With 1 toss of a pair of dice what is the probability of rolling...?
545123
12 to 1  less than 2 .....12 to 5 for number over a 7
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The probability of getting heads on one toss is 1/2 (or 50%). To figure out the probability of getting heads on 5 consecutive tosses you just multiply the probability of a single toss 5 times:\n\n1/2 x 1/2 x 1/2 x 1/2 x 1/2 = 1/32\n\nThe probability of getting all tails is the same, for the same reasons: 1/32.\n\nTherefore, the probability of getting either all heads or all tails is simply 1/32 + 1/32 = 2/32 = 1/16
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In the experiment of tossing a coin ten times.\nTotal number of possible choices = 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 \n\n      = 1024 \n => n = 1024\n\nLet A be the event of getting number of heads is double the number of tails.\n\nFor event A\n------------\nNumber of favourable choices  =  Number of chances of getting heads is double the number of tailes.\n\n                  = No chances for occurance of event A\n                  = 0\n           =>  m  = 0\n\nProbablity of occurance of an event A =  Number of favourable choices / total number of possible choices\n     => P(A) =  m/n\n     => P(A) =  0/1024\n     => P(A) =  0\nTherefore the probability is zero.\n\n\n\nFor all about probabilty:\n\nhttp://www.futureaccountant.com
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In 36 possibles outcomes\nprobability of doubles\n1-1\n2-2\n3-3\n4-4\n5-5\n6-6\n6 in 36 or 1 in 6 or 16,666%\nOf the sum being 4\n1 +3\n2+2\n3+1\n3 in 36 or 1 in 12 or 8,333%\nIf it is combined probability you can sum the two\n16,66+8,33=25%
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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
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a) Both: .1(.02) = .002\n\nb) Exactly one:  .02(1-.1) = .018 + .1(1-.02) = .018 + .098 = .116\n\nc) (1-.02)(1-.1) = (.98)(.9) = .882\n\nThey add up to one.  I'm not a probability person, but I think this is correct.
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The answer is one chance in 2^5  or 1 in 32.\n\nNotice that the chances are the same as flipping five heads in a row.
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In the experiment of rolling two dice\n\nTotal number of possible choices = 36 ==> n=36\n\nLet "A" be the event of rolling the dice such that the sum on the dice is 7\n\nFor Event "A"\n\nNumber of favorable choices = 6  ==> m = 6\n[(1,6), (2,5), (3,4), (4,3), (5,2), (6,1)]\n\nProbability of getting a sum of 7 on rolling the dice \n==> Probabilty of occurance of event "A" = m/n\n==> P(A) = 6/36\n==> P(A) = 1/6\n\nhttp://www.futureaccountant.com/\nhttp://www.schoolingkids.com/probability/
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Probability of getting one head = 1/2\nProbability of getting two heads = (1/2) * 2 = 1/4\nYou have a 25% chance of getting a head two times...\nAnd 12.5% chance of getting a head 3 times...\nAnd 6.25% chance of getting a head 4 times...\n.\n.\n.\nAnd 0.09765625% chance of getting a head 10 times!!! :)
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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
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try\nhttp://www.peterwebb.co.uk/probability.htm
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Pick one knight anywhere.\nIf you pick either of the knights directly adjacent to him, you are done.  This happens 2/24 of the time.\n....OOXXOO...\n\nIf you pick a knight that is separated by exactly one seat (2/24 of the time), then there are 3 out of 23 seats that are adjacent.\n...OXOXO....\nIf you pick a knight that is more than one seat away (20/24), then there are 4 out of 23 seats that are adjacent.\n...OXO...OXO...\n\nSo the probability is:\n1/12 + 1/12 * 3/23 + 10/12 * 4/23\n\nAdding it up:\n23/276 + 3/276 + 40/276, or 66/276 = 33/138\n\nThis is approximately 23.91%
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I can't believe that this question is being asked in the sixth grade.  Maybe half of my college students could answer this.\n\nThe best way to answer this question is to make a table.  Make the rows the results of the first die and the columns the results of the second die.  Then put the product of the two roll inside of the table.\n            1   2   3   4   5   6\n          ------------------\n        1| 1   2   3   4   5   6\n        2| 2   4   6   8  10  12\n        3| 3   6   9  12  15  18\n        4| 4   8  12  16  20  24\n        5| 5  10  15  20  25  30\n        6| 6  12  18  24  30  36\n\n(Sorry I can't get the columns to even out!)\n\nNotice that there are 36 possible rolls of the two dice.  To get the two probabilities, just count the number of times you get the event in the table and divide by 36.\n\nProbability of an even product = 27/36\nProbability that the product is less than or equal to six = 14/36\n\nI see a lot of incorrect answers up there.  I teach statistics at the college level.  Trust me, this is right.
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In Theory, it is 1/6.\n\nIn Practice, the dice will have some bias.\n\nIf you observe the dice that has its numbers carved out, you will noticed that the number 1 is a bigger circle and the number 6 has smaller circles hopefully to remove the same amount of material from the dice so that the weight is evenly distributed and the center of gravity is maintained at the true center of the dice.\n\nFurthermore, the table on which the dice roll may be biased as well, uneven surface, minute sharp points, etc. Also, your hand has sweat and different amount of stickiness when you release the dice. The dice may also experience uneven wear and tear with usage. You can go on and probe all the "real" world imperfections and these may add up to skew the probablities of the dice and bias the results.\n\nSo how likely is likely?\n\nIn the real world, it is unlikely to be as likely as you would likely like it to be.
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Zero probability. Anything more than that is technically *possible*.
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Flipping a coin 4 times means that there can be 16 different posibilites of Heads and Tails occuring (2^4 = 16).\n\nThus by dividing 1 (max. probability) by 16 we get .0625. So each combonation has a probability of 6.25%. Now there is only one possibilty (TTTT) where there are no heads thus the possibility of no heads is simply .0625 or 6.25%\nWhere as one head has the following possibilites:\n\nTTTH\nTTHT\nTHTT\nHTTT\n\nThus multiplying .0625 by 4 we get .25 which is possibilty of only 1 head occuring in four flips.\n\nSo in conlusion:\n\nFor four flips of a coin:\n\nProbability of only one head occuring = .25 or 25%\nProbability of no head occuring       = .0625 or 6.25%
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If you have two events "A" and "B"\n\nThe probability that\n\na) Both would occur = P(A intersection B)\nb) Only one would occur = P (A intersection B') + P (B intersection A')\nc) At least one of them would occur = P (A union B)\nd) None of them would occur = P (A' intersection B')\n\nTry this site for a good explanation on probability\n\n\nhttp://www.futureaccountant.com/probability/\nhttp://www.schoolingkids.com/