Everytime I open a box I have 5% chance of getting something, on average how many boxes would I have to open?
107013
Maybe there's a trick to this, but if the probability is\n5%, then if (for example) you opened 100 boxes, you would\nhave "gotten something" 5 times (ie. 5% of 100 = 5).\n\nThat seems like you'd get something, on average, once\nevery 20 boxes.
754620
you can only get 6 - 36.you can only get 6 one way, and 36 one way.  you get 7 six ways, and 35 six ways.  figure out all the ways to get each number, including 19, and add them all up.  the odds of getting 19 is the number of ways to get 19 over the total number of outcomes
552270
would calculate your chance of me calculating a statistics question for you at 1234.367:1
108367
if they are in the box they could fetch $20-30.If they are treasure hunts or redlines or prototypes,You will get 4 times that.
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?
244013
First find the number of ways to have 5 spades in your hand.\nEach hand consists of 13 cards.  \nThat means 5 will be spades, and 8 will not be spades.\nThere are 13 spades in a deck and 39 cards that are not spades.\nThe number of ways of choosing 5 spades from 13 is\nC(13,5)\nThe number of ways of choosing 8 cards that are not spades from 39 is \nC(39,8)\nHence, the number of ways of having 5 spades is \nC(13,5)*C(39,8) = 1287*61523748 = 79181063676\nWe divide this number by the total possible number of 13 card hands, which is C(52,13) because you are choosing 13 cards from 52\nC(52,13) = 635013559600\nThus, the answer is\nC(13,5)*C(39,8)/C(52,13)\n= 79181063676/635013559600\n= .124691925832\nso there is approximately a .125 chance of being dealt 5 spades
1177996
The probability of winning n times is the ratio of combinations of winners and losers among your tickets to the combinations of tickets drawn:\n(winners and losers of your 5 tickets) / (total drawings)\nC(5, n) * C(1995, 20 - n) / C(2000, 20), or\n[ 5! / (n! * (5 - n)!) ] * [1995! / ((20 - n)! * (1975 + n)!)] / [2000! / (20! * 1980!)]\n\nThe probability of winning 0 times is approximately 95.094%, so the probability of winning at least once is about 4.906%.\n\nThe probabilty of winning 1 time is approximately 4.812%.\nThe probabilty of winning 2 times is approximately 0.093%.\nThe probabilty of winning 3 times is approximately 0.001%.\nThe probability of winning 4 or 5 times is miniscule.\n\n[Edited below to add comment]\nThe misconception many people are having is that it's a straight 2000 tickets drawn each time.  In a drawing like this, once the first of the 20 winning tickets is picked, there are only 1999 tickets left from which to draw.  After the second winner is chosen, the remaining tickets become 1998.  You have to use combinations to get the correct probability, here.
908589
Well, unfortunately, in order to have a good chance of winning the big prize, you'll probably have to spend more than the prize itself.  \n\nTo calculate the odds of winning, we start by finding the odds of getting one number.  In your scenario, it seems that there are 80 numbers, of which you've picked 40.  Neither the actual numbers themselves, nor the order they are selected makes any difference, since, presumably, each number has an equal chance of being picked.  For the first number, our odds are 40/80, or 1/2, since we have picked 40 numbers out of 80 total.  Not too bad, huh?\n\nWell, here's where it gets rough in a hurry.  We now need to get the second number.  The odds of us having this number have changed slightly, since one number is now off the board, to 39/79 (we have 39 numbers left, the board has 79 left).  This number must then be multiplied to the previous number, as only half of the situations have us selecting the first number correctly.  This comes out to 0.2468, or a little less than 1/4.\n\nContinuing along, each successive number has slightly decreasing odds, and must also be multiplied by the earlier odds as well.  So the odds on the third number are 40/80 * 39/79 * 38/78.  And so on, until we've calculated the odds of picking the 20th number (21/61) and multiplied it to the other 19 values.\n\nThe final value is a miniscule .00000003899, or 1 in 25.7 million.  By purchasing the second ticket, your odds increase to 2 in 25.7 million, with three tickets, the odds are 3 in 25.7 million, etc.  So to even have half a chance of winning, you will need to have a staggering 12.8 million tickets.  This may seem ridiculously high, but it's how lotteries, kenos, and other similar games of chance are able to do so well.\n\nHope this helps.
1122155
not sure what box method is...the answer is\n(2x+5)*(x-3)
245215
There are five tosses with two choices each.\nEach toss is a 50% chance of coming up heads regardless of the results before it or after.\nThe probability of getting a certain number of heads would be calculated as such: number of heads/number of choices.\nWe are looking for two heads and there are ten choices (heads and tails of each coin)\nThe probability of it would be 2/10 or 20%\n\nIt is correct in that each toss is 50% chance, but in this case we are speaking of of the chances of a group of coins having a certain number turn up heads.  We are also speaking of the probability of this before any actions are taken.  The odds will change with each coin toss till the last one which will be 50%
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
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
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%
1120383
Fine.\n\nYou need the find the average and SD of the box.\n\nAverage = (0+0+1)/3 = 1/3\nSD = (1-0)sqrt{(1/3)(2/3)} = 0.471\n(using the shortcut SD = (big-small)*sqrt(frac of big * frac of small))\n\nThe expected number of 1's is the number of draws times average of the box, so 600(1/3) = 200.\n\nHowever, we would expect to be off of the expected by a standard error or so.  The standard error of the number of 1's is \nsqrt(number of draws)*sd of the box, so sqrt(600)*0.471 = 11.55.
24421
NONE.Well... how many did u get?
380874
Or 7:3 if the number of coins
329705
Five boxes full of 5 apples each-gives you 30 objects..the 25 apples and the 5 boxes