If a coin is flipped for times, whats the probability of no heads; one heads?
466671
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%
1076427
Probability of 10 heads, 1/2^10 = 1/1024\nProbability of 9 heads    10/2^10 = 10/1024\nProbability of 8 heads    45/2^10 = 45/1024\n\nProbablity of at least 8 heads = \nProbablity of 10 heads +  Probability of 9 heads + Probability of 8 heads  = 56/1024 = 7/128.
1092672
Its 1:220\n\n(4/52)(3/51) = 1/221 which is the probability\nThus the odds are 1/220 or 1:220
887044
to calculate just time the total of first choice and the total of second choice or until the total of n choice.\nor you can try this: if you toss n coins total probability is 2^n\nif you toss n dice total probability is 6^n\n\nor if you wanna know the probability just arrange it and try this\nex: the probability if you take two alfabets from A - I are\nAA,AB,AC,ETC\nBA,BB,BC,ETC\nETC\n\n: )
633099
The triangle shows the coefficients in the expansion of (a+b)^n, for n incrementing by 1 in successive rows of the triangle. \n\nStart top row with single 1\n\nNext row (a + b) records the probabilities when a coin is tossed once (1 chance in 2 of heads, 1 chance in 2 of tails).\n\nNext row (a^2 + 2ab + b^2) records the probabilities when a coin is tossed twice (1 chance in 4 of 2 heads, 2 chances in 4 of one head and one tail, 1 chance in 4 of 2 tails).\n\nNext row (a^3 + 3a^2b + 3ab^2 + b^3) records the probabilities when a coin is tossed three times (1 chance in 8 of 3 heads, 3 chances in 8 of two heads and one tail, 3 chances in 8 of one head and two tails, 1 chance in 8 of 3 tails). etc.\n\nThe mathematical properties of the triangle then strike you and provide an alternative way of constructing it.
152796
try this site...\n\nhttp://www.futureaccountant.com/probability/
540053
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
19818
try\nhttp://www.peterwebb.co.uk/probability.htm
1015803
The basic equation of probability is P(A)= Limit(NA/N) as N approaches infinity. Where \nA is an event\nNA is the number of occurrence of event A over a population sample N \nN is the to total number of samples (N does not have to infinite but sufficiently large)\n\nFor example if a pole is taken  of a population size of 1000 people by asking them the same question and 400 of them answered with “No” then P(“Yes”)=600/1000 and P(“No”)=400/1000 \n\nAs soon as you asked a question and included “A or B” statement you have opened … No that is okay, just more info.\n\nWhen A and B are two independent events then the probability that events A and B both occur is: P(A and B) = P(A) x P(B).  The outcome of one toss of a coin is independent from your next toss. \n\nWhen A and B are mutually exclusive, then the probability of A or B is simply: \nP(A or B) = P(A) + P(B).\nThe outcome of one toss of a coin will yeld either heads (H) or tail (T) since both H and T are mutually exclusive.\n\nI hope that helps
761738
Probability(1st person picks any bill)=8/8\n\nProbability(2nd person picks the same bill)=1/8\n\nProbability(3rd person picks the same bill)=1/8\n\nThe events are independant as the outcome of one event will not affect the outcome of another. Thus allowing multiplication of probabilities.\n\nProb(All 3 draw same bill) = 8/8*1/8*1/8\n\n=1/64
248227
50% or .5 or 1/2\nThere are two possibilities when you flip the coin, and there's only one way it will be tails.\nTherefore, the probability is 1/2.
817621
If you tossed a coin, you have an equal chance of it landing heads as you do tails.  If you tossed it multiple times and it came up tails more often, then while you have the same raw possibility of it coming up tails next time, the likelihood (assuming a balanced coin) is that you are about to face a time of heads coming up more frequently.\n\nStocks are not coins.  It isn't just chance that sets a price up or down on any given day.  Still, if the company has been on the top ten biggest gainers list for three days, I'd start "betting" down because a "correction" is coming.  That is the closest thing to the 'coin toss' kinds of probabilities as you will get.  If you are trying to tie "probability concepts" to "being a good stock market analyst", I can tell you right now, you won't be a good stock market analyst.  Sorry.  As much as we popularly talk about the stock market being a crap shoot, it isn't.
917829
When you are computing the probability of a given event such as 2 heads, 2 tails, or one of each, you are essentially counting the number of ways a given event can occur and dividing this number by the number of all possible outcomes.\n\nIf you toss 2 coins once ( or one coin twice) there are 4 possible outcomes:\n\nHH = both heads\nHT = 1st heads, 2nd tails\nTH = 1st tails, 2nd heads\nTT = both tails\n\nAssuming the coins are "fair" each outcome is equally likely.\n\nThere is one way in four that two heads (or 2 tails) can happen so the probability for each of these outcomes is 1/4\nBut there are two ways you can have one head and one tail and so the probability of this happening is 2/4 =1/2. The sum of all the probabilities of the distinct outcomes will always equal 1. \n1/4 +1/4 + 1/2 =1\n\nHow about if you have 3 tosses? The 8 possible outcomes are as follows:\n\nHHH \nHHT\nHTH\nTHH\nHTT\nTHT\nTTH\nTTT\nWe see there is only one way to have 3 heads (or 3 tails) and so the probability of 3 heads (or 3 tails) = 1/8\nHowever there are 3 ways you can have 2 heads and one tail, so the probability of this happening is 3/8. And there are also 3 ways you can have one head and 2 tails, so 3/8 is the probability for this too.
771356
you flip it whenever multiplying or dividing by a negative.  The first case would NOT flip it.  The second case would flip it.
479364
Yes i can say there are so many events that you can believe 50/50 chances.\n\nAs you said 50/50 is calculating for 100. means you have 2 chances.\n\n1. Toss the coin: to get head or tail you have 50/50 chance\n2. If a women is giving birth: the baby may be BOY/GIRL 50/50 Chance\n\n3. for this question : someone give the answer: which is correct/not correct   50/50 chance.\n........... \nyou can find somany events which are having 50/50 chances
262867
Considering  possible events are\nChoosing heads \nChoosing tails \n\nprobability of chosing heads is 0.5\n\nConsidering possible events are\nChoosing heads \nChoosing tails \nChoosing Nothing\n\nprobability of chosing heads is 0.33\n\nConsidering possible events as only heads because next is tails\nprobability of chosing heads is 1.\n\nprobability is always a value betweem 0 and 1.
698606
Hmm, interesting question, actually, and a practical application of math.  Hey, all you folks who ask what this stuff is good for, here is a real-life situation where you can use it!\n\nGetting to the problem.  Obviously, the contestant to go last spins again if his score is lower than the best score.  If tied, there is a spin-off where his winning chance is 50%; so, if there is a tie, he should respin if he has 45 or less (giving him at least 11 out of 20 good numbers), not spin if he has 55 or more (giving him at most 9 out of 20 good numbers - better to go to the spin-off, where he has a 50% chance), while 50 is a "push".   \n\nNow, for the contestant to go 2nd to last: First, let's assume the first player busts (of course, the second player respins if his score is lower than the first player), which essentially makes this a two-player game.  Based on what number the second player stops on, this is the winnning percentage I calculate for the last player:\n\n100 - 4.875%\n95 - 14.375%\n90 - 23.375%\n85 - 31.875%\n80 - 39.875%\n75 - 47.375%\n70 - 54.375%\n65 - 60.875%  \n60 - 66.875%\n55 - 72.375%\n50 - 77.375%\n\nSo, you may think the second player should respin whenever his odds are under 50%, but that is actually incorrect.  For example, if the second player has 70, giving him a 45.625% chance of winning, respinning would give him a 70% chance of going bust - he is better off stopping with the 45.625% chance.  To determine whether or not respinning improves the second player's chances, a second calculation is needed that considers each of the 20 scores after the second spin - some of them bust the player but the others improve his chance of winning.  Based on that, here is a new table, giving the last player's chance of winning if the second player respins with the score given:   \n\n95 - 95.24375%\n90 - 90.9625%\n85 - 87.13125%\n80 - 83.725%\n75 - 80.71875%\n70 - 78.0875%\n65 - 75.80625%  \n60 - 73.85%\n55 - 72.19375%\n50 - 70.3125%\n45 - 69.68125%\n\nSo, comparing the two tables, the critical points are 55 and 60.  If the second player has 60, he should stop - the last player's chance of winning improves if the second player spins.  At 55, however, the second player should spin, improving his odds by about 0.2%.\n\n\nMoving on to the three player game... now it gets really messy.  The player to go first needs a score that can beat the other two players.  Does he stop on 55 again, hoping the other two players both bust?  He'll need a better score than that to have a 50% chance of winning, but, as was noted in the two-player analysis, stopping with a 30% chance can be better than making a second spin.  As for how to do a complete analysis, uh, beats me.  This may require a computer to do all the math.\n\n* * * * * *\n\nThinking about the three-player game, maybe it isn't as messy as I initially thought - the only additional consideration for the first player, as opposed to the second player, is that either player will stop if the first player's score is beaten.  This can be calculated from the first table.  For this table, listed is the chance of the first player losing (i.e. either player beats him) if he stands on the score given:\n\n100 - 9.51% \n95 - 26.68%\n90 - 41.29%\n85 - 53.59%\n80 - 63.85%\n75 - 72.31%\n70 - 79.18%\n65 - 84.69%\n\nSimilar to the two-player game, I can now calculate the odds if the first player decides to respin.  Since I’m getting a little tired crunching these numbers, for this last table I’ll only list the critical ones:\n\n70 - 83.4%\n65 - 82.3%\n\nIn other words, the first contestant should respin on 65, stop on 70.\n\nIn conclusion, I think the optimum strategies are.....\n\nFirst player: spin on 65, stop on 70\n\nSecond player: spin on 55, stop on 60 (assuming his score at least matches the first player)