Probability of choosing 2 students...?
825072
x = prob of choosing a senior who has a sibling\ny = prob of choosing senior's sibling from junior class\nZ = prob of having chosen a sibling pair\n\nZ = x.y\n   = 60/800 .  1/1000\n   = .000075
264930
The answer to a) is exactly the same as the answer to b), since all children should be equal. The answer to a) & b) is 0.25*0.25=0.125\n\nc) The answer is 0.25*0.25*0.75*0.75*0.75 * 10 = 135/512 We multiply 0.25*0.25*0.75*0.75*0.75 to get the probability of first 2 being blood type O and the others not. We multiply by 10 because there are 10 possible combinations of this. (Such as the last 2 being blood type 0 and the first 3 not...etc)\n\nd) Exactly none: 0.75^5 = 243/1024\nExactly one: 0.25 * 0.75^4 * 5 = 405/1024\nExactly two: 135/512 as above\nExactly three: 0.25^3 * 0.75^2 * 10 = 45/512\nExactly four: 0.25^4 * 0.75 * 5 = 15/1024\nExactly five: 0.25^5 = 1/1024
67102
43 out of 400.\nThere is 120*120 pairs of minuites within which in which they can arrive resturent. And to be within 3 minuite, the diffrence of this of this  two pair have to be 3 or less. If the first one of this is between 4 to 117, there will be 7 possibility of  second one.\nAnd teher will be 6 for 3 and 118, 5 for 2 and 119,and 4 for 1 and 120.\nso total possibolity is 114*8+6*2+5*2+4*2 that is 828\nSo the probability is 828/14400=43/200.
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%
279356
This is always so much fun, isn't it.  This is probably why the previous answerer is now looking for satisfying employment in the food-service sector.\n\nBy the way, you forgot to mention if they have to specifically be in that order or in any order, though.  It's a very different answer.\n\nOkay, what's the probability of drawing one green marble each time you draw, if you put the marbles back each time?  It's the partial over the sum, or in other words 4 out of 6, or 2/3.  Similarly the probability of drawing a red marble is 2 of 6, or 1/3.\n\nSo if you specifically want the order to be green, green, red, then you just multiply the probabilites each time.  2/3 * 2/3 * 1/3 is  4/27.\n\nHowever, suppose it can be in any order?  Here's where it gets tricky.  First we figure how many ways there are to draw two greens and a red -- we can draw (g,g,r) or (g,r,g) or (r,g,g) -- so three ways.  The probability is the same for each time so we add them together, essentially 3 * 4/27 is 12/27, or a little less than 50%.\n\nPart b) is harder, since the probabilities change with each draw.  The first time we draw it's a 2/3 chance of getting green, but after that there are only 3 green marbles out of 5, so it's now 3/5 instead of 4/6.  So if we want to draw two green and one red IN THAT ORDER it's 2/3 * 3/5 * 1/2 = 1/5.\n\nIf you don't care the order, then figure out each of the ways to draw 2 greens and one red (same as above) and add the probabilities from each time.  I'll leave it to you for the answer.\n\nActually there are very few interesting people in the world who like statistics.  But even the boring ones can usually find very good jobs.  Also it's been a while since my stats classes so if I'm wrong, please don't come after me with a baseball bat, thanks  :)
284048
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 5\n\nFor Event "A"\n\nNumber of favorable choices = 4  ==> m = 4\n[(1,4), (2,3), (3,2), (4,1)]\n\nProbability of getting a sum of 5 on rolling the dice \n==> Probabilty of occurance of event "A" = m/n\n==> P(A) = 4/36\n==> P(A) = 1/9\n\nhttp://www.futureaccountant.com/\nhttp://www.schoolingkids.com/probability/
1161943
Let's deal with "b" first because it makes the general solution clear.\n\nFor an unbiased coin the probability of "Heads" (H) = probability of "Tails"(T) = 1/2 = 0.5\n\nIf n = 3, then you interested in the probabilty of two people having heads and one tails; or two people having tails and one heads.\n\nThe pobability of a single event where two people have heads and one has tails = 0.5 x 0.5 x 0.5= 0.125. But this can happen 3 ways: THH, HTH, HHT and therefore the probability that 2 people have heads and one has tails = 3 x 0.125 = 0.375. Of course the probability that two will have tails and one heads will also be 0.375. So the probability that a leader would have been chosen after one iteration (vote) = 2 x 0.375 = 0.75. (Note: if 3 coins are tossed there are only 4 possibilities: all heads, all tails, two tails and one head, or two heads and one tail.)\nThe probability that a leader is not chosen the first time = 1-0.75 = 0.25. And the probablty that a leader will not be chosen after two iterations is (0.25)^2 = 0.0625. And therefore the probabilty a leader will be chosen within two iterations = 1-0.0625= 0.9375 or nearly 94%. Now we'll deal with the general case.\n\na) For arbitrary n, and a single iteration, the probability of choosing a leader = 2n(0.5)^n. And the probability of not choosing a leader is 1-2n(0.5)^n. There is a difference between the probability of choosing a leader "in" the kth iteration and "by" the kth iteration. The probability that a leader will not have been chosen after k iterations is \n(1-2n(0.5)^n)^k. And the probabilty a leader will have been chosen is simply 1 minus this value.\n\nSample calculation: assume n = 20 and k=50\n\nProb. leader not chosen = (1-40(0.5)^20)^50 = 0.9981\n\nAnd the probability a leader will have been chosen = 1-0.9981 = 0.0019 which is less than 0.2 percent or 1 in 500. This shows that as n increases, the probability that one person will have a different result than everyone else becomes very small and enormous number of iterations would be required to have a reasonable chance of choosing a leader.
242094
In the experiment of testing for the number of students out of 37 who consider statistics exciting\n\nNo. of trials made \n==> No. of students tested = 37\n==>n = 37\n\nLet the event of a student tested being intersted in the subject of statistics be a success\n\nProbability of success in each trial\n= probability that a student tested is interested in the subject of statistics\n==> p = 0.75\n[Given that 75% of the population of all business statistics 1 students consider statistics exciting]\n\nTherefore, \nq = 1 - p ==> q = 1 - 0.75 ==> q = 0.25\n\nLet "x" indicate the number of successes in "n" trials\nsince the number of students who are interested in statistics can be either 0, 1, 2, .... 37\nThe values "x" can carry are 0, 1, 2, .... 37\n==> "X" is a discrete random variate with range = {0, 1, 2, .... 37}\n\nThe probability distribution of "x" is a binomial distribution..\n\n[Since there a number of trials all of which are identical and independent with only two possible outcomes in each trial either a success i.e. the student being interested or a failure i.e. the student not being interested, with the probability of success in each trial remaining the same all throught]\n\nThe probability function of a binomial distribution\n==> P(X=x) = nCx * p^x * q^(n-x)\n\nTherefore, for the above distribution\nP(X=x) = 37Cx * (0.75)^x * (0.25)^(37-x)\n==>P(X=x) = 37Cx * (0.75)^x * (0.25)^(37) * (0.25)^(-x)\n==>P(X=x) = 37Cx * (0.75)^x * (0.25)^(37) * 1/(0.25)^(x)\n==>P(X=x) = 37Cx * (0.75/0.25)^x * (0.25)^(37)\n==>P(X=x) = 37Cx * (3)^x * (0.25)^(37)\n\nTherefore, \nProbability of observiing 36 or more of the students who consider statistics to be an exciting subject\n==> P(X >= 36) = P(X=36) + P(X=37)\n\n\nP(X = 36) = C(37,36) * (3)^36 * (0.25)^(37)\nP(X = 37) = C(37,37) * (3)^37 * (0.25)^(37)\n\n\n\nhttp://www.futureaccountant.com/probability/study-notes/\n\nhttp://www.schoolingkids.com/
242656
This one is easy-assume a 50% chance each time, so its 1/2 X 1/2 X 1/2 X 1/2, so 1 chance in 16.  There is another way of figuring this, using combinations, like sn777 did, but sn777 made a logical error.  There are these possible combinations: 4 m, 3m 1F, 2m, 2F, 1m 3F, and 4F. So just looking at that, you would think its 25%. BUT, there is only one combination of 4M, while there are 4 combinations that make 3m 1F, and 6 combinations that make 2m and 2F, etc-so the total possible combinations of 4 children is 16, and 4M is only one of those.
430637
First choose which number it is.  There are 6 numbers, so there are C(6,1) = 6 choices.\nThe probability of that number is 1/6\nSince there are three\n(1/6)(1/6)(1/6) = 1/216\nNow multiply this times 6\n6/216 = 1/36
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.
435037
try this site for help on probabiliity\n\nhttp://www.futureaccountant.com/probability/
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
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.
138694
each child can be a male or a female \ntherefore the total no of events possible is 2^6=64\n\nthe question says that you should have atleast 3 girls \ni.e there can be three, four ,five or all six girls .\n\nthe number of ways of choosing 3,4,5,6 girls from 6 children is the sum\n6C3 + 6C4 + 6C5 + 6C6 (here C stands for combinations )\n\ni,e\n\n(6*5*4)/(3*2*1) + (6*5*4*3)/(4*3*2*1) + (6*5*4*3*2)/(5*4*3*2*1) + 1\nthis reduces to \n20+15+6+1\n=42\n\ntherefore the probability is \n42/64
777500
1) \n    a) The probability that the first one is not right is 7/8, the next is\n6/7 since there are 6 chances of getting the wrong one out of the remaining 7. The next is 5/6, then 4/5, 3/4,2/3,1/2. The probability of all these events happening is\n 7!/8! = 1/8\nMy colleagues who answered prior to my answer fail to\nrealize the difference between dependent and independent\ndistribution. In this case we have dependent distribution since each time a letter and envelope is matched we have 1 less\nof each to choose from.\n\n\nb) The probability that at least one envelope right is the additive\ninverse of the probabilty of no envelope getting it right based on\nall probabilities which 1. So 1 - 1/8 = 7/8.\n\n2)  The probability that no 2 have the same birthday is  computed like this. Based on a 365 day year just to make it reasonable. \n  First person - any day. \n Second - 364/365 'cuz 1 day is eliminated by the first one\nThird -363/365   'cuz 2 days are eliminated by the first two\n etc.\n The probability of no two the same is therefore\n (364)(363)(362)...(346)/365^19 = .588
631161
This is a binomial probability.\nP(success) = 1/5, P(failure) = 4/5.  There are 6 events here, so P(2 successs) = (1/5)(1/5)(4/5)(4/5)(4/5)(4/5).\nP(at least 2 successes) = P(2 successes) + P93 successes) + . . . . P(6 successes) or 1 - P(O successes) + P(1 success) + P(2 successes) which you already computed above.\nLet me know if you don't understand.