Math probabilty help?
492063
Count the total number of groups:\n9 choose 5 = 9*8*7*6/(4*3*2*1) = 9*2*7 = 126.\n\nb) Count the number of groups where 3 women where chosen and 2 men.\n(4 choose 3) * (5 choose 2) = 4 * 10 = 40.\n\nThe prob that a random committee has exactly 3 women = 40/126\n\nc) If there were no women then the number of ways this can happen is:\n(4 choose 0) * (5 choose 5).  As you could probably already see, there is only 1 way to have a committe of 5 men and 0 women.  The prob = 1/126.\n\nf) >=3 women is the same as saying 3 or 4 women as 4 women is the most you can have.\n\n(4 choose 3)*(5 choose 2) + (4 choose 4) * (5 choose 1)=\n4*10 + 1*5 = 45.  Prob = 45/126.
936495
This is the opposite of the probability of getting all no's.  The probability of getting all No's is (9/10)^7.  The probability of getting at least one yes is 1-(9/10)^7.\n approximately 0.522
628830
Total no of outcome : 2^11 = 2048 \n(A coin tossed 11 times)\nNo of favourable outcome : 11C4 = 330\n(Exactly 4 heads)\nSo probablility would be 330/2048 = 0.1611
898818
There are sixty-six possible outcomes in the first two scenarios, and fifty-five possible outcomes in the third (Mary does not win):\n\nWhat is the probability a married couple wins the prizes? \n9.09% (6/66 x 100)\n\nWhat is the probability one prize goes to a man and one to a woman? \n54.5% (36/66 x 100)\n\nIf Mary does not win a prize, then what is the probability that John will? \n18.2% (10/55 x 100)\n\n(or 0% if John never made it to the party!)
1113295
(1, 2, 3,4) or (-1, -2, -3, -4) or (1, 2, -3, -4) or (1, -2, -3, 4)\nYou also have\n(-1, -2, 3, 4), (-1, 2, -3, 4), (-1, 2, 3, -4), and (1, -2, 3, -4)\nSo there are 8 possibilities.\nEach having a probability of 1/8 or occurring.\nFor (1,2,3,4) and (-1,2,-3,4), the prob. of an even natural number is 2/4 = 1/2\nMultiply this by 1/8 to get the conditional prob.\n(1/2)(1/8) = 1/16\nFor each of the following: (1,2,-3,-4), (1,-2,-3,4), (-1, -2, 3, 4), and (-1, 2, 3, -4), the prob. of an even natural number is 1/4\n(1/4)(1/8) = 1/32\nFor (-1,-2,-3,-4) and (1,-2,3,-4), the prob. of an even natural number is 0.\n0(1/8) = 0\nAdd all these together.\n2(1/16) + 4(1/32) + 2(0)\n= 1/8 + 1/8\n= 2/8\n= 1/4\nThis doesn't match your solution, but you can see if this makes any sense to you.
672123
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.
249369
Is the term you are looking for "probability"?\n\nI hope this helps.
524692
Its the product of the probabilities of selecting each letter in order.\n\nProb(M) * Prob(a) * Prob(t) * Prob(h)\n2/11 * 2/11 * 2/11 * 1/11 = 8/14641\n\nIts really unlikely, about 5 in 10,000.
515436
plz complete the question. i like to solve the questions on probability, but yur question is incomplete.
476052
it's pretty complicated, and I'm too far removed from my math courses to keep it all straight.  But it varies based on what your point is, simply because some outcomes are more likely.\n\n4 has 3/36 odds of being the point\n5 has 4/36\n6 has 5/36\n8 has 5/36\n9 has 4/36\n10 has 3/36\n\nBut you can't roll 7.  So if we assume the point is 5, for each roll there is 32/36 odds that we don't roll 5, but 6/36 odds we roll a seven.  Reconcilling that involves equations that I've long since forgotten about.\n\nI think maybe [((8/9 - 1/6)^11)*(1/9)] would be the odds you'd roll 11 times without getting 7 or 5 and get a 5 on the 12th roll.  That'd be about .0214.  That you'd follow that up with a 7 or 11 would be .0214*(8/36), or .00535\n\nSite I link to be low discusses the outcomes a bit more formally than I'm cabaple of doing these days.
967346
This probability problem can be used using Binomial Distribution.  \n\nGiven a number of 'x' success, the probability in a binomial experiment is:\n\nP(x) = [n choose x]*p^x*q^(n-x)\n\nwhere:\n\nn choose x is a combination problem: n!/[(n-x)!x!]\np = probability of success\nq = probability of failure\nn = total selection\nx = successes\n\nIn this problem, we are looking for at least one case has a success (which is one of the highest 9 numbers).  In essence, we are solving for:\n\nP(x=1) + P(x=2) + P(x=3) + P(x=4) + P(X=5).  \n\nHowever it would be easier to find the probability of the failures and subtract from the total probability because we only have to calculate 1 probability.  This would be:\n\n1 - P(x=0) is what we are looking for.  Now all we need is to calculate the probabilty.\n\nFor P(x=0), we have the following:\n\nn = 5\nx = 0\np = (9/26) = probability of success\nq = (17/26) = probability of failure\n\nPlugging into P(x), we get:\n\nP(x=0) = (5!/5!)(9/26)^0(17/26)^5\n\nSolving, you get P(x=0) = 0.12\n\nTherefore, \n\n1 - P(x=0) is ------>\n\n1- 0.12 = 0.88\n\nBelow is a website devoted to Binomial Distribution exercises\n---------\n\nHope this helps
19818
try\nhttp://www.peterwebb.co.uk/probability.htm
885691
Part (a) of problem 1 is pretty easy.  There are 10 marbles total, and 6 of them are red.  On your first draw, you have a 6/10, or 0.6 (60%) probability of drawing a red sock.  In other words, there are 6 ways of drawing a red sock out of a total of 10 possibilities.\n\nFor part (b), you need to know that the probability of two independent events happening is given by the product of the probabilities of each of the individual events.  In this case, the two independent events are that the first marble drawn is red AND the that the second marble is also red.  We already calculated the probability for drawing a red marble on the first draw in part (a).  After the first draw, the box contains 9 total marbles, 5 of which are red, so the probability that the second marble is red is 5/9.  The probability that both are red is given by the product, 6/10 * 5/9 = 30/90 = 0.3 (30%).\n\nThe reasoning for part (b) is the same as for part (a).  The probability that the first marble drawn is white is 4/10, and the probability that the second marble drawn is also white is 3/9.  The probability that both are white is given by the product, 4/10 * 3/9 = 12/90 = 0.133... (13.333...%).\n\nFor part c, there are two ways one can get a red and a white marble: either the first marble is red and the second white, OR the first marble is white and the second red.  When there is more than one way to achieve the desired result, the probability of getting that result is given by the SUM of the probabilities of each of the different ways to achieve that result.  In this case, the probability of drawing red-white is 6/10 * 4/9 = 24/90, while the probability of drawing white-red is 4/10 * 6/90 = 24/90.  The total probability of getting a red and a white marble is then 24/90 + 24/90 = 48/90 = 0.53333....\n\nUsing the same reasoning, you should now be able to solve parts (a) and (b) of the second problem.  You should find for part (a) that the probability is equal to 0.42424242...., and that the probability for part (b) is 0.09090909....\n\nFor part (c), you would use a similar logic as used for part (c) of the first problem.  The probability of getting a matched pair (either both white or both blue) is given by the sum of the probabilities found in parts (a) and (b).  You should find that the probability is 17/33 = 0.51515151....\n\nFor the third problem, you need to know that the expected "payoff" of a game is given by the sum of the products of the prize value times the probability of winning that prize.  In this case, the probabilities of winning the various prizes are:\n\n$100 : 1/500\n$50 : 3/500\n$25 : 5/500\n\nSo the expected payoff (i.e., the "mean" or "average" winnings for a single ticket) is given by:\n\nP = $100*(1/500) + $50*(3/500) + $25*(5/500)\nP = $100/500 + $150/500 + $125/500\nP = $375/500\nP = $0.75
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.
624712
This problem is solvable.\nProbability of event A is written P(A).\n\nP(A)= Total # of outcomes that correspond to event A (divided by -or- over) Total # of outcomes in sample\n\nAnswer:\nP(A) = 12/20 = 3/5
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.
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