another hard math probablity?
813078
P(2 are women) = \nP(1st is woman) * P(2nd is woman) = \n6/10 * 5/9 = 30/90 = 0.333\n\nP(at least 1 is woman) =\n1 - P(both are men) =\n1 - P(1st is man)*P(2nd is man) = \n1 - (4/10)*(3/9) =\n1 - 12/90 =\n78/90 = 0.867\n\nI'm a little rusty on my "given that..." types of problems, and I'd hate to possibly steer you wrong on your remaining two questions.  But let me just say that I *THINK* the P of your 3rd question is 5/9; and I think the P of the 4th question is the same as the P of your 3rd question.\n\n~ ♥ ~
696155
Whats the probability that the problem is *not* solved? Tom doesn't solve it (0.75) and Alice doesn't (0.5) so its 0.75*0.5 = 0.375. So the probability it is solved is 1-0.375 = 0.625.\n\nNow, we are given that it is solved. What is the probability that Tom solves it? 0.25. \nSo:\nP(Tom | solved) = P(Tom and solved) / P(solved) = P(Tom) / P(solved) = 0.25/0.625 = 0.4.\n\n(edit - note that the reason the below answer is not quite correct is because it is possible for *both* of them to solve it - so the probability Tom solves it when solved + probability Alice solves it when solved should *not* equal 100% - it should be more.)
735064
That sounds familiar... Think I'm working on it now! But I don't remember q. It is n in my book. In which case, q (if I'm thinking right) is the number of samples. Then again what's up with the ^'s? If np was > 10, then it meant that the distribution was approx. normal. You might be ahead of me right now...
249369
Is the term you are looking for "probability"?\n\nI hope this helps.
872621
well in my skool we take it as {<and rule>} and {<or rule>example:there is 5 white marble and 5 black marbles in a bag...u pick 2 randomly without replacing them in the bag.....wat is the probability that u get 2 white marbles==>this is a combination 2 the way it is solved is multiplying the the 2 combined probabilities\nanswer:pr:(white AND white)=(5/10)*(4/9)\n\nnow the set is this way....wat is the probability that u get 2 marbles of different colors in the first 2 picks===>set rule\n\nanswer:there is 2 combinations in which u can get 2 different colors there are {white then black} OR {black then white} u find the 2 combinations then ADD them\npr(WandB OR BandW)={(5/10)*(5/9)}+{(5/10)*(5/9)}
152796
try this site...\n\nhttp://www.futureaccountant.com/probability/
660150
That would not be possible without the sample values or some other information.  You can make some statements about possible sample means.  For example, since the standard error of the sample mean would be 0.29/sqrt(25) = 0.058, then\n\nThe probability that the sample mean will be between 3.02-0.058 and 3.02+0.058 is 68%.\nThe probability that the sample mean will be between 3.02-0.116 and 3.02-0.116 is 95%.\n\nand so on.
227479
IMO, fb2kool is on to something.  Luck is used to describe a short series of events that appear to deviate from the average (when they simply contribute to the average).  The strange thing about probability is that the "average" never seems to happen.
1119536
1) If no one is injured, then no subs ever get to play.  The probability is (.9)^11 = .3138106 or about 31.38%.\n\n2) (.9)^9 * (.1)^2 = .003874205 or about 0.39%\n\n3) "At least" is best calculated as the complement of "no substitutes were needed."  In the first calculation we determined the probability of not needing any substitutes.  So the probability of needing at least one sub would be .6861894 or about 68.62%\n\n4) The trick here is that the two injured could both be players, or one could be a player and the other could be the sub.  (They could not both be subs, however.)  The calculation in #2 (above) is part of this--the probability that two of the players were injured.  For the other part we need to know the probability that one player is injured (while 10 are not) and that the sub is also injured.  .1 * (.9)^10 * .15 = .0052302.  So the overall result for either of these outcomes (add this result with the result from #2) is 0.00910438 or about 0.91%.
1082362
Sometimes - rarely - there are.  I have very occasionally come to the end of the puzzle with two numbers that could go anywhere in two squares, ie, 2 and 6 might be interchangeable.  Then I think I'm wrong, and will do it all again to be faced with the same outcome!
779526
How are you even sure that such a function exists? the numbers look very random.
693179
according to string-theory other universes do exist in what is called a "landscape of probabilities", but the probability of one being parallel as in just like ours, is extremely low.\n\nand 60 words are way too few to explain that, so try to be realistic next-time.
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.
955894
x1*p(x1) + x2*p(x2)\n\nx1 = 1 for  event dots> 3,\nx2 = 4 for event  dots<=3.  \nThese are equally probable because there are 6 dots on the die.  \n\n1*(1/2) + 4*(1/2)
448085
Okay...this may get a little confusing. :) \n\nThe first thing that you need to do is figure out whether the two event listed in each number are independent of each other or dependent.  In this case, the events are dependent.  For example, if you have these 8 marbles in a bag and you pick a green marble out of it...without replacement you would just set that green marble aside.  Now the chances of getting a green or white marble are different because you have changed the total number, does that make sense?\n\nNow, because the events (let's just work with number 1 for example but the same thing can be done to number 2) are dependent, you must use the multiplication rule.  In math terms this says that "The chance that two things both happen equals the chance that the first happens multiplied by the chance that the second happens pretending the first has happened".  In simpler terms this just means that you take the chance that the first will happen and multiply it by the chance that the second thing will happen.  \n\nSo in number 1, the first one you select is a green marble and the second one is a white marble.  The chances of getting the green marble are 3/8 because there are 3 green marbles and 8 marbles all together.  The chances of getting the white marble second are 5/7.  There are 5 white marbles in the bag, but because you have already taken the green marble out you have changed the total number of marbles - hence the 8 in the denominator changes to a 7...does that make sense?\n\nSo your answer to the first one would be (3/8 x 5/7).  I'm not for sure what the numerical answer to that would be, but you can do that calculation on your own! :) \n\nThe second problem can be done the same way, except instead of having a white marble second you would have another green marble.\n\nI really hope that this helps!! :) :) Good luck! :D
815085
Not every problem should be solved with an elephant gun.\nJust ask yourself a different question. What is the probability that x2 is closer to x1 than to x3. since the answer to that question is the same as the question you ask and the only 2 possibilities are certain\nand mutually exclusive, the answer should be the same, 1/2.\n\nWhat about if x2 is exactly in the middle you ask?\n\nIt's like asking what is the probability that a random number in an interval is the midpoint. 1/infinity=0
514908
This means "not very likely" (85% unlikely, 15% likely) if it's even possible.