If situation A has 3 possible outcomes, and sitation B has 4 possible outcomes, how many total combinations can there be if there is at least 1 but no more than 2 outcomes of each situation occur?  Both situations are occuring simultaneously and one of each must occur, and one or two outcomes are possible at the same time.
132786
nCr means from n objects, choose r of them...this is known as a combination in statistics, and is given by the formula\n\nnCr = n!/(r!(n-r)!)\n\nwhere the ! means factorial, given by the product of all the numbers from 1 to that number.\n\nThis is constrasted with the permutation operation nPr.  In the permutation operation, the order that the objects are arranged matters so the formula is different.  We aren't worried about order of occurance in this problem so we use nCr.\n\nSo for situation A we can have 2 of them or 1 of them occuring from a total of 3 possibilities.  This means we have (3C2+3C1) possibilities.\n\nFor situation B we have the 4 objects to choose from and we need to pick either one or two of them, so we have:\n(4C2+4C1) possibilities.\n\nSince A and B are independent events, the total number of ways to have any of our possibilities is the product of the A and B possibilities\n\nTotal number = (3C2+3C1)*(4C2+4C1)\n\nYou can do the math.
367257
a) There are 16 combinations and 4 of them include exactly 3 heads (HHHT, HHTH, HTHH, THHH).  So the probability is 4/16 or 1/4.\n\nb) This question results in the same probability (4/16 = 1/4), except it asks for it as 'odds' which is the probability of success divided by the probability of failure.  That is 1 to 3 odds.\n\nc) The odds against heads are just the reverse.  Probability of failure divided by probability of success.  That is 3 to 1 odds against.
434991
In this problem we have a few combinations to calculate.  A 5-member community with more boys that girls can have the possible scenarios:\n\n5 boys, 0 girls\n4 boys, 1 girl\n3 boys, 2 girls\n\nNow we need to calculate the combinations of this problem.  A combination (n choose k) is defined by the following:\n\n[n choose k] = n!/((n-k)!k!) where:\n\nn = # of possible chocies\nk = # of successes\n\n----------\n\nFor 5 boys, 0 girls the following combination is:\n\n[7 choose 5], because we have 7 boys to choose from out of 5 possible slots.  This becomes:\n\n[7 choose 5] = 7!/(2!5!) = 7*6/2 = 21 ways\n\n----------\n\nFor 4 boys, 1 girl, we need to calculate 2 separate combinations.  For the boys, we have [7 choose 4] and for the girls we have [6 choose 1]\n\n[7 choose 4] = 7!/(4!3!) = 7*6*5/6 = 35 ways\n[6 choose 1] = 6!/(5!1!) = 6 ways\n\nSince the selection of boys and girls are indpendent of each other, we can multiply the separate combinations to get the total combinations.\n\n4 boys, 1 girl = 35*6 = 210 ways\n\n-------------------\n\nThe same reasoning can be made for 3 boys, 2 girls.  For the boys, we have [7 choose 3] and the girls, we have [6 choose 2]\n\n[7 choose 3] = 7!/(4!3!) = 7*6*5/6 = 35 ways\n[6 choose 2] = 6!/(4!2!) = 6*5/2 = 15 ways\n\nThus, 3 boys, 2 girls = 35*15 = 525 ways\n\n---------\n\nTherefore, the total # of possible cominations is: 21+210+525 = \n\n756 possible ways\n\n----------\n\nHope this helps
79509
In the experiment of testing for the number appearing on the spinner \n\nTotal number of possible choices\n= Number of ways in which a number can appear on the spinner \n= C(15,1)\n=15\n[Any one of the 15 numbers may appear]\n\nLet "A" be the event of the spinner landing on 7\n\nFor Event "A"\n\nNo. of favorable choices \n= No. of ways in which the spinner would stop at 7\n==> m(a) = 1\n\nProbability of the spinner landing on 7\n==> Probability of occurance of event "A" = m/n\n==> P(A) = 1/15\n=0.067 (approximately)\n\nhttp://www.futureaccountant.com/probability/study-notes/\n\nhttp://www.schoolingkids.com/
499352
I'll do some for you to show how they are done. In all cases probability P = #number of good throws/ # of all throws.\nAlso 1-P = #number of bad throws/ # of all throws\nOften you can arrive at a solution by considering the dices one by one.\n1) First dice can be everyting, second dice can be one of the five numbers not yet thrown, third dice can be one of the four numbers not yet thrown, etc. This leads to P = (6/6) * (5/6) * ... *(1/6) =720/46656 = 0.0154\n2) Divide the dice in three pairs, named A,B and C. The probability to throw a pair of two's with pair A is 1/36. Idem for B and C. But we have 6 posibilities to arrange the pairs ABC, ACB, ... And we have 6x5x4x3x2x1 possibilities to arrange the dice into the three pairs. Therefore P = 6 x 720 x (1/36)^3 = 0.0926\n3) First dice 1, second dice 1, third dice 1 gives a probability of (1/6)^3 = 1/216. But there are 20 possibilities to arrange the dice.\nP=20/216 = 0.0926
304487
A coin is flipped three times.\n\nA is the event that there is a head on the first flip.\nB is the event that there is a head on the second flip.\nC is the event that there is exactly one head in three flips.\n\nP(A) = 0.5\nP(B) = 0.5\nP(C) = 3/8 = 0.375 (If you flip a coin three times, there are 8 possible outcomes; of those, three result in exactly one head.  You can also use the binomial here.)\nP(A|C) = P(A and C)/P(C)\n= (1/8)/(3/8) (Because only HTT will allow A and C to both happen, which is one outcome out of eight.)\n= 1/3\nP(B|C) = 1/3 (Same reasoning as above.)\n\nSo A and B are independent, but are not independent of C.\n\nP(A and B|C) = P(A and B and C)/P(C)\n\nNow if A and B, happens, then the first two flips were heads, meaning that there are at least two heads.  But if C happens, there can only be one head, so\n\nP(A and B and C) = 0\n\nSo,\n\nP(A and B|C) = 0\n\nHowever P(A|C)*P(B|C) = (1/3)^2.\n\nSo I guess that the equality doesn't hold.  And that also means that there is no positive constant K that will make it hold since K would have to be zero in this case.
326188
The probability of A is 26%.  Probability of B is 42%.  Both A and B is 7%.  Probability of neither = 100% - [(26%+42%)-7%] = 39%.  Or in your format, 0.39.
390779
It depends if one person can be in a room or if all 4 people can be in the same room.  If only one person can be in a room at any given time, then it would be \n4*3*2*1=24 different ways\n\nIf everyone can go in the same room, then there are 4^4=256 different ways.
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.
666229
The Cube has 43,252,003,274,489,856,000 different possible configurations.\n\nI cheated and looked up the answer, but to figure it out you'll want to use combinatorics.  I'll have to work on that part of it...\n\nIf you're taking the cube apart and counting how many different ways you can put it back together, this is a relatively simple problem (although it doesn't match the above answer).\n\nThe center cubes don't move at all, so there is only one possible configuration of them.\n\nThere are 12 edge cubes, which each have 2 outside edges and therefore two orientations for each location.  The number of possible combinations of these edges is (2*12)*(2*11)*...*(2*2)*(2*1) = (2^12)*12!\n\nThere are 8 corner cubes, which each have 3 outside edges and therefore three orientations for each location.  The number of possible combinations of these corners is (3*8)*(3*7)*...*(3*2)*(3*1) = (3^8)*8!\n\nMultiply the number of possible combinations of edge cubes by the number of possible combinations of corner cubes and you end up with 519,024,039,293,878,300,000 combinations.  \n\nThis is different than the answer I looked up because there are possible combinations when you take the cube apart that cause it to no longer be solvable.  I can't think of an easy way to find the number of combinations that keep solvability.
126071
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
116373
Solution: 2 to the 5th power \nor written out it looks like this: \n2 X 2 X 2 X 2 X 2 = 32 outcomes\n\nBasically, you place what each position's outcomes can be as the base number. In this case it is 2 because every position can only be either a T or an F exclusively. There are 5 questions. That is why you raise it to the 5th power.\n\nIt is hard to show how to work this one with the The counting principle or the tree diagram method with the limitations of this text processor. I wish I had a whiteboard to show you. LOL\n\nSo that you can see it better, I have a link to a video that covers this topic. It is free to watch on the net - http://www.rogerbacon.org/~jgutzwiller/mathVideos/fundamental%20counting%20principle.asx\n\nOne method you can use is on the video at time mark 2:40. It is a tree diagram.\n\nAnother method you can use is the fundamental counting principle which is at time mark 5:50.
38610
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.
189143
3 - You're guaranteed either 2 of one color and 1 of the other or 3 of the same color.
133513
I'm using && for intersection and || for union.\n\n1. P(A && B) = P(A) + P(B) - P (A || B) = 1/4\nP(A|B) = P(A&&B)/P(B) = (1/4)/(5/8) = 2/5\nP(B|A) = P(B&&A)/P(A) = (1/4)/(3/8) = 2/3\n\n2. (i) 1 because if A is a subset, then whenever A occurs, B must occur\n(ii) 0 because if A happens then B can never happen
103962
In the experiment of choosing the 3 digits\n\nTotal no. of possible choices \n= (no.of ways in which the first digit can be chosen) * (no.of ways in which the second digit can be chosen) * (no.of ways in which the third digit can be chosen)\n==> n = 9 * 10 * 10\n==> n = 900\n\n[Since the first digit cannot be zero, there would be 9 possibilities for choosing it, 10 for second and third]\n\nLet "A" be the event that all the 3 digits are different and "B' the event that the 3rd digit is different from the first 2 digits.\n\nFor Event "A"\n\nNo.of favorable choices\n= (no.of ways in which the first digit can be chosen) * (no.of ways in which the second digit can be chosen) * (no.of ways in which the third digit can be chosen)\n= 9 * 9 * 8\n==> m(a) = 648\n\n[Since the three digits to be chosen should be different, there would be 9 possibilities for choosing the first digit (since 0 is excluded), 9 possibilities for the second (excluding the one chosen for the first and including 0) and 8 for the third (excluding the ones chosen for the first and the second and including 0)\n\nProbability that all the digits are different\n==> Probabilty of occurance of event "A" = m(a)/n\n==> P(A) = 648/900\n==> P(A) = 18/25\n\nFor Event "B"\n\nNo.of favorable choices\n= (no.of ways in which the first digit can be chosen) * (no.of ways in which the second digit can be chosen) * (no.of ways in which the third digit can be chosen)\n= 9 * 10 * 8\n==> m(a) = 720\n\n[Since the first two digits can be any of the ten and the last should be different from the first two, there would be 9 possibilities for choosing the first digit (since 0 is excluded), 10 possibilities for the second (any digit can be chosen including 0) and 8 for the third (excluding the ones chosen for the first and the second)\n\nProbability that the 3rd digit is different from the first two\n==> Probabilty of occurance of event "B" = m(b)/n\n==> P(A) = 720/900\n==> P(A) = 4/5\n\n\nhttp://www.futureaccountant.com/probability/study-notes/\n\nhttp://www.schoolingkids.com/
635566
The first number has to be (x+6) and the second number is (x).  There are four possible combinations: 6,0 (60); 7,1 (71); 8,2 (82); 9,3 (93).  60 plus 8 (68) does not equal 06, or 6.  71 plus 8 does not equal 17.  82 plus 8 does not equal 28.  93 plus 8 does not equal 39.\nUnfortunately, this problem is impossible.\nSorry I couldn't be of more help.