Hi there,\n  In any random set of playing cards, what is the \n  probability of a king and a queen being next to each\n  other?  I get an answer of 8/13. Does anyone agree?\n                                       Cheers
77491
try this site\n\nhttp://www.futureaccountant.com/probability/
96640
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.
52279
Hearts is a game involving chance but also considerable skill. If someone were very proficient at keeping track of the cards that had been played, they could probably win 30+ times in a row. Because of this, I don't think it would be able to calculate a probability. But congrats, though, 13 times is tres awesome!
266527
I hope you mean 52 cards.\n\nAnyway, to figure out this problem, you would need to find the number of combinations of 5 taken from 52. The equation for that is:\n52!/(52-5)!5!\nThis is equal to 52!/47!5!\nWhich is equal to: (48x49x50x51x52)/(2x3x4x5)\nThere are a total of 2,598,960 poker hands. \n\nIt's best to find the number of ways to get three kings, then two queens, then to multiply the two. \nThere are four possible kings and combinations of three. \n4!/(4-3)!3! This equals 4. \n\n4!/(4-2)!2! This equals 6. \n6x4=24\nThere are 24 possibilities of three kings and two queens. \n\n(c) is a little harder to explain. For every card Ace thru King there are 12 other cards it could make a full house with. The king can make 24 with a queen, but it can also make 24 with ace, two, three, four etc. To find the total number one must multiply 24 by 12. This equals 288. Now we must multiply that number by the total number of different kinds of cards in the deck. That would be 13. 288x13 = 3,744.\n\nThere are 3,744 ways to make a full house. \n\nTo find the probability, one must use the answer from question (a) and put the answer to question (c) over it. \n\n3744/2598960 =\n\n6/4165.\n\nOr an approximately 0.144 % chance of being dealt a full house.
276543
1) the pdf is actually a binomial.\nIf you consider only 1 man, what is the probability that he will be paired to his wife out of the 10 women? \nIt is 1/10\n\nLet X be the number of paired couples, paired meaning they are paired to their own spouse.\nTherefore X~B(10,1/10)\n\nSo if we want to find the probability that there is only 1 pair, using the formula of a binomial dist,\n\nP(X=1)= [10 choose 1] (1/10) (9/10)^9\n= 0.3874\n--------------------\n\n2) To find the mean and variance, using the Binomial parameters formula, we compare what we got in part 1 with the general formula,\nX~B(n,p)\nX~B(10,1/10)\n\nSo our mean = (n) multiply (p)\n= 10 x (1/10)\n= 1\n\nVariance = (n) multiply (p) multiply (1-p)\n= 10 x (1/10) x (9/10)\n9/10\n----------------------\n3) P(X >= 3) = 1 - P(X <= 3) \n= 1 - [ P(X=0) + P(X=1) + P(X=2) + P(X=3) ]\nreferring to the above eg in part 1, its easy to derive\n= 1 - [ 0.0349 + 0.3874 + 0.1937 + 0.0574]\n= 0.3266
457875
Boy, this is tricky.  I think the probability is (8! * 41c2 * 10c2)/49!, where 41c2 = 41!/39!2! and 10c2 = 10!/8!2!  In this number, 8! represents the correct picks, 41c2 the incorrect picks, and 10c2 the number of ways to order the picks.\n\nFor exactly seven correct numbers in 10 picks , only a minor change is needed.  It becomes (7! * 42c3 * 10c3)/49!.  Follow this pattern for 6 of 10, 5 of 10, etc
526485
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%
522108
if you are lucky you will pick 8 balls of the same colour. \nand the question is how lucky are you??
77842
In the experiment of choosing a committee of 6 from a group of 8 women and 7 men,\n\nTotal no. of possible choices \n= no. of ways in which the 6 members of the committee can be chosen from the total 15\n==> n = C(15,6)\n==> n = 15*14*13*12*11*10/6*5*4*3*2*1\n==> n = 7*13*11*5\n==> n = 5005\n\nLet "A" be the event of choosing the committee with 3 women and 3 men\n\nFor Event "A"\n\n---------- Men----Women---Total\nAvailable---8-------7------15\nTo Choose---3-------3-------6\nChoices----C(8,3)-C(7,3)--C(15,6)\n\nNo. of favorable choices\n= (no. of ways in which the 3 men can be chosen from the total 8) * (no. of ways in which the 3 wommen can be chosen from the total 7) \n==> m(a) = C(8,3) * C(7,3)\n==> m(a) = 8*7*6/3*2*1 * 7*6*5/3*2*1\n==> m(a) = 8*7 * 7*5\n==> m(a) = 1960\n\nProbabilty of choosing the committe with 3 women and 3 men \n==> Probability of occurance of event "A" = m(a)/n\n==> P(A) = 1960/5005\n==> P(A) = 392/1001\n\n\nhttp://www.futureaccountant.com/probability/study-notes/\n\nhttp://www.schoolingkids.com/
516885
C is at the end, so its just the other 8 positions to play with.\n\nCount how many permutations of those 8 there are.\n\nThen count how many there are where A and B stand side by side.\n\nThen subtract those out.\n\nEdit:  Poster below me is way off.  without any restrictions, 9 men could be arranged in over 360000 ways.   Not 81.\n\nBasically, you can take 'C' out of it.  There are 8 spots left, and 8 men.  Spot 1 can hold any of the 8.  spot 2 any of the remaining 7 and so on, giving 8X7X6X5X4X3X2 which is 8! possibilities.\n\nThat's 40320.\n\nOf those 40320, some involve A and B being side by side.  How many?  Well, they could be in slot 1&2, 2& 3 .... up to 7&8.  And they could be A-B or B-A.  That's 14 possibles.  For each of those 14, the other 6 spots could be filled 6! ways.\n\nSo there are 14 X 6! = 14 X 720 = 10800 combos with C in the 9 spot and A & B side by side.\n\nTherefor, of the 40320 with C in the 9 spot, 40320-10800 = 30240 possible arrangements have C at the end, but A and B NOT side by side.
416044
Figure the probability of each outcome first:\nP(red marble) = 4/18 = 2/9\nP(green marble) = 3/18 = 1/6\nP(white marble) = 5/18\nP(black marble) = 6/18 = 1/3\n\nThe expected winnings would just be:\n2/9 * $5 + 1/6 * $4 + 5/18 * (-$3) + 6/18 * (-$2)\n\nThis works out to an expected winning of $0.277778.\n\nAn alternate way to figure this is:\n[(4 * $5) + (3 * $4) + (5 * -$3) + (6 * -$2) ] / 18\n= (32 - 27) / 18\n= $5/18 = $0.277778\n\nThe game is better than fair, since you don't have to pay anything but you are expected to win more than $0.  (Technically it is unfair since the expected return is not equivalent to what you pay, but I'm going to keep playing it because it is unfair to the house, not to me!)
285500
Your logic is flawed. Say you always pick card #2. The first time, you'll have 14.2857...% (that's the same as saying one seventh) of the chances of winning the car.\n\nThe second time, with a new set of cards, your chances that the card #2 is the winner are again 14.2857...%, and so will every new time, because your chances are always 1 out of 7. You are picking one card out of seven cards that you are given.\n\nYour reasoning would be right if you could be assured that the car card will always be a different one (that for example, it will be in the first case #4, then #6, and so on). In that case, after the 7th. guess (if not earlier), you'd get the car. But the way you explained the game, nothing prevents the car from being always for example at card #5. If that is the case, always trying card #2 will make you fail every time.
178452
try this site for help on probability\n\nhttp://www.futureaccountant.com/probability/
267609
Players - There are 22 players, each with a box numbered between 1 and 22.  Each day a player is chosen at random to play the game.  That person's space is filled the following day, and each person remains in the game as a box holder until they are selected to play the game themselves.\n\nThe Game - The player for that day does not know how much is in their box.  The board shows the amounts of money in play.  They are 1p, 10p, 50p, £1, £5, £10, £50, £100, £250, £500, £750, £1000, £3000, £5000, £10000, £15000, £20000, £35000, £50000, £75000, £100000, and £250000.\nThe player chooses boxes one at a time, and their values are eliminated from the game board to show the possible amounts available still to win.\nAfter 5 boxes, the phone will ring and the banker will offer an amount of money based on the possibilities still remaining.  Following this, the banker will make an offer after every 3 boxes have been chosen.\nAfter each offer, the player must decide whether to accept this offer, or eliminate more boxes in the hope that he will knock out the low amounts, and therefore generate a higher offer from the banker.  Ideally, he will have the £250000 in from of him, meaning it cannot be eliminated, and will generate a high offer to the end of the game.\nThe question "Deal Or No Deal" is asked each time, hence the title of the show.\nIf the player accepts the offer at any time, that is the amount they go home with.  The game is then played to it's conclusion to see how much they would have won if they had carried on to the end.  The offers continue to be made by the banker as if the player had not made a deal at an earlier time.
114576
This isn't probability - it's permutations.  Often times teachers will teach permutations during probability units because you use the one in the other.  \n\nAnyway, the best solution is to make an organized list.  For example, if you choose color one as the background, then how many stripe combos can you have?  This ends up being a simpler problem.  If you can keep making these problems simpler and simpler, get the answers then combine them all together, it will help you solve the hard ones.
568976
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?
58219
Ah, a variation on the Monty Hall problem:\n\nhttp://en.wikipedia.org/wiki/Monty_hall_problem\n\nIt is a probabilistic curiosity, this one. The highly unintuitive answer is that you actually **can** improve your chances by changing your mind.\n\nYour problem appears to only be different in that it starts with 5 boxes. The above link should give you sufficient information to solve the problem.