a coin is held above a horizontal table.what initial conditions will enure equal probability of heads&tails?
474729
That the coin is dropped directly on it's side, there is no wind, and both sides of the coin are perfectly equally weighted.\n\nHowever, due to chaos, just dropping the coin from any reasonable height will net an equal division over a large enough sampling of tests.
657146
Time for Pascal's Triangle.\n.......1\n......1 1\n.....1 2 1\n....1 3 3 1\n...1 4 6 4 1\n..1 5 10 10 5 1\n1 6 15 20 15 6 1\n\n1 chance in 64 of 6 Heads and 0 Tails\n6 chances in 64 of 5 Heads and 1 Tail\n15 chances in 64 of 4 Heads and 2 Tails\n20 chances in 64 of 3 Heads and 3 Tails\n15 chances in 64 of 2 Heads and 4 Tails\n6 chances in 64 of 1 Heads and 5 Tails\n1 chance in 64 of 0 Heads and 6 Tails\n\nChance of exactly 3 tails: 20 in 64\nChance of 3 or more tails: 42 in 64\nChance of more than 3 tails (4 or more) : 22 in 64\n\nAnyone who think the chance of 3 tails in 6 tosses is 50% is generalising wildly.\n\nThe chance of 1 tail in 1 toss is 50%\nThe chance of 1 head & 1 tail in 2 tosses is 2 chances in 4\nThe chance of 2 heads & 2 tails in 4 tosses is 6 chances in 16. (37.5%)\nThe chance of 3 heads & 3 tails in 6 tosses is 20 chances in 64. (31.25%)\nThe chance of 4 heads & 4 tails in 8 tosses is 70 chances in 256. (27.34%)\nThe chance of 5 heads & 5 tails in 10 tosses is 252 chances in 1024. (24.61%)\n\n(read the middle number in each row of Pascal's Triangle, and extend it downwards)\nThe bigger the number of tosses, the probability of tossing exactly half of the tosses as tails keeps on falling. This should not surprise us. There are so many more possible outcomes.
407787
You should get steady-state probabilities of 45%, 35%, and 20% for states 1, 2, and 3 respectively.\n\nYou need to solve the system:\n0.8x + 0.2y + 0.1z = x\n0.1x + 0.7y + 0.3z = y\n0.1x + 0.1y + 0.6z = z\n\nOr equivalently in standard form:\n\n-0.2x + 0.2y + 0.1z = 0\n 0.1x - 0.3y + 0.3z = 0\n 0.1x + 0.1y - 0.4z = 0\n\nYou can see that one of the equations is redundant (the first equation is the opposite of the sum of the other two equations, for example) so you can get rid of one. It doesn't matter which. This will always be the case. You can also pick an arbitrary value for any of the variables if you think it will help you out. In this case I'll let z = 10 and drop the first equation. This leaves us with:\n\n0.1x - 0.3y = -3\n0.1x + 0.1y = 4\n\nTake the difference to get 0.4y = 7, y = 17.5, x = 22.5.\n\nBut now x + y + z = 50 and it's supposed to equal 1. This is no problem, as we can just divide everything by 50 to get: x = 0.45, y = 0.35, and z = 0.2.\n\nIf it seems like I pulled a lot of things out of nowhere here, take a look at some theory regarding the homogeneous system Ax = 0 when A is singular, particularly the part where it mentions that if you've got a solution x to the homogenous system, then any scalar multiple of x is also a solution (this explains why we can just divide everything by 50 at the end and still have a solution to the original system). Consult a linear algebra book for more details.
963120
A.  The probability that it will BE solved is 1 - prob that it WON'T be solved.  That is 1.00 - (1.00 - .015 [Tom doesn't]) * (1.00 - 0.50 [Alice doesn't]).  \n\n = 1.00 - (.85) * (.50).  = 1.000 - .425 = .575\n\nB.  If Tom solves it then it is 150 / 575 = 6/23
297895
We have to insure that the coin is fairly weighted. (if one side is heavier then the other, it could land wrong.)  The horizontal surface must be completely flat.  The coin should be dropped on edge, in a repeatable fashion. (So that whatver drops it isn't angling it or affecting it.
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.
339708
Use a Normal table to find that the value of z so that Probability that Z < z = .90 is z = 1.2816.\n\nNow set z = 1.2816 = (x-100)/15, and solve for x:\n\nx = 1.2816*15 + 100 = 119.224
600596
There are 6 possible outcomes for the roll. There is only one outcome when the number is 4. Therfore the chances of getting a 4 are 1/6.
251771
Suppose P(k) is the probability that k of the N pots are occupied. Considering the process to begin on the second throw, there are N possible states (k=1,2,3...N). The initial probability distribution (after the first rock is thrown) is:\n\nP(1)=1, P(2)=P(3)=...=P(N)=0\n\nEach time a rock is thrown at random, the probability distribution undergoes a constant linear transformation (this is a Markov process):\n\nP --> P'(k) = (k/N)*P(k) + ((k-1)/N)*P(k-1)\n\nTo answer the question it's necessary to calculate P(N) after r transitions, equate it to .95 and solve for r. That can be done in principle for various specific values of N. For example for N=2, it's not to hard to see that:\n\nP(1) = (1/2)^r\nP(2) = 1 - (1/2)^r\n\nSolving (1/2)^r = 1-.95 = .05 gives about r = 4 (i.e. 5 actual rocks).\n\nI can also solve the case N=3 exactly. In this case, the transition P -> P' caused by each throw is: \n\nP'(1) = (1/3)P(1)\nP'(2) = (2/3)P(1) + (2/3)P(2)\nP'(3) = P(3) + (1/3)P(2)\n\nThis has solution (P(k) after r transitions):\n\nP(1) = (1/3)^r\nP(2) = (2^(r+1) - 2)/3^r\nP(3) = 1-P(1)-P(2)\n\nSo the exact answer in this case is the solution of:\n\nP(1) + P(2) = (2^(r+1)-1)/3^r = (2^(r+1)-1)/3^r = .05\n\n==> r is between 7 and 8 (i.e. between 8 and 9 actual throws).\n\nIf there's an expression for P(N) for general values of N, it's probably relatively complicated. \n\nHowever for large N it might be a pretty good approximation to treat the occupation states of each pot as independent (even though they're not really). Taking r as the actual number of rocks thrown, the probability that any given pot is occupied is exactly:\n\nP = 1 - ((N-1)/N)^r\n\nIf these events were independent, the probability that all pots are occupied would be:\n\nP^N = [1-((N-1)/N)^r]^N = .95\n\n==>\n\nr = ln(1- .95^(1/N))/(ln(N-1)-ln(N))\n\nFor N=3 this gives r = 10 which is a bit larger than the exact value, but it's probably more accurate for larger N.
90357
F- even and odds.  There are 3 even sides and 3 odd sides. So the likelihood is the same for both.
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
284480
Each coin, of course, will come up head or tail. There are eight possible distinct outcomes:\n\nHHH\nHHT\nHTH\nHTT\nTHH\nTHT\nTTH\nTTT\n\nAs you can see, three heads and three tails come up once each, two tails and a head come up three times, and two heads and a tail come up three times.
402919
ok i'm not doing your homeowkr for you, but i'll help.  you're unclear about some things:  number cube?  you mean a six-sided die?  in that case, one roll can give you: 1,2,3,4,5, or 6.  which ones are perfect squares?  let's see: 1^2=1, 2^2=4, 3^3=9>6 therefore just 1 and 4 are perfect squares.  so what's the probability or rolling a 1 or a 4 out of the possibilities 1,2,3,4,5, or 6?\n\nprobability=what you want/what you CAN get.
834182
sample space=8\nnumber of outcomes with at\nleast 2 heads=4\n\nP(at least 2 heads)=4/8=1/2\n\nodds:even money\n\ni hope that this helps
292028
For and event, odds are ususally expressed as the number of possible outcomes that qualify as a success compared to the total number of possible outcomes: 1 in 6, 3 in 10, etc.\n\nProbability is the same thing expressed as a fraction:\n1/6 = .66... 3/10 = .3\n\nWhen you select k objects from a set of n objects, the number of possible ways that you can do this is n!/(k!(n-k)!).\n\n\nIn all of your problems, your event is selecting a set of 3 objects from a set of 16. The number of possible outcomes of this is: \n16!/(3!13!) = 16x15x14/6 = 8x5x14 = 560.\n\nThe number of outcomes that are considered to be a success are:\n\n3 forks (pick 3 forks from the set of 7) = 7!/(3!4!) = 7x6x5/6 = 35\n\n2 forks 1 knife  (pick 2 forks from the set of sevan and 1 knife from the set o 5) = 7!/(2!5!) x 5 = 7x6/2 X 5 = 21x5 = 105\n\n3 spoons (Select 3 spoons from the set of 4) = 4!/(3!1!) = 4\n\n1 fork, 1 knife, 1 spoon  (Select 1 fork from the set of seven, 1 knife from the set of 5 and one spoon from the set of 4) = 7!/(1!6!) x 5!/(1!4!) x 4!/(1!3!) = 7 x 5 x 4 = 140\n\n1. P(3 forks) = 35/560\n2. P(2 forks, 1 Knife) = 105/560\n3. P (3 spoons) = 4/560\n4. P(1 fork, 1 knife, 1 spoon) = 140/560
1056545
Find the probability of each hand and add them up.  The hands would be: straight, flush, full house, 4 of a kind (straight flush would be included in straights AND flushes, don't count them twice).\n\n\nThere are 52C5 = 2598960 total hands, that's the bottom number.\n\nThere are 4x4x4x4x4 = 1024 ways to get an A-2-3-4-5 straight\nThere are the same for each of the ten possible straights up to 10-J-Q-K-A so 10x1024 = 10240 possible straights.  40 of these would be flushes too.\n\nDo this for the other hands then add them up. Put the sum over the total.
1179837
a) Try counting the possibilities.\n     There is only one way to roll a 12...(6, 6)\n     and there are 36 different possible rolls.\n     so the chances are 1/36\n\nb) There are 2 possibilities for this one (3, 6) and (6, 3)\n     so it's 2/36 or 1/18\n\nThe difference between the book the you is notation.\n1:35 means that for every one success there are 35 failures.\n1/36 means that there is one success for every 36 tries.