a) If a card drawn from a pack of cards is black,find the probability that the card drawn is ace of spade.\nb) A geology professor has five silicates ,seven pyrites and \neight carbonates  in a rock collection.He picks six rocks at random for a student to analyse.Find the probability that the professor picked~~~~~\n(i) two silicates and one pyrite~~~~~(ii) two silicates~~~~~(iii) four pyrites.
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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
398829
1) this operation is made up of two steps, that means the probablility of the whole operation = the multiplication of the probablility of every step\n a) probaility of a 1 from the first box = 1/6, the other box = 1/7 , so P(Both will be 1's) = 1/7 * 1/6\n\nb) p(Exactly one of the will be a 2) = 1/6 * 6/7 + 1/7 * 5/6 \n\nc) P(Both will be even) = 3/6 * 3/7 \nd) P( Exactly one of them will be even) = 3/6 *4/7 + 3/6 * 3/7\n\ne) P(At least one of them will be even) = 1- both of the odd = 1- (3/6 * 4/7)\n\n>> as for the sample space it would take me forever to write it down! sorry\n\n2) Not sure about this one\n3) a) 5%\n    b) 16% >> dont know wut details to include its just a law
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.
101591
Total no of items\n = 7 forks + 4 spoons + 5 knives \n = 16  \n\nIn the experiment of drawing 3 items from among 16\n\nTotal no. of possible choices \n= no. of ways in which the 3 items can be drawn from the total 16\n==> n = C(16,3)\n==> n = 16 * 15 * 14/3 * 2 * 1\n==> n = 560 \n[On the assumption that the drawing is being made without replacement]\n\nLet \n"A", "B", "C" and "D" be the events of the three items drawn being 3 forks, 2 forks and a knife, 3 spoons and 1 fork, 1 knife and 1 spoon respectively\n\nFor Event "A"\n\n              Forks      Knives    Spoons     Total\nAvailable       7           5        4          16\nTo draw         3           0        0          3\nChoices       C(7,3)      C(5,0)    C(4,0)     C(16,3)\n\nNo. of favorable choices\n= No. of ways in which the 3 forks can be drawn from the total 7\n==> m(a) = C(7,3)\n==> m(a) = 7 * 6 * 5/3 * 2 * 1\n==> m(a) = 35\n\nProbability that the three items drawn are forks = m(a)/n\n==> P(A) = 35/560\n==> P(A) = 1/16\n\n\nFor Event "D"\n\n              Forks      Knives    Spoons     Total\nAvailable       7           5        4          16\nTo draw         1           1        1          3\nChoices       C(7,1)      C(5,1)    C(4,1)     C(16,3)\n\nNo. of favorable choices\n= (No. of ways in which 1 fork can be drawn from the total 7)* (No. of ways in which 1 knife can be drawn from the total 5) * (No. of ways in which 1 spoon can be drawn from the total 4)\n==> m(d) = C(7,1) * C(5,1) * C(4,1)\n==> m(d) = 7 * 5 * 4\n==> m(d) = 140\n\nProbabilty for the 3 items drawn to be 1 fork, 1 kinfe and 1 spoon = m(d)/n\n==> P(D) = 140/560\n==> P(D) = 1/4\n\nYou should be able to solve the two parts in between for Events "B" and "C".\n\n\nhttp://www.futureaccountant.com/probability/study-notes/\n\nhttp://www.schoolingkids.com/probability/
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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
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i don't believe the person above me got the second part right.  he is right about the beginning - the first probability is 2/5:\n\npossible numbers to draw: 2 4 6 11 13 15\n\nso you can choose any of these six out of the fifteen given (6/15 = 2/5).  the next number you draw must be also a "winning" number.  BUT, you have to have an even number less than 7 AND an odd number greater than 10.  this gives us two scenarios:\n\nIF the first number picked was even and <7, then the possible numbers you can pick from are ONLY: 11 13 15, since picking another even number <7 will not meet your requirements.  this gives us a probablity of 3/14, since there are 3 "winning" numbers out of the 14 numbers left.\n\n\nIF the first number picked was odd and >10, then the possible numbers you can pick from are ONLY: 2 4 6, since picking another odd number >10 will not meet your requirements.  this gives us a probablity of 3/14, since there are 3 "winning" numbers out of the 14 numbers left.\n\nSO, finally, suppose you go with the first scenario.  the probability would be: (6/15)*(3/14) = 3/35  going with the second scenario will give you the same thing, 3/35.  thus adding these two, since you can have either scenario occur and still meet your requirements, gives you: 3/35+3/35 = 6/35.\n\nAnd that would be the odds: 6/35.
12357
P(queen)= 4/52 = 1/13\nP(red card)= 26/52 = 1/2\nP(black face card) = 6/52 = 3/26
399784
1) \n    a) The probability that the first one is not right is 7/8, the next is\n6/7 since there are 6 chances of getting the wrong one out of the remaining 7. The next is 5/6, then 4/5, 3/4,2/3,1/2. The probability of all these events happening is\n 7!/8! = 1/8\nMy colleagues who answered prior to my answer fail to\nrealize the difference between dependent and independent\ndistribution. In this case we have dependent distribution since each time a letter and envelope is matched we have 1 less\nof each to choose from.\n\n\nb) The probability that at least one envelope right is the additive\ninverse of the probabilty of no envelope getting it right based on\nall probabilities which 1. So 1 - 1/8 = 7/8.\n\n2)  The probability that no 2 have the same birthday is  computed like this. Based on a 365 day year just to make it reasonable. \n  First person - any day. \n Second - 364/365 'cuz 1 day is eliminated by the first one\nThird -363/365   'cuz 2 days are eliminated by the first two\n etc.\n The probability of no two the same is therefore\n (364)(363)(362)...(346)/365^19 = .588
636314
1)There are thirtysix possible combination of dice of which 10 are equal to or greater than 9. Therefore the probability is 10/36 or 27.8%.\n2) There is nine socks and only one with the correct number at that point. Therefore the odds are one in nine. \n3) x=-7/3\n4) x=-1\n5) x=-3
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x1*p(x1) + x2*p(x2)\n\nx1 = 1 for  event dots> 3,\nx2 = 4 for event  dots<=3.  \nThese are equally probable because there are 6 dots on the die.  \n\n1*(1/2) + 4*(1/2)
115728
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  :)
356939
Really Good Question.\n\nThe answer is 3 socks.\n\nExplanation : Taking the case of least probability, suppose the first sock is brown and the second is grey, then the third has to be either brown or grey as there are only two colour sock. So the purpose is solved by drawing 3 socks.
7321
try\nhttp://www.peterwebb.co.uk/probability.htm
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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
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)*.
441206
1. There are a total of 11 coins \nA. 2/11\nB. 0\nC. 6/11\n\n2. \nA. there are 2 possibilities (5,6) and (6,5) out of 36 poswsibilities (6*6).\n1/18\n\nB. 6 possibilities: (1,6) (2,5) (3,4) (4,3) (5,2) (6,1)\n6/36 = 1/6\n\nC. There are no possibilites, the Max sum of two numbers can be 12 (two sixes)\n0\n\n\nGood luck.
239953
Start with the first box. The probability of pulling a white card is 5/9 and a black card is 4/9.\n\nNow, if the first box resulted in a white (a 5/9 chance), then the probability of pulling a white card in the second box is 8/18. (Overall probability of these events is 5/9 * 8/18 (0.247))\n\nIf the first box resulted in a black (a 4/9 chance), then the probability of pulling a white card in the second box is 7/18. (Overall probaibily of these two events is 4/9 * 7/18 (0.173).\n\nEither of the two above events will result in a final white card (and are the only ways a final white card could happen). So simply add the two probabilities together...\n\n0.247 + 0.173 = 0.42 or 42%