A card is drawn at radom from a well-shuffled deck of 52 cards. What is the probability of drawing a face card
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Face: 12/52 = 3/13\n4: 4/52 = 1/13\n\nSo, add them up and get 4/13
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If they are all the same size die (same number of faces), take the number of faces of one die and raise it to the power of the number of dice. If the dice are of different sizes, multiply the number of their faces together.
933079
There are 2 black 2s in a deck out of 52. The odds of selecting a black 2 is 2/52. \nThe odds of NOT selecting a black 2 is therefore 50/52.
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Is the term you are looking for "probability"?\n\nI hope this helps.
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?
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
307770
Shuffling is pretty complex as you probably mean a riffle shuffle or something like that.  But let's assume you're shuffling by spreading out all the cards on the table, and randomly picking them.\n\nAssuming you want the cards in a specific order, then for each card, you'd have to randomly select the "right" card.  For the first card, you have 52 to choose from, and a 1/52 chance of getting it right.  For the second card, you now have 51 to choose from and thus a 1/51 chance of getting it right...on down to the last card.\n\nSo your odds are:\n\n1/52 * 1/51 * 1/50 * 1/49 * ... * 1/3 * 1/2 * 1/1\n\nor 1/52! or about 1/8*10^69.  To put that in perspective, lets say all 6 billion people on earth are shuffling one trillion times a second (man is that fast shuffling!) for a billion years, you're still not even in the ball park of being close (1/10^30) of having one person get it right.
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To have four sixes in a hand of five means there are only 48 cards left to choose from as the last card. So 48.
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In the experiment of rolling two dice\n\nTotal number of possible choices = 36 ==> n=36\n\nLet "A" be the event of rolling the dice such that the sum on the dice is 5\n\nFor Event "A"\n\nNumber of favorable choices = 4  ==> m = 4\n[(1,4), (2,3), (3,2), (4,1)]\n\nProbability of getting a sum of 5 on rolling the dice \n==> Probabilty of occurance of event "A" = m/n\n==> P(A) = 4/36\n==> P(A) = 1/9\n\nhttp://www.futureaccountant.com/\nhttp://www.schoolingkids.com/probability/
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In the experiment of rolling two dice\n\nTotal number of possible choices = 36 ==> n=36\n\nLet "A" be the event of rolling the dice such that the sum on the dice is 7\n\nFor Event "A"\n\nNumber of favorable choices = 6  ==> m = 6\n[(1,6), (2,5), (3,4), (4,3), (5,2), (6,1)]\n\nProbability of getting a sum of 7 on rolling the dice \n==> Probabilty of occurance of event "A" = m/n\n==> P(A) = 6/36\n==> P(A) = 1/6\n\nhttp://www.futureaccountant.com/\nhttp://www.schoolingkids.com/probability/
319861
a)  There are 26 black cards and out of the 26 only one is the ace of spades.\n1/26\n\nb)number of ways of picking any 6 rocks is\nC(5+7+8,6) = C(20,6) = 38760.  This will be the denominator for each part.\nnumber of ways of picking 2 silicates, one pyrite and 3 carbonates is\nC(5,2)*C(7,1)*C(8,3) = 10*7*56 = 3920\nProbability is 3920/38760 = .1011\nFor two silicates and 4 pyrites of carbonates:\nC(5,2)*C(15,4) = 10*1365 = 13650\nProbability = 13650/38760 = .3522\nFor 4 pyrites and 2 silicates or carbonates:\nC(7,4)*C(13,2) = 35*78 = 2730\n2730/38760 = .0704
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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.
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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.
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In solving problems in probability, you must determine first if it is binomial (with replacement) or hypergeometric (without replacement).\n\nsince you did not specify that the card you'd drawn was not replaced in the deck of cards, the distribution is hypergeometric.\n\nTo solve this problem, you can use any of the ff method.\n\nfirst method\n\nfor 2 spades:\nP(spade on the first draw)=13/52=1/4\nP(spade on the second draw, provided you picked a spade on the first draw)=12/51\n\nTherefore\nP(2 spades)=1/4 * 12/51 = 1/17 = 0.058823529...\n\nfor 2 aces:\n\nP(ace on the first draw)=4/52=1/13\nP(ace on the second draw, provided you picked an ace on the first draw)=3/51\n\nTherefore\nP(2 aces)=1/13 * 3/51 = 1/221 = 0.004524886...\n\n\nSECOND METHOD:\n\nUsing combination\n \nfor 2 spades;\nThere are 13C2 ways to pick two spades\nThere are 52C2 ways to pick two cards\n\nP(2 spades) = 13C2 / 52C2 = 1/17 = 0.058823529...\n\n\nfor 2 aces\n\nThere are 4C2 ways to pick two aces\nThere are 52C2 ways to pick two cards\n\nP(2 aces) = 4C2 / 52C2 = 1/221 = 0.004524886...\n\n\nNOTE: Both solutions I have shown is applicable only for hypergeometric. That is, there was no replacement done during the experiment.
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54 - if you count the jokers\n+12\n+190\n---------\n256\n----------\n\nor     52\n     + 13 - counting the number 12 as two seperate numbers\n     +190\n    ---------\n        2 5 5\n   ------------\n\nor 54+13+190= 257\n\nor Cards 52 + numbers on clock(counting each single digit) 15+ 190 = 257 with jokers in a pack of cards = 259\n\nso the answer is either 257, 259, 254, 256, 86 or 88\n\noooo! My head hurts!\n\n=kizzy=
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Depends on the game played.\n\nAnd if you mean pre or post draw, here is the predraw\nHE PROBABILITY OF BEING DEALT SPECIFIC HANDS BEFORE THE DRAW\n(52-card deck - without Joker)\n\nThe probability that you will make... Expressed in percent (%) is... The odds against it are... Number of possible combos\nStraight Flush 0.00* 64,973 to 1 40\nFour-of-a-kind 0.02 4,164 to 1 624\nFull House 0.14 693 to 1 3,744\nFlush 0.20 508 to 1 5,108\nStraight 0.39 254 to 1 10,200\nThree-of-a-kind 2.11 46.3 to 1 54,912\nTwo-Pair 4.75 20.0 to 1 123,552\nPair of A,K,Q,J\n   (Openers)  \n13.00  \n6.69 to 1  \n337,920\nShorts (10's - 2's) 29.25 2.42 to 1 760,320
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This is always so much fun, isn't it.  This is probably why the previous answerer is now looking for satisfying employment in the food-service sector.\n\nBy the way, you forgot to mention if they have to specifically be in that order or in any order, though.  It's a very different answer.\n\nOkay, what's the probability of drawing one green marble each time you draw, if you put the marbles back each time?  It's the partial over the sum, or in other words 4 out of 6, or 2/3.  Similarly the probability of drawing a red marble is 2 of 6, or 1/3.\n\nSo if you specifically want the order to be green, green, red, then you just multiply the probabilites each time.  2/3 * 2/3 * 1/3 is  4/27.\n\nHowever, suppose it can be in any order?  Here's where it gets tricky.  First we figure how many ways there are to draw two greens and a red -- we can draw (g,g,r) or (g,r,g) or (r,g,g) -- so three ways.  The probability is the same for each time so we add them together, essentially 3 * 4/27 is 12/27, or a little less than 50%.\n\nPart b) is harder, since the probabilities change with each draw.  The first time we draw it's a 2/3 chance of getting green, but after that there are only 3 green marbles out of 5, so it's now 3/5 instead of 4/6.  So if we want to draw two green and one red IN THAT ORDER it's 2/3 * 3/5 * 1/2 = 1/5.\n\nIf you don't care the order, then figure out each of the ways to draw 2 greens and one red (same as above) and add the probabilities from each time.  I'll leave it to you for the answer.\n\nActually there are very few interesting people in the world who like statistics.  But even the boring ones can usually find very good jobs.  Also it's been a while since my stats classes so if I'm wrong, please don't come after me with a baseball bat, thanks  :)