probabilty with permuations and combinations?
547623
the sample space will be 7776 elements (6*6*6*6*6 five dice each with 6 possibilities)\n\nthe elements range from 11111, 11112, 11113... to 66666\n\nthe six possibilities for Yahtzee are \n11111\n22222\n33333\n44444\n55555\n66666\n\nwhich is 6 out of 7776 \n\nor 1 out of 1296 = 1/1296
767048
Random chances are the same as probabilities. If you are so interested, you should look into Quantum Theory. It describes it in depth.
1009251
Ahh!  True, 6 consecutive flips of a coin in which each lands heads has a (1/2)^6 probability of happening.  But what are the odds that the seventh flip will be heads, too?
114803
If you have two events "A" and "B"\n\nThe probability that\n\na) Both would occur = P(A intersection B)\nb) Only one would occur = P (A intersection B') + P (B intersection A')\nc) At least one of them would occur = P (A union B)\nd) None of them would occur = P (A' intersection B')\n\nTry this site for a good explanation on probability\n\n\nhttp://www.futureaccountant.com/probability/\nhttp://www.schoolingkids.com/
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
106376
Also, if you agree that the formula for the number of combinations of n objects taken r at a time is\n\nn!/(r!(n-r)!)\n\nand that there is only one way to get a combination of n objects taking all n at a time, you'd get\n\nn!/(n!(n-n)!)  or n!/(n!0!), or canceling, 1/0! which must equal 1.\n\nSo 0! must equal 1.  More of a demonstration than a proof but it's easy to understand.
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
392328
You use combinations to determine how many different combinations you can make out of a number of different objects at a certain number. For example, You have 10 people to be chosen for a 3-person commitee. How many different 3-person commitees can be formed? This means that out of 10 people, you need to choose 3 different people, and how many different combinations out of that. Think of combinations this way, as it alters the definition of it: In fighting games like Mortal Kombat and Tekken, you use "combo's" to fight successfully. You can do multiple different combo's to defeat your opponent. Now, Combo's is just a short term for combinations.\n\nBasic probability fraction is the chances of something happening, not how many different combinations can happen.\n\nHope this helps.\n\nPS Penguins!!!
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
460997
If you want a number less than three,then the likely numbers are:1 and 2\nThe probability of throwing a 1 is 1/6\nThe probability of throwing a 2 is 1/6\nThese two events are mutually exclussive,therefore sum up their respective probabilities. \n1/6+1/6=1/3\nSo,the probability of obtaining a number less than 3 is 1/3
154434
The problem cannot be attended to without knowing the total number of possibilities.\n\nhow many sections are there on the spinnner..\n\n\nhttp://www.futureaccountant.com/probability/\nhttp://www.schoolingkids.com/
761738
Probability(1st person picks any bill)=8/8\n\nProbability(2nd person picks the same bill)=1/8\n\nProbability(3rd person picks the same bill)=1/8\n\nThe events are independant as the outcome of one event will not affect the outcome of another. Thus allowing multiplication of probabilities.\n\nProb(All 3 draw same bill) = 8/8*1/8*1/8\n\n=1/64
506414
I answered this one for you earlier...\n\nYou have X ~ Bin(6, 1/6)\n\nas you have 6 trials, with the proability of success (a four) being 1/6\n\nYou then simply need to stick this in the formula:\n\nP(X = x) = nCx × p^x × (1-p)^(n-x)\n\nIn this case you want the probability of 2 fours:\n\nP(X = 2) = 6C2 × (1/6)² × (5/6)^4 = 0.2009
855670
Boy, this is tricky.  I think the probability is (8! * 41c2 * 10c2)/49!, where 41c2 = 41!/39!2! and 10c2 = 10!/8!2!  In this number, 8! represents the correct picks, 41c2 the incorrect picks, and 10c2 the number of ways to order the picks.\n\nFor exactly seven correct numbers in 10 picks , only a minor change is needed.  It becomes (7! * 42c3 * 10c3)/49!.  Follow this pattern for 6 of 10, 5 of 10, etc
733920
a) There are 16 combinations and 4 of them include exactly 3 heads (HHHT, HHTH, HTHH, THHH).  So the probability is 4/16 or 1/4.\n\nb) This question results in the same probability (4/16 = 1/4), except it asks for it as 'odds' which is the probability of success divided by the probability of failure.  That is 1 to 3 odds.\n\nc) The odds against heads are just the reverse.  Probability of failure divided by probability of success.  That is 3 to 1 odds against.
342259
6! or 720 arrangements.\n\n2 * 4! or 48 with vowels at the ends, so it's a 1/15 chance.\n\nEdit: Yeah, I messed up.  Forgot the *2.
1028806
I'll assume you already know what permutations and combinations are and how they work.  Now lets say you're conducting an experiment where the result of the experiment can either be a success or a failure (this is called a Bernoulli trial in probability).  To ensure the accuracy of your results, you are going to repeat the exact same experiment three times, and you will look at the total number of successes in those three Bernoulli trials.  If we know the probability that each individual trial will be a success, we can calculate how many successes we can expect to get in three trials.  This is called a binomial distribution.  Let's look at all possible outcomes:\n\nYou could have all successes:\ns s s\nYou could have all failures:\nf f f\nThe first trail may be a success and the other two failures:\ns f f\nAnd so on... here's the list of all possible outcomes:\ns s s\ns f f\ns s f\ns f s\nf s s\nf s f\nf f s\nf f f\n\nLet's say you want to know what's the probability that you will get exactly two successes and one failure from your three trials.  The number of possible outcomes from this list that have exactly two successes in it is (3 choose 2)=3 out of the eight in the list.  Now if "p" is the probability that each individual trial is a success, and if "q" is the probability that it is a failure (meaning q = 1 - p), then the density function for this binomial distribution is:\nf(x) = (3 choose x) * p^x * q^(3-x)\nSince we want to know the probability of getting exactly two successes, we plug in two for x:\nf(2) = (3 choose 2) * p^2 * q^(3-2)\nf(2) = 3 * p^2 * q\nDepending on what p and q are, the probability of getting two successes in your three experiments can change.\n\nThe binomial distribution is one example of many discrete distributions that use combinations in its formulas.  Permutations have other applications, but this answer is getting long, so I'll leave it at that.