What is the probability that from Iraq's 55 Most Wanted Card Deck, you draw 14 cards and not one is Hussein.
435663
The first card you draw has a 54/55 chance it's not Saddam Hussein (I'm assuming he's the only Hussein in the deck). The second card has a 53/54 chance; the third 52/53 and so on.  When 14 of these are multiplied, everything cancels out to 41/55, or about 74.5%.
67102
43 out of 400.\nThere is 120*120 pairs of minuites within which in which they can arrive resturent. And to be within 3 minuite, the diffrence of this of this  two pair have to be 3 or less. If the first one of this is between 4 to 117, there will be 7 possibility of  second one.\nAnd teher will be 6 for 3 and 118, 5 for 2 and 119,and 4 for 1 and 120.\nso total possibolity is 114*8+6*2+5*2+4*2 that is 828\nSo the probability is 828/14400=43/200.
507905
if the 1st card was replaced: (1/13)(1/13) = 1/169\n\nif the 2nd card was not replaced: \n given that the first card is an ace: (1/13)(3/17)=3/91\n  \n given that the 1st card is not an ace: (1/13)(4/51)=4/663
32846
P(queen)= 4/52 = 1/13\nP(red card)= 26/52 = 1/2\nP(black face card) = 6/52 = 3/26
1113134
We'll need two things to calculate this: the number of ways of getting a hand with exactly one pair and the total number of hands.\n\n1) The total number of hands with one pair is of the form\naabcd.\nThe number of ways of choosing the rank of card "a" is 13.  And within that rank, there are 6 different ways to choose the 2 cards.\nFor choosing "b", we have 48 cards to choose from, basically those that are not of the same rank as "a".  For "c", there are 44 ways, and for "d" there are 40 ways.  \nHowever, we are multiple counting things here, because I could choose b = 5 of clubs, c = 3 of hearts, d = king of diamonds, but I could also have chosen b = 3 of hearts, c = king of diamonds, and d = 5 of clubs.  So to account for this, we have to divide by 3!.  This gives:\n6 * 13 * 48 * 44 * 40 /6\n= 1 098 240\n2) The total number of hands is simply \n52C5\n= 52*51*50*49*48/5!\n= 2 589 960.\n\nSo the probability of getting exactly one pair is\n1 098 240 / 2 589 960\n= 9152 / 21583\n\nThe odds are\n9152 : (21583-9152)\n9152:12431
630441
Let us make the following assumptions\nAn average person meets n donkeys in his/her lifetime\nAn average person buys m lottery tickets\n\nMy search for carnivorous donkey resulted in 143,000 hits on google.\nMy search for winning the lottery resulted in 14,900,000 hits on google.\nLets assume that this roughly the ratio of meeting carnivorous donkeys to winning the lottery.\n\nSo your chance of being eaten alive by a donkey as against winning a lottery is = (n*143,000)/(m*14,900,000)\n\nSo unless you meet 104 donkeys for every lottery ticket you buy, your chances of winning the lottery are possibly higher :-)\n\nI decided to solve this problem mathematically since this is a math forum. Any other place, I would say, you need to check with Shrek on how he won the princess (a lottery for him?) and avoided getting eaten by the Donkey!!
1061162
it is a conditional probablility problem.  if you select a defective on your first pick (3/12), there are only 11 left, 9 of which are not defective.  so the probablility of not picking a defective on your second pick given that you picked a defective the first tome is 9/11.
375114
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.
940139
26/52, 1/2 or 50%, half of a deck of cards are red, the other half black
985314
its 1/36 chance of any two particular numbers coming up on each roll of the dice.  (1/6 is for one die)
869566
12 in 51\n\nDraw one card any card.\nNow what is the chance that the second card is of the same suit.\nSay the first card is a heart.  There are 12 hearts left in a deck that now have 51 cards so 12/51.
492063
Count the total number of groups:\n9 choose 5 = 9*8*7*6/(4*3*2*1) = 9*2*7 = 126.\n\nb) Count the number of groups where 3 women where chosen and 2 men.\n(4 choose 3) * (5 choose 2) = 4 * 10 = 40.\n\nThe prob that a random committee has exactly 3 women = 40/126\n\nc) If there were no women then the number of ways this can happen is:\n(4 choose 0) * (5 choose 5).  As you could probably already see, there is only 1 way to have a committe of 5 men and 0 women.  The prob = 1/126.\n\nf) >=3 women is the same as saying 3 or 4 women as 4 women is the most you can have.\n\n(4 choose 3)*(5 choose 2) + (4 choose 4) * (5 choose 1)=\n4*10 + 1*5 = 45.  Prob = 45/126.
968185
First the probability that a family of 4 is all girls is:\n1/16 (and the chance that they are not is 15/16).\n\nRather than figuring the chance of at least one family having this arrangement, figure the probability that *none* of the families do.  Then subtract from 1.\n\nThe answer will be 1 - (15/16)^10\nor about 47.6%\n\nIt's not a simple multiplication by n... if it were, then with 16 families you should have a 100% chance.  And with 32 families you would have a 200% chance.  That logic would be correct if you couldn't repeat a combination, but you can.  Just because family A has four boys, for example, doesn't mean family B can't also have four boys.\n\nLabel the 10 families as A, B, C, D, ... , I, J.  The probability that A has four girls is 1/16.  The probability that B has four girls is 1/16, etc.\n\nThe chance that *none* of them have four girls is 15/16 x 15/16 x 15/16 x ... x 15/16.  This is (15/16)^10 or about 52.4%.  Therefore the chance that at least one of them has four girls is 100 - 52.4 or 47.6%
360679
1, 2, 3, 4 are not greater than 4.  There are 6 numbers on a die.  4/6 or 66.6% chance that the number is not greater than 4.
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!!
830442
52 Cards in a Playingcard-deck
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.