What are your opinions on probability?
402601
Fascinating subject:  On the balance of probabilities, if the universe is infinite, someone identical with you, typed the same question, on an identical computer, on an identical forum and is being answered by someone identical with me.    Mind blowing if you think about it.........
153163
Chance .. the luck of the draw in life! We praise it when it goes our way and curse it when it doesn't .. some avoid risk and others gamble at every chance .. There is no difference between a fat chance and a slim chance .. Good luck in life!
1097980
0.8037 or 80.37%\n\nThis is why:\n\nP(E)=1-P(not E); so the probability of at least 2 people having the same number is 1- the probability that each person has a unique number.\n\nThe probability that each person has a unique number:\n\nFirst person has 100 choices out of 100 numbers, so 100/100.\nSecond person can't choose the first person's number, so only 99 choices out of 100, so 99/100;\nThird only 98/100;\n. . .\n18th person, has (100-17)/100=84/100\n\nTherefore the probability that none of these people have the same number is (100/100)•(99/100)•(98/100)•(97/100)•. . .•(84/100) ≈19.63%\n\nSo the probability that at least two people have the same number is 100%-19.63%=80.37%.
1167496
I find interesting the use of probability to explain god. I mean, they use well known (and limited) data to explain how unlikely (even if possible) it is for life to start as a mere accident. Even though, these people seem to forget to apply the same logic and known data to explain god, if the experiment is a fair experiment, that is, using known data in both experiments, the probability that a being like god actually exists is 0. ZERO probability.
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.
442652
In a way I could say, but I think a lot of these people got the right idea already.  I don't REALLY believe in luck.  I think that everything that just happens, happens.  Like when you play a card game you don't get lucky to win or don't even have to have any skill, it is just a matter of probability.  The ones we consider "lucky" just happen to be going in the right direction.  And card players we call "skillful" just know how to hit the odds just right.  It's all about the odds.  Like when someone randomly throws a basketball and makes it in the goal, we would say "That was just luck".  It is all a matter of probability and how we use it.
691596
themselves and chance
90357
F- even and odds.  There are 3 even sides and 3 odd sides. So the likelihood is the same for both.
1058790
The gambler's fallacy applies to independent random events, like coin tossing or die rolling. It's the belief that the occurrence of independent random events in the past can help you predict whether those events will occur more or less frequently than expected in the future. It's a fallacy. There's still only once chance in two that a coin will land heads-up no matter how many times it's landed tails-up recently.\n\nBlackjack is a random game, with hidden information, but which cards will be drawn in the future is influenced by which cards have been drawn in the past. In one deck, for instance, there are only four jacks, so if a jack has already been drawn, the chance of another jack appearing before the next shuffle has decreased.\n\nDon't count on being able to count cards in a casino, though. They shuffle together several decks and reshuffle drawn cards into the decks before the decks are exhausted.
627413
OK.  Here are the possibilities.  It is important to keep the order of the dice in mind, so you can think of this as rolling a die 5 times.\n\nSSSSD\nSSSDS\nSSDSS\nSDSSS\nDSSSS\n\nNow each of these will have the same probability, so if I can find the probability of one of them, I can take that and multiply by 5 since there are five combinations.\n\nSSSSD can also occur in different ways.  If S=1, then D could be 2, 3, 4, 5, or 6.  So if S=1, then the probability of this combination is (1/6)^4(5/6).\n\nNow if S=2, then D can be 1, 3, 4, 5, or 6.  So again, (1/6)^4(5/6).\n\nSo hopefully you can see that there are 6 ways that you can do this.  So the probability of SSSSD is 6(1/6)^4(5/6).\n\nSo overall, the probability of getting four of a kind (without getting five of a kind) is\n5*6*(1/6)^4(5/6) = 150/6^5.\n\nSo it looks like your method does work.  Good job; this isn't an easy on.  There is always more than one way to look at probabilities.
31132
A lot of people refuse to believe that statistics can apply to them.  If you focus on anecdotal evidence, you can ignore broader trends.  If you understand the ubiquity of probablity, you'd be less likely to engage in dangerous behaviors where the odds are agaist you (even though you might know some of the rare exceptions to the rule in your everyday life).
152796
try this site...\n\nhttp://www.futureaccountant.com/probability/
933392
Probability in statistics would agree with your claim to a point of certainty
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  :)
714109
When there are 80 numbers with equal chance the probability is 1 / 80 for each number.\nThe chance of 20 out of 40 numbers is 1 out of \n40 x 39 x 38 x... x 22 x 21 = 40! / 20! = 3 x 10^29\nSince you have all the 80 numbers, the chance is 4 x 10^27.\nThe very, very, very most probability is that you waist your money again\n\nTh
277670
probability is the # of favorable outcomes over the # of possible outcomes:\n\nprobability=\n# of favorable outcomes\n------------------------------\n# of possible outcomes\n\nfavorable outcomes would be the # of possible resultis in your favor\n\nexample:\ni have 5 cards- a joker, a jack, a queen, a king, and a ace. there is a 1 in 5 chance (1/5) that you will pull a jack...there is a 3 in 5 chance (3/5) that you will pull a royal card (jack, queen, king)\n\nin both examples, the first number (1 and 3) are favorable outcomes
540053
In the experiment of tossing a coin ten times.\nTotal number of possible choices = 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 \n\n      = 1024 \n => n = 1024\n\nLet A be the event of getting number of heads is double the number of tails.\n\nFor event A\n------------\nNumber of favourable choices  =  Number of chances of getting heads is double the number of tailes.\n\n                  = No chances for occurance of event A\n                  = 0\n           =>  m  = 0\n\nProbablity of occurance of an event A =  Number of favourable choices / total number of possible choices\n     => P(A) =  m/n\n     => P(A) =  0/1024\n     => P(A) =  0\nTherefore the probability is zero.\n\n\n\nFor all about probabilty:\n\nhttp://www.futureaccountant.com