The following stats questions is on Bayes and introduction to odds?
460739
Here are the formulas:\nIf P(E) = a/b, then the odds for E are a to (b-a)\nTo find the odds against E just reverse the order.\nOdds against E are (b-a) to a\n\nIf the odds of E are a to b, then P(E) = a/(a+b)\n\nFor example,\nif P(R) = 1/10, then the odds for R are\n1 to (10-1) or\n1 to 9.  \nThus, the odds against R are 9 to 1.\n\nAnother example,  say the odds of R are 5 to 3.  \nThe probability of R is 5/(5+3) = 5/8.\nThe probability of not R would then be \n1 - 5/8\n= 3/8
470321
P(A or B) = P(A) + P(B) - P(A and B)\n\nP(A) = probability of trying out for basketball\nP(B) = probability of trying out for baseball\n\nNow, P (A and B) = 0, since Jamie will not try out for basketball and baseball\n\nP(A or B) = 1 (or 100%), since Jamie will definitely try out for basketball or baseball\n\nSo, P(A) + P(B) = 100%\nP(B) is more than 42% or more (since you have to consider that Jamie will try out for other positions for baseball).\n\n100% - 42% = 58%\nP(A) has to be 58% or lesser. The only answer that is less than 58% is option A (40)
633099
The triangle shows the coefficients in the expansion of (a+b)^n, for n incrementing by 1 in successive rows of the triangle. \n\nStart top row with single 1\n\nNext row (a + b) records the probabilities when a coin is tossed once (1 chance in 2 of heads, 1 chance in 2 of tails).\n\nNext row (a^2 + 2ab + b^2) records the probabilities when a coin is tossed twice (1 chance in 4 of 2 heads, 2 chances in 4 of one head and one tail, 1 chance in 4 of 2 tails).\n\nNext row (a^3 + 3a^2b + 3ab^2 + b^3) records the probabilities when a coin is tossed three times (1 chance in 8 of 3 heads, 3 chances in 8 of two heads and one tail, 3 chances in 8 of one head and two tails, 1 chance in 8 of 3 tails). etc.\n\nThe mathematical properties of the triangle then strike you and provide an alternative way of constructing it.
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.
121605
I'm assuming that you mean introductory statistics.  This online textbook is very good, it makes good use of hyperlinks and also provides exercises and answers for each chapter.  It also has links to useful sites and books for the various topics.  \nhttp://davidmlane.com/hyperstat/\nHope this is helpful :-)
553417
if there are only 10 numbers to chose from, say 0-9 then the odds are 10*10*10, if only nine number say 1-9 then it is 9*9*9. that help ya?
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%.
469430
the probility of a particular event is basicly "the amount of favorable outcomes over the total amount of outcomes. for example, \n\nwhats the chances of trowing a 3 with a dice?\n\ntotal favorable outcomes= 1 (because the is only one 3 on a dice)\ntotal outcomes= 6 (because there a six outcomes posible with a dice)\n\ntherefore the probibility of trowing a 3 with a dice is:\n=1/6\n\nthere are other formulars (eg. the  binomial theorm) for more complicated probibilities but this is the basics.
249369
Is the term you are looking for "probability"?\n\nI hope this helps.
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.
242624
http://en.wikipedia.org/wiki/Mean_%28statistics%29\n\nThis is a link from Wikipedia. It shows lots of examples along with explanations.
158411
Start from www.statistics.com or search "statistics books" in Yahoo Answers. Greetings.
435037
try this site for help on probabiliity\n\nhttp://www.futureaccountant.com/probability/
1015803
The basic equation of probability is P(A)= Limit(NA/N) as N approaches infinity. Where \nA is an event\nNA is the number of occurrence of event A over a population sample N \nN is the to total number of samples (N does not have to infinite but sufficiently large)\n\nFor example if a pole is taken  of a population size of 1000 people by asking them the same question and 400 of them answered with “No” then P(“Yes”)=600/1000 and P(“No”)=400/1000 \n\nAs soon as you asked a question and included “A or B” statement you have opened … No that is okay, just more info.\n\nWhen A and B are two independent events then the probability that events A and B both occur is: P(A and B) = P(A) x P(B).  The outcome of one toss of a coin is independent from your next toss. \n\nWhen A and B are mutually exclusive, then the probability of A or B is simply: \nP(A or B) = P(A) + P(B).\nThe outcome of one toss of a coin will yeld either heads (H) or tail (T) since both H and T are mutually exclusive.\n\nI hope that helps
584200
1) the pdf is actually a binomial.\nIf you consider only 1 man, what is the probability that he will be paired to his wife out of the 10 women? \nIt is 1/10\n\nLet X be the number of paired couples, paired meaning they are paired to their own spouse.\nTherefore X~B(10,1/10)\n\nSo if we want to find the probability that there is only 1 pair, using the formula of a binomial dist,\n\nP(X=1)= [10 choose 1] (1/10) (9/10)^9\n= 0.3874\n--------------------\n\n2) To find the mean and variance, using the Binomial parameters formula, we compare what we got in part 1 with the general formula,\nX~B(n,p)\nX~B(10,1/10)\n\nSo our mean = (n) multiply (p)\n= 10 x (1/10)\n= 1\n\nVariance = (n) multiply (p) multiply (1-p)\n= 10 x (1/10) x (9/10)\n9/10\n----------------------\n3) P(X >= 3) = 1 - P(X <= 3) \n= 1 - [ P(X=0) + P(X=1) + P(X=2) + P(X=3) ]\nreferring to the above eg in part 1, its easy to derive\n= 1 - [ 0.0349 + 0.3874 + 0.1937 + 0.0574]\n= 0.3266
43772
Hi! I'm a statistician and I use http://www.statsoft.com as well as cheap textbooks from Half Priced Books in my area. Both seem to help. If you have a specific question, feel free to email me. Don't know how much good I'll be because I use fancy software for my analysis, but I'll try.
262867
Considering  possible events are\nChoosing heads \nChoosing tails \n\nprobability of chosing heads is 0.5\n\nConsidering possible events are\nChoosing heads \nChoosing tails \nChoosing Nothing\n\nprobability of chosing heads is 0.33\n\nConsidering possible events as only heads because next is tails\nprobability of chosing heads is 1.\n\nprobability is always a value betweem 0 and 1.