is important to keep the order of the dice in mind, so you can think of this as rolling a die 5 times.\n\nSSSSD\nSSSDS\nSSDSS\nSDSSS\nDSSSS\n\nNow each of these will have the same probability, so if I can find the probability of one of them, I can take that and multiply by 5 since there are five combinations.\n\nSSSSD can also occur in different ways.  If S=1, then D could be 2, 3, 4, 5, or 6.  So if S=1, then the probability of this combination is (1/6)^4(5/6).\n\nNow if S=2, then D can be 1, 3, 4, 5, or 6.  So again, (1/6)^4(5/6).\n\nSo hopefully you can see that there are 6 ways that you can do this.  So the probability of SSSSD is 6(1/6)^4(5/6).\n\nSo overall, the probability of getting four of a kind (without getting five of a kind) is\n5*6*(1/6)^4(5/6) = 150/6^5.\n\nSo it looks like your method does work.  Good job; this isn't an easy on.  There is always more than one way to look at probabilities.
1094028
Most of them got it right.\n\nThis is another way to think about the result.\n\nFirst player = T or H doesn't matter\n\nSecond Player = Must be that same as the First Player\n                        =  only 2 possibility (H or T)\n                        = 1/2\nThird Player = Must be the same as First Player\n                    = again, only 2 possibility ( H or T )\n                    = 1/2\n\nIn order for the Second Player and Third Player to have the same result as First Plaer = 1/2 x 1/2\n                                 = 1/4.
403576
In game one between A and B, the outcomes are:  A wins, B wins, or they draw, equaling 3 outcomes.\n\nIn 14 games, there are 3^14 outcomes.
496704
Assuming that the 1, 2 and 3 are supposed to be consecutive:\n\nFirstly, for three throws of the die, the probability of throwing 1, 2 and 3 in order is (1/6)^3 = 1/216\n\nIf you throw the die 10 times then there are 8 independent chances of throwing 1, 2, 3. That's because if you have not thrown 1, 2, 3 by the time you get to the 9th throw, you cannot do it because you only have two throws left. \n\nTherefore the chance of throwing 1, 2, 3 with 10 throws of a die are (10 - 2) * (1/216) = 8/216 which is equivalent to 1/27.\n\nIf you prefer to express that as a decimal (not recommended!) then the answer is 0.0370 to 3 s f
236313
Wouldn't the probability than 5 of the 38 thought it was exciting be\n\n38C5(.79^5)(.21^33) = 6.63x10^-18 : basically impossible?  Even if you added in the probabilities of 4, 3, 2, 1, and 0, it's still tiny.\n\nSo I would infer that either the 79% is wrong or the sample was not independently chosen.
1136735
P(Last is W) = P{(2nd is W and 1st is W) or (2nd is B and 1st is W)}\n= P(2nd is W and 1st is W) + P(2nd is B and 1st is W) (since these are mutually exclusive events)\n= P(2nd is W)P(1st is W|2nd is W) + P(2nd is B)P(1st is W|2nd is B) (both use the general multiplication rule)\n= (6/10)(7/11) + (4/10)(6/11)\n= (42+24)/110\n= 66/110\n= 0.6\n\nWhich is the probability of getting a white from either box, which I think makes sense, since both have the same proportion of whites.
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
656922
Okay, intially, you pick one marble, out of 12. and there are 3 yellows, then your initial chances are 1/4 of picking a yellow. Now, you've picked one, so there are 11 marbles left in the bag. There are 4 blue marbles in the bag, so you have a 4/11 chance of picking a blue ball at tthis point. \nTo Calculate Probability based  on an event that 1) Has already happened ( the yellow ball was picked first) and that is an event depending on the results of the , you simply multiply the chances. ( 1/4*4/11) or 4/44, or 1/11 chances that that will happen.
1037406
1st cube: 0, 1, 2, 3, 4, 5\n2nd cube: 0, 1, 2, 6, 7, 8\n\nThe 6 on the 2nd cube can be inverted to make 9.\n\nBTW, no cube can be left out. This is the only way to make sure of that.
941928
if you are lucky you will pick 8 balls of the same colour. \nand the question is how lucky are you??
957682
half it then times it by 2! that will give u the answer
630810
a) probability of rolling a 2 is 1 out 6. for 5 balanced dice\n(1/6)(1/6)(1/6)(1/6)(1/6) = (1/7776)\n\nb) isnt this the same as part a?
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.