you are using a deck of cards without jokers, you draw two cards without replacement. find P(heart|queen)?
547199
I'm assuming that you want the probability of drawing a heart followed by a queen.\nP(heart)=(13/52)=(1/4)\n\nAssume that the 1st card wasn't a queen, so (4/51) \n1st card is a queen, so (3/51)\nP(queen) = (4/51) or (3/51)\n\nor = add\n\nso P(queen) = 4/51 + 3/51 = 7/51\n\nP(heart) and P(queen) = 1/4 and 7/51\n\nand = multiply\n\nso P(heart/queen) = 1/4 x 7/51 = 7/204\n\nHope this helps\n\nTake care
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.
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
1121165
There is definitely something wrong with the wording of the puzzle, because you can just turn over the [A] card. Anyways, I doubt your problem is that easy. I'm guess that you're talking about the reverse of both Aces.\n\nSince each card card will have only a number and a letter, there are only 4 possibilities:\n\n1)\nFront: [A] [K] [2] [7]\nBack: [2] [7] [A] [K]\n\n2)\nFront: [A] [K] [2] [7]\nBack: [7] [2] [A] [K]\n\n3)\nFront: [A] [K] [2] [7]\nBack: [2] [7] [K] [A]\n\n4)\nFront: [A] [K] [2] [7]\nBack: [7] [2] [K] [A]\n\nSince there are actually 2 Aces, you could play a trick question and flip over the other ace as well. So even if there is a 2 behind the [A], there might be another Ace behind the [2]. The easy way is to flip over the [A] and the [2].\n\nHowever, you can do it 3 other ways. By flipping over exactly 1 face card (A or K) and 1 numbered card (2 or 7), you can determine exactly what's behind all four cards.\n\nFor example, if you flip over the [7] and the [A]:\nIf a 7 is behind the [A], then the 2 is behind the [K], and vice versa. If the A is behind the [7], then the K is behind the [2], and vice versa.\n\nOnce the combination is found, you will know if either 2 is paired with either Ace. (If they are randomly sorted, the probability will be 3 out of 4).
629566
Was it the card game "pairs"?
339421
Board games and card games.
302246
if the order you draw them is important its a permutation.\n\nthe answer is basically; a 1 in 55 chance for the first card, a 1 in 54 chance for the second card (you've taken the first card out) and a 1 in 53 chance of getting the final card. all three events are independant of eachother. so the answer is 55 x 54 x 53 = 157410\nso 1 in 157410 chance.\n\nif the order you draw them is unimportant them its a combination.\nfor the first card there is a 3 in 55 chance\nthe second card has a 2 in 54 chance\nand the final card has a 1 in 53 chance\nthe answer is: (3/55)  x (2/54) x (1/53) = 1 in 8745
632203
a. first draw doesnt matter since it's replaced; you're back at 3 in 10 chance\n\nb. two ways this could happen: first, 3/10 chance first ball is red, leaving you 2/9 chance then of second ball being red on second draw.  second, 7/10 chance you draw a white ball, leaving you with 3/9 chance of second ball being red.  (3/10*2/9)+(7/10*3/9) which leaves you with 3/10 chance. again.\n\n\nEDIT:\n\nsurprises me how many people can't do math.\n\nfor A), the second draw is completely independent from the first.  Doesn't matter if I took a dump in the bowl in the first draw; as long i fished it out, you're left again with the same start on the second draw-3 reds and 7 whites.  your probability doesn't change because of what happened in the first draw as the conditions are RESET.\n\nfor B), let's run through allll possible scenarios\nNumber the Balls\n\n1 - red\n2 - red\n3 - red\n4 - white\n5 - white\n6 - white\n7 - white\n8 - white\n9 - white\n10- white\n\nPossible draws (X denotes red second ball outcome)\n1,2 - X\n1,3 - X\n1,4\n1,5\n1,6\n1,7\n1,8\n1,9\n1,10\n(2 fav outcomes if ball 1 is drawn first)\n\n2,1 - X\n2,3 - X\n2,4 \n...\n2,10 \n(2 fav outcomes if ball 1 is drawn second)\n\n3,1 - X\n3,2 - X\n3,4 \n...\n3,10 \n(2 fav outcomes if ball 1 is drawn second)\n\n4,1 - X\n4,2 - X\n4,3 - X\n4,5 \n...\n4,10\n(3 fav outcomes if ball 4 is drawn first)\n\n...\n\n10,1 - X\n10,2 - X\n10,3 - X\n...\n10,8\n10,9\n(3 fav outcomes if balls 5-10 are drawn first)\n\n\nthere are 90 outcomes\nof which there are 2+2+2+3+3+3+3+3+3+3 favorable outcomes = 27.\n\n27/90 = 3/10\n\n\nSecond edit: guy below, 2*3 + 7*3 = 6 + 21 = 27, not 28
302269
Bicycle is the brand name of the cards. There were 53 which means he was cheating and died for it. (Sorry to spoil your fun with the answer :)
642736
a)P(full house) = P(XXYYY or XYXYY or XYYXY or ...), where each combination keeps order in mind.\n\nnow\n\nP(XXYYY) = P(22333) + P(22444) + P(22666) + ...\nP(22333) = (4/44)(3/43)(4/42)(3/41)(2/40) , which you would find is the same as P(22444) and P(22666) and whatever combination you can think of.  There are 11*10 = 110 ways to select the two different types of cards(11 because 2, 3, 4, 6, 7, 8, 9, J, Q, K, and A), So\n\nP(XXYYY) = 110(4/44)(3/43)(4/42)(3/41)\n*(2/40)\n\nYou would see that P(XXYYY) = P(XYXYY) = P(XYYXY) =..., and there are 5C2 = 5!/(2!3!) = 10 ways to arrange 2 X's and 3 Y's, so\n\nP(Full house) = 1100(4/44)(3/43)(4/42)(3/41)\n*(2/40) = 165/67880 = 0.00243\n\nb) P(straight) = 0 because a straight has to have either a 5 or 10 in it.\n\nc) P(flush) = P(All H's) + P(All S's) + P(All C's) + P(All D's)\n\nP(All H's) = (11/44)(10/43)(9/42)(8/41)\n*(7/40)\n\nAnd you will see that P(All H's) = P(All S's) = P(All C's) = P(All D's), so\n\nP(flush) = 4(11/44)(10/43)(9/42)(8/41)\n*(7/40) = 3/1763 = 0.0017\n\nSorry about having to line break sometimes; Yahoo Answers apparently doesn't like when you have a whole bunch of numbers all in a row.\n\nedit: swalker5037, I'm trying to figure out why our answers differ slightly.  One thing I notice is that in a) you are computing P(PTTT|P), meaning the probability that the final four cards are PTTT given that the first is P.  Now in order to find P(PPTTT), you need to compute P(PTTT|P)P(P), which I'm not sure you did.  However if that were the case then your answer would be much more different from mine.  In c), we have the same reasoning, but we have slightly different answers.\n\nAnyway, close enough.
307982
i dont know any card tricks sorry
539344
I assume the question you are asking is how many possible five card hands contain at least one pair.\n\nThere are 13 face cards from which the pair can come.  Once you have chosen from among the 13, there are (4C2) ways to choose the two suits.  There are 52-4=48 remaining cards, and you need to choose 3, so there are (48C3) possibilities for the other three.\n\nThe answer is 13 x (4C2) x (48C3) possible five card hands containing at least one pair.
785346
Yes, 26 red and 26 black.\n\n4 suits: spades (black) clubs (black) hearts (red) and diamonds (red).\n\nThere are four each of Queens, Kings, Jacks, 4s and Aces (one of each suit e.g Queen of Spades, Queen of hearts etc.).\n\nThere are 13 cards of each suit.  Ace, 2, 3, 4,5,6,7,8,9,10, Jack, Queen, King.  \n\n(Usually 2 is counted as low and Ace is high... this is the case in most types of poker as you mentioned and many other games - there are however some exceptions and some games in which a player may choose high or low).
821305
a) effective or not - 2 choices - binomial\nb) smoke or note - 2 choices -binomial\nc) answers could be 1,2,3,4,5,6 . . .therefore NOT binomial
294632
oh let me see if i can remember. First you shuffle all the cards. Then you put cards into a pile. 2 players= 30 cards\n3-4 players= 25 cards, 5-6 players= 20 cards. The pile is then placed on the left side of the person. The cards are face down and one card is faced up right on the top of the pile. Then once everyone had a pile then deal out 5 cards to each player. The youngest starts first. All the extra cards go into a draw pile. You have 4 piles in the center of the table and each person has 4 discard piles. To start the piles in the center you either have to have a 1 or a Skip-bo card. Skip-bo card are wild card and can be used for any number. The cards in the center go up numericlly Ex. 1,2,skip-bo, 4 .... 12. Once a pile hits 12 then it is removes and shuffled in to the draw pile. A person turns ends when they can't play any more cards in there had and they discard to there discard piles. If a person runs out of cards before their turn is over then they are allowed to pick up 5 new cards. The point of the game is to remove all the cards from you pile that you were given at the beginning. The first person to play all of ther cards in the pile wins then game. Oh i forgot about the discard pile. You can play your cards in the discard pile. and once ypu play a card in the discard pile it cannot be moved to make room for other cards. Just play on top of the other cards that are there. And you can only play the top card on the discard pile.\n\nI hope that this isn't to confusing!!
30239
You can't just draw one? It's 10 boxes across the top and down the side, starting with 0 and going to 9 (and no, the corner doesn't share both sides). All you need are a pen and a ruler.
65434
The question is WRONG. If you have length =6 and width =7, P=26 and A=42.