An honest coin is tossed ten times. The probability that the number of heads is double the number of tails is?
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
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.
1136735
P(Last is W) = P{(2nd is W and 1st is W) or (2nd is B and 1st is W)}\n= P(2nd is W and 1st is W) + P(2nd is B and 1st is W) (since these are mutually exclusive events)\n= P(2nd is W)P(1st is W|2nd is W) + P(2nd is B)P(1st is W|2nd is B) (both use the general multiplication rule)\n= (6/10)(7/11) + (4/10)(6/11)\n= (42+24)/110\n= 66/110\n= 0.6\n\nWhich is the probability of getting a white from either box, which I think makes sense, since both have the same proportion of whites.
821305
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
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.
664305
For that to happen, 1 must come on each dice.\n\nThus P(Sum is 2) = P(1 appears)*P(1 appears)\n\n=(1/6)^2 = 1/36
245215
There are five tosses with two choices each.\nEach toss is a 50% chance of coming up heads regardless of the results before it or after.\nThe probability of getting a certain number of heads would be calculated as such: number of heads/number of choices.\nWe are looking for two heads and there are ten choices (heads and tails of each coin)\nThe probability of it would be 2/10 or 20%\n\nIt is correct in that each toss is 50% chance, but in this case we are speaking of of the chances of a group of coins having a certain number turn up heads.  We are also speaking of the probability of this before any actions are taken.  The odds will change with each coin toss till the last one which will be 50%
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
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.
1022123
A bit is a one or a zero, so decide on a convention such as "Heads is 1 and tails is 0" then flip a coin.\n\nIf you need to produce a random bit in some other fashion, you will need to provide more details about your problem.
214891
If your probability of winning is .45, the probability of not winning is .55.  The probability of not winning in two plays is (.55)^2 = .3025 so the probability of winning at least once in two tries is .6975 - NOT 1.  It is not a guarantee.\n\nThe thing that gets you on the doubling business is that the casino has a heckuva lot more money than you.  It can last until you go broke.  If that isn't true, and you have as much as Bill Gates, you have no reason to gamble and you find other things to do with your money.
557990
This is a binomial problem.  The number of trials is 4 and the probability of success is 1/6.  No more than twice means either 0, 1, or 2 sixes.  So\n\n4!/(0!4!)(1/6)^0(5/6)^4 + 4!/(1!3!)(1/6)^1(5/6)^3 + 4!/(2!2!)(1/6)^2(5/6)^2\n\n= 0.482 + 0.386 + 0.116\n\n= 0.984
232542
I did this on my calculator because you have to do 75 calculations (which would take forever by hand).\nLess than 75 means it landed tails 0, 1, 2,..., or 74 times.  The probability of tails is .5 and the probability of not tails is also .5\nSay you want tails to come up k times where k = 0,1, 2,...,74\nThe probability of x = k is\nP(X=k) = C(n,k)(.5)^k(.5)^(n-k) Add the exponents on .5 to get\n= C(n,k)(.5)^n where n = 200 (because that is the amount of trials)\nk goes from 0 to 74, so we need to take the sum from 0 to 74\nThe sum (k=0 to k=74) of (C(200,k)(.5)^200\n=.000144637632\nI see no other way of doing this except with a calculator or computer.  (I'm sure your teacher does not expect you to do 75 calculations by hand and then add them all up.)
622584
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.
1186248
In this case 'N', which is the sample size, is 120.  The probability of an event occuring is equal to the number of times it occurred in the sample divided by N, so:\n\nprobability of X=0:  20/120 = 1/6 ~ 0.1667\nprobability of X=1:  65/120 = 13/24 ~ 0.5417\nprobability of X=2:  25/120 = 5/24 ~ 0.2083\nprobability of X=3:  10/120 = 1/12 ~ 0.0833\n\nAlso the average value of X would be 1.2082 using this probability distribution through an infinite number of line checks.
220111
Greetings:\n\nHaving selected one of the four, the probability that person #2 will have a different date of birth is 364/365 (if we seek no repeats, #2 can have any 364 of 365 possible b.days).  Similarly, having selected two, the probability that #3 will have a unique d.o.b. is 363/365 (any 363 of 365 possible dates okay).  Finally, for #4, the probability is 362/365.  Hence by the multiplication property, the probability that all four will have different b.days is given by, (364/365)(363/365)(362/365) = (364)(363)(362)/(365)^3 = 0.984 to three places.\n\nRegards,\n\nNikkor180
412043
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.