A deck of cards has 4 yellow, 6 violet, and 5 white cards. you pick 4 cards from the deck. cards are not returned to the deck after they are picked. What is the probability that the firdt card id white, the second card is violet, the third card is violet, and the fourth card is yellow
120375
(5/15)*(6/14)*(5/13)*(4/12) = 600/32760 = 5/273. This is correct because the order you picked cards matters.
300955
OK.  Here are the possibilities.  It is important to keep the order of the dice in mind, so you can think of this as rolling a die 5 times.\n\nSSSSD\nSSSDS\nSSDSS\nSDSSS\nDSSSS\n\nNow each of these will have the same probability, so if I can find the probability of one of them, I can take that and multiply by 5 since there are five combinations.\n\nSSSSD can also occur in different ways.  If S=1, then D could be 2, 3, 4, 5, or 6.  So if S=1, then the probability of this combination is (1/6)^4(5/6).\n\nNow if S=2, then D can be 1, 3, 4, 5, or 6.  So again, (1/6)^4(5/6).\n\nSo hopefully you can see that there are 6 ways that you can do this.  So the probability of SSSSD is 6(1/6)^4(5/6).\n\nSo overall, the probability of getting four of a kind (without getting five of a kind) is\n5*6*(1/6)^4(5/6) = 150/6^5.\n\nSo it looks like your method does work.  Good job; this isn't an easy on.  There is always more than one way to look at probabilities.
224823
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.
31877
Here are some code snippets that might be useful.\nNote: I wrote these down, did _not_ compile them, or test them.\n\nThe deck of cards is a array of 52 unsigned integers,\nThey can take a value of 0-52.\n\n//Create an array to hold the deck:\n\nunsigned int Deck[52];\n\nfor(unsigned int i = 0; 1 < 52; i++)\n{\n   Deck[i] = i;\n}\n\n// Shuffle the deck\nsrand(); // Initialize the random number generator\n\nunsigned int temp;\n\n// The shuffle works this way. First round, every card gets a chance\n// to be in slot 51. Second round, each of the remaining cards gets a \n// chance to be in slot 50, third round etc.\n\n\n// Note: This method allows an array value to be swapped with itself, i.e. not moved, which is what you want to get all possible shufflings of the deck.\nfor(unsigned int j = 52; 0< j; j--)\n{\n   // Generate an index in the range 0-j\n   unsigned int k = rand( void ) % (j);\n\n   temp = Deck[k];\n   Deck[k] = Deck[j-1];\n   Deck[j-1] = temp;\n}\n\n///////////////////////////////////\n\nunsigned int GetFaceValueOfCard(unsigned int n)\n{\n  return Deck[n] % 13 + 1;    // Returns a value 1-13\n}\n\n// If you want to get fancy:\n\nchar GetSuitOfCard(unsigned int n)\n{\n   char Suit[4] = "HDCS";\n   return Suit[n/13];\n}
453882
The answer to this question is (b) 2/7.\n\nWhen removing two dogs there are four possible scenarios:\n\n(1) Remove poodle, then remove terrier.\n(2) Remove poodle, then remove another poodle.\n(3) Remove terrier, then remove another terrier.\n(4) Remove terrier, then remove poodle.\n\nIn each case, take the probability of the first even happening and multiply it by the probability of the second even happening. For instance:\n\nProbability that a poodle is first selected = 5(# of poodles originally) / 7(total # of dogs).\nProbability that a terrier is then selected = 2(# of terriers left) / 6(total # of dogs after a poodle is removed).\nMultiply 5/7 and 2/6 to get 10/42.\n\nRepeat this with each of the remaining three cases to get\n(1) 10/42\n(2) 20/42\n(3) 2/42\n(4) 10/42\n\n(Note that these add up to 1, which means that we have all of the possible cases)\n\nNow, just add the probabilities of the two cases with a terrier being selected second to get:\n\n10/42(first case) + 2/42(third case) = 12/42\n\nReduce 12/42 down to 2/7 to get your answer.\n\nI hope this helps, good luck with future probability problems!
340459
There are 10 different digits\nso p(x)=1/10=0.1\nwhere p(x) means the probability of number 'x' being selected.\neg probability of the number 3 being selected,\np(3) = 0.1
303664
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
89291
This cannot be ascertained without knowing how many are being selected in all...\n\n\nhttp://www.futureaccountant.com/probability/\nhttp://www.schoolingkids.com/
188192
There are 5 ways of getting a 6 (1-5, 2-4, 3-3, 4-2, 5-1) and two ways of getting 11 (6-5, 5-6), so the probability is 7/36.
647018
1) Let's first consider not the combination but the permutations without repetition (considering the order).\n\nSuppose to have 39 numbered cards in your hands and shuffle them well. \nIf you get one card and you drop it on the table you have 39 different possibilities, obviously.\n\nNow, from the remaining 38 cards in your hands, If you drop another card on the right side of the previous card, you'll have 38 different possibilities for just that second choose, but If you consider the order of both cards you'll have exactly 39*38 = 1482 different possibilities that the 2 card will be ordered and choosen in that way.\n\nIf you get a 3rd card from the remaining 37 in your hand you'll have on the table in the right side of previous exraction, a group of 3 ordered cards choosen by 39*38*37 = 54834 different ways will be on the table.\n\nRecursively go to choose until 6 ordered cards will be the table and you'll have 39*38*37*36*35*36 = 2349088560 possibile ways if you consider the order. \n\n2) Now let's see what happend when the order of the selection matters as you asked in your question. \nSo let's make this consideration:\n\na) If you have one card (A) you don't mind about ordering problems:  {A} = (A)\n\nb) if you have two cards (A, B) you have to consider the couple (A, B) = (B, A) so we write {A, B} are commutative;\n\nc) if you have three cards "A", "B" and "C" you can group (A, B) as one element {A, B} because order doesn't matter (point b).\n\nSo we can consider the 3 cards (A, B, C) as:\n ( {A, B}, C ) = ( C, {A, B} ) as we made in the point b.\nYou could also group (A, {B, C}) = ( {B, C}, A)  or {A, C} and then swapping with B.\n\nSo with 3 elements we have 3 different ways to create couple to swap so we can create 3*2 = 6 different combination.\n\nIf we have 4 cards we know it's possible to choose 4 different ways to create list of 3 cards. Let's stop one each time with your finger and you can see that you can group easily the remaining other 3's.\nSo the possible combination with 4 cards is 4 times the possible combination I have with 3 cards. So it's  4 * (3 * 2).\n\nWith 5 cards you'll have 5 times the number of combinations we found with 4 cards, so it's  5 * (4 * 3 * 2).\n\nFinally, with 6 cards in a table arrange in 6 * 5 * 4 * 3 * 2 = 720 different ordered ways.\n\nNow, your question was how to do if the order of selection matters and in the previous example we find that the number of possibilities we have when we select 6 cards ordered is this number 39*38*37*36*35*34 = 2349088560. And we know now that this number is 720 times greater than the possibilities wether the order of selection doesn't matter. (720 = 6 * 5 * 4 * 3 * 2).\n\nIn other words, 39 * 38 * 37 * 36 * 35 * 34 is the number of permutations without repetition of 39 elements grouped by a list of 6 elements considering the order.\n\nAnd now we know with 6 elements you can have 6*5*4*3*2 possible different positions of those elements. \n\nSo Just divide 39*38*37*36*35*34  by 6*5*4*3*2 and you'll have the result of the combinations.\n\nSo to see how many combination of 6 cards bye a set of 39 just divide 39*38*37*36*35*34 / 6*5*4*3*2 = 3262623.\n\nthe number 6*5*4*3*2 could be wrote "6!" using the the factorial function.\n\nThe factorial function of positive integer n, is defined as the product of first n numbers:\n\nn! = 1 * 2 * 3 * ... * n\n\nand is a convention that the product of no numbers at all is 1.\n\n0! = 1\n\n\nThe factorial is also defined recursively by the follow definition:\n\n(n+1)! = (n+1) * n!\n\n\nSo if you have n object and you want to know how many permutation (the order matters) without repetition (each object can be chosen only once) the objects makes with a number k of selection the formula is:\n \n  n! / (n-k)!\n\nso it's the product of k terms starting from n and descending until reached (n-k+1).\n\nn*(n-1)*(n-2)*.
95173
There are 8! different arrangements (permutations) of the 8 friends.\n\nThere are 4! different arrangements of the 4 boys.\n\nThere are 5 different locations where the leftmost boy can sit when the 4 boys sit together.\n\nTherefore, there are 5*4! (or 5!) arrangements of the boys where all the boys sit together.  But there are also 4! arrangements of the 4 girls in the remaining slots.  So there are 5!*4! arrangements where the 4 boys sit together.\n\nSince there are 8! arrangements overall, the probability of all boys sitting together is (5!*4!)/8!.\n\nSimilarly, if the boys sit to the left and the girls sit to the right, then the probability is (4!*4!)/8!.\n\nInterestingly, this last answer is also the same probability as boy-girl-boy-girl-boy-girl-boy-girl and also the probability of girl-boy-girl-boy-girl-boy-girl-boy.  Therefore, the probability of alternating seats is (2*4!*4!)/8!.\n\nRecall that n! (read n-factorial) is n*(n-1)*(n-2)*...*3*2*1.
114495
Assuming that all the gumballs are unique, then there are 8*5 = 40 different ways to draw two gumballs at random, one from each bag.\n\nJust to clarify, though - the concept illustrated here is NOT probability, but an illustration of the fundamental counting principle. Probability can be derived from this, but it's not the main thing illustrated here.
379121
1.Probability of getting a number from 1-4 on throwing the dice\n         = 4/6\n\n2.Probability of a correct throw = 1/4\n\n3.Total number of questions = 160\n\nYour Answer = (4/6 * 1/4) raised to the power 160
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.
178452
try this site for help on probability\n\nhttp://www.futureaccountant.com/probability/
192530
First choose which number it is.  There are 6 numbers, so there are C(6,1) = 6 choices.\nThe probability of that number is 1/6\nSince there are three\n(1/6)(1/6)(1/6) = 1/216\nNow multiply this times 6\n6/216 = 1/36
163702
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.