90ballsof red,blue,white,&10 ofmix colors, how many balls does it take to make at least 10 of the same color?
345437
Wait, I'm confused. Are we saying: 90 Red Balls, 90 Blue Balls, 90 White Balls, 10 Multicolored Balls? If that is the case, and you are asking how many must be taken to guarantee that there will be at least 10 of one variety, then the answer is 37. It is possible to pick 9 of each of the four varieties. The next ball would therefore guarantee at least 10 of the same type.\n\nADDED:\nSo then, it's 30 Red, 30 Blue, 30 White, and 10 Multicolor? Then the answer is still 37, for the reason I described before. Unless...are those 10 other balls all the same? Do you mean 10 balls with mixed colors on them, or 10 balls, each a different, random color, but not red, blue, or white? Because if that's the case, it's 38. It is possible to pick 9 of each of the three colors, and all 10 balls of different colors, and you would then have 37. The next ball (38th) would have to be the 10th of either red, blue or white.
631136
Total 20 nines (2 in 99) are there in 0 to 100.\n\n9, 19, 29, 39, 49, 59, 69, 79, 89, 90, 91, 92, 93, 94, 95, 96, 97, 98, 99
619988
Each time you pot a red, you must choose another colour.\n\nOnce all the reds are potted, you must work through the colours in the correct order.\n\nRed balls = 1 point each\n\nYellow = 2\nBrown = 3    THESE 3 (yellow, brown & green) I get mixed up\nGreen = 4\n\nBlue = 5\nPink = 6\nBlack = 7
651368
Answer: 4\n\nPut 30 kids in each room. So, after 10 rooms are filled, you have out (30*10) kids in the 10 rooms. So, you are left with 6. If you put 1 kid in 6 rooms, to make the number of kids in these 6 rooms to 31, you are left with 4 rooms with 30 kids.
392841
fill 7 and tip into 10\nfill 7 again and tip into ten leaving 4\ntip out ten and put 4 into 10\nfill 7 and tip into 10 leaving 1\ntip out 10 and pour in 1\nfill 7 again and pour into 10 leaving 8
725814
This is a very difficult counting problem.  I will show one possible grouping and count the number of ways it can be done.\n\nTo put just 1 boy in group 1, there are 10 ways to pick the boy and "combination of 15 things taken 4 at a time" ways to pick the girls.  This is 10 * (15*14*13*12 / 1*2*3*4) = 10*1365 = 13650 ways to pick 4 girls and 1 boy.\n-----------------\nThere are many grouping scenarios; let us start with 1 boy, 1 boy, 1 boy, 2 boys, 5 boys.\n\nWe know there are 13650 ways to pick a single boy for group 1.\n\nTo then pick a single boy for group 2 from the remaining children, there are 9*(11*10*9*8 / 1*2*3*4) = 9 * 330 = 2970 ways.\n\nIf a single boy is then selected for group 3, there are 8 * (7*6*5*4 / 1*2*3*4) = 8*35 = 280 ways.\n\nGroup 4 must have at least 2 boys.  There are 7 boys left, and we must choose 2 of them.  So there are 7*6/1*2 = 21 ways to choose those boys with the remaining girls.\n\nThe last group would have 5 boys.\n\nSo just for this one possible way of grouping - 1 boy, 1 boy, 1 boy, 2 boys, 5 boys, there are \n13650 * 2970 * 280 * 21 * 1 = 238,378,140,000 ways to arrange them.\n\nTHIS MAY NOT BE THE CORRECT FIGURE FOR THIS GROUPING SCENARIO - THERE MAY BE DUPLICATES, IN WHICH CASE THIS FIGURE IS TOO HIGH.  I did not know how to remove them from the figure.  The 13650 is correct for the first, but then we need to look at how many ways are there to grab 4&1, 4&1, 3&2, 0&5 from the remaining children.\n\n\nWe must also account for other grouping scenarios, such as 2, 2, 2, 2, 2 boys; 1,1,2,2,4 boys; 1,1,1,3,4 boys; 1,2,2,2,3 boys, and some others.  For each of these grouping possibilities, the above such calculations would have to be done.  The sum of all the grouping possibilities is the total.\n\nSo this is a very difficult counting problem.  I would think you should search for more tools from your professor/teacher, or book.  There may be some more advanced way of saving time rather than go through each grouping scenario one at a time; though that will give the correct answer.
885691
Part (a) of problem 1 is pretty easy.  There are 10 marbles total, and 6 of them are red.  On your first draw, you have a 6/10, or 0.6 (60%) probability of drawing a red sock.  In other words, there are 6 ways of drawing a red sock out of a total of 10 possibilities.\n\nFor part (b), you need to know that the probability of two independent events happening is given by the product of the probabilities of each of the individual events.  In this case, the two independent events are that the first marble drawn is red AND the that the second marble is also red.  We already calculated the probability for drawing a red marble on the first draw in part (a).  After the first draw, the box contains 9 total marbles, 5 of which are red, so the probability that the second marble is red is 5/9.  The probability that both are red is given by the product, 6/10 * 5/9 = 30/90 = 0.3 (30%).\n\nThe reasoning for part (b) is the same as for part (a).  The probability that the first marble drawn is white is 4/10, and the probability that the second marble drawn is also white is 3/9.  The probability that both are white is given by the product, 4/10 * 3/9 = 12/90 = 0.133... (13.333...%).\n\nFor part c, there are two ways one can get a red and a white marble: either the first marble is red and the second white, OR the first marble is white and the second red.  When there is more than one way to achieve the desired result, the probability of getting that result is given by the SUM of the probabilities of each of the different ways to achieve that result.  In this case, the probability of drawing red-white is 6/10 * 4/9 = 24/90, while the probability of drawing white-red is 4/10 * 6/90 = 24/90.  The total probability of getting a red and a white marble is then 24/90 + 24/90 = 48/90 = 0.53333....\n\nUsing the same reasoning, you should now be able to solve parts (a) and (b) of the second problem.  You should find for part (a) that the probability is equal to 0.42424242...., and that the probability for part (b) is 0.09090909....\n\nFor part (c), you would use a similar logic as used for part (c) of the first problem.  The probability of getting a matched pair (either both white or both blue) is given by the sum of the probabilities found in parts (a) and (b).  You should find that the probability is 17/33 = 0.51515151....\n\nFor the third problem, you need to know that the expected "payoff" of a game is given by the sum of the products of the prize value times the probability of winning that prize.  In this case, the probabilities of winning the various prizes are:\n\n$100 : 1/500\n$50 : 3/500\n$25 : 5/500\n\nSo the expected payoff (i.e., the "mean" or "average" winnings for a single ticket) is given by:\n\nP = $100*(1/500) + $50*(3/500) + $25*(5/500)\nP = $100/500 + $150/500 + $125/500\nP = $375/500\nP = $0.75
1102914
10\nThey are factors of 60 (in order from largest to smallest).
627773
1,2,5,10,20,50,100.\n\nSo that's 7.
171898
9!!\n\nsuppose he took out 4 from the green and for from the blue, noe he has to take at least one to ensure he has a blue one...
351129
Take the number of red blocks and divide by the total number of blocks.\nnumber of red = 18\nTotal number of blocks = 27+18+12+43 = 100\n18/100\n= 9/50
1065918
To get your answers, imagine what has to happen.  \n\na) First, you have to pick 1 of the 4 white balls out of the 12 balls there.\n\n2nd, you have to pick 1 of the 3 white balls left out of the 11 balls there.\n\nSo, in order to get a red and then a red, you need to get 4 out of 12 followed by 3 out of 11 or 4/12 times 3/11 or 12/132 = 1/22.\n\nb) First, how many ways can 5 people line up?  5!.  How many ways are correct (assuming they all have different names, only 1).  So, the prob is 1/5! or 1/120.
41863
To expand on the answer above, RJ has 10 choices for the first feature. For each of the 10 first shows, there are 9 remaining tapes from which to choose for the second feature. For each of those 90 combinations there are 8 tapes left from which to choose the third feature, so 720 total.\n\nAny time you have n objects, and you pick k objects out of that set, _and_ the order that you pick them matters, the number us n!/(n-k)!  In this case it's 10!/7! = 10x9x8=720.
347016
Normally 336, but anywhere between 300 and 500 is acceptable. Different balls have different numbers of dimples.
529774
okay so if you just subsitute the symbols w/the numbers then your answer would be 35 x 96 + 6640 = 10,000.
899962
Let x equal the cost of each ball.........\n\n5x </= to 20\n\nNow try your hand at solving it.
182612
As fewofmany mentioned, it's just three numbers smashed together, so you could conceivably write a function that compares them:\n\nColorX = 00FF00\nColorY = 00EE00\n\nThen, this means that ColorX has a Green component that is 255, and ColorY has a Green component that is 238. The two colors are identical in their Red and Blue components.\n\nIt's conceivable that you could then write a simple comparison check... Take the two colors' Red values & subtract them, and you'll end up with a measure of how similar their Red components are.  Do the same with Green, and you'll get a similarity measure for Green.  Finally, do the same for Blue, and you'll get a measure of the similarity of the Blue components.\n\nAdd up those "similarity" values, and you'll get some number.  In this case, the number is (0 - 0) + (255 - 238) + (0 - 0) = 17.  It seems that all you'd need to do is prevent that number from falling below a certain threshhold.\n\nI hope that makes sense...