The odds against R are O( R ) = P(R^C)/ P( R )/ = [1- P( R )]/ P( R ).\nThe odds in favour of R are 1/O( R ).\n(i)Calculate the odds against R if P( R ) = 1/10, 1/3, 1/2, 2/3, 9/10.\n(ii)What are the corresponding odds in favour? (we sometimes say 'odds on). \n(iii)For the following odds for an event R, find P ( R ).\nOdds = evens,  2 to 1 against,  3 to 2 on,  100 to 30 against, 100 to 1 on.
208179
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
497178
Well, unfortunately, in order to have a good chance of winning the big prize, you'll probably have to spend more than the prize itself.  \n\nTo calculate the odds of winning, we start by finding the odds of getting one number.  In your scenario, it seems that there are 80 numbers, of which you've picked 40.  Neither the actual numbers themselves, nor the order they are selected makes any difference, since, presumably, each number has an equal chance of being picked.  For the first number, our odds are 40/80, or 1/2, since we have picked 40 numbers out of 80 total.  Not too bad, huh?\n\nWell, here's where it gets rough in a hurry.  We now need to get the second number.  The odds of us having this number have changed slightly, since one number is now off the board, to 39/79 (we have 39 numbers left, the board has 79 left).  This number must then be multiplied to the previous number, as only half of the situations have us selecting the first number correctly.  This comes out to 0.2468, or a little less than 1/4.\n\nContinuing along, each successive number has slightly decreasing odds, and must also be multiplied by the earlier odds as well.  So the odds on the third number are 40/80 * 39/79 * 38/78.  And so on, until we've calculated the odds of picking the 20th number (21/61) and multiplied it to the other 19 values.\n\nThe final value is a miniscule .00000003899, or 1 in 25.7 million.  By purchasing the second ticket, your odds increase to 2 in 25.7 million, with three tickets, the odds are 3 in 25.7 million, etc.  So to even have half a chance of winning, you will need to have a staggering 12.8 million tickets.  This may seem ridiculously high, but it's how lotteries, kenos, and other similar games of chance are able to do so well.\n\nHope this helps.
363434
1) AA*aa = 100% Aa\n\n2 & 3) Aa*Aa = 50% Aa, 25% aa (also 25% AA)\n\nFor 4-6, figure the probability of the A gene result, then multiply that by the probability of the B gene result.\n\n4) Aa*Aa = 50% Aa (.5); Bb*BB = 50% BB (.5)\n.5*.5 = .25 = 25% AaBB\n\n5) 50% Aa; 50% Bb = 25% AaBb\n\n6) 25% AA; 50% BB = .25*.5 = .125 = 12.5%
388050
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
194660
ok well. when youre doing ANDs (as in whats the chance of him going AND gettin his autograph), its just multiplying the probabilities.\n\nso 80% * x% = 5.6%\n\nso the x% would be 5.6% / 80% = 0.07 = 7%. \n\nso there is a 7% chance of him getting his favourite players autograph.\n\nhope this helped. let me know if you need more help
299433
This problem is solvable.\nProbability of event A is written P(A).\n\nP(A)= Total # of outcomes that correspond to event A (divided by -or- over) Total # of outcomes in sample\n\nAnswer:\nP(A) = 12/20 = 3/5
239953
Start with the first box. The probability of pulling a white card is 5/9 and a black card is 4/9.\n\nNow, if the first box resulted in a white (a 5/9 chance), then the probability of pulling a white card in the second box is 8/18. (Overall probability of these events is 5/9 * 8/18 (0.247))\n\nIf the first box resulted in a black (a 4/9 chance), then the probability of pulling a white card in the second box is 7/18. (Overall probaibily of these two events is 4/9 * 7/18 (0.173).\n\nEither of the two above events will result in a final white card (and are the only ways a final white card could happen). So simply add the two probabilities together...\n\n0.247 + 0.173 = 0.42 or 42%
77491
try this site\n\nhttp://www.futureaccountant.com/probability/
342063
Hmm, interesting question, actually, and a practical application of math.  Hey, all you folks who ask what this stuff is good for, here is a real-life situation where you can use it!\n\nGetting to the problem.  Obviously, the contestant to go last spins again if his score is lower than the best score.  If tied, there is a spin-off where his winning chance is 50%; so, if there is a tie, he should respin if he has 45 or less (giving him at least 11 out of 20 good numbers), not spin if he has 55 or more (giving him at most 9 out of 20 good numbers - better to go to the spin-off, where he has a 50% chance), while 50 is a "push".   \n\nNow, for the contestant to go 2nd to last: First, let's assume the first player busts (of course, the second player respins if his score is lower than the first player), which essentially makes this a two-player game.  Based on what number the second player stops on, this is the winnning percentage I calculate for the last player:\n\n100 - 4.875%\n95 - 14.375%\n90 - 23.375%\n85 - 31.875%\n80 - 39.875%\n75 - 47.375%\n70 - 54.375%\n65 - 60.875%  \n60 - 66.875%\n55 - 72.375%\n50 - 77.375%\n\nSo, you may think the second player should respin whenever his odds are under 50%, but that is actually incorrect.  For example, if the second player has 70, giving him a 45.625% chance of winning, respinning would give him a 70% chance of going bust - he is better off stopping with the 45.625% chance.  To determine whether or not respinning improves the second player's chances, a second calculation is needed that considers each of the 20 scores after the second spin - some of them bust the player but the others improve his chance of winning.  Based on that, here is a new table, giving the last player's chance of winning if the second player respins with the score given:   \n\n95 - 95.24375%\n90 - 90.9625%\n85 - 87.13125%\n80 - 83.725%\n75 - 80.71875%\n70 - 78.0875%\n65 - 75.80625%  \n60 - 73.85%\n55 - 72.19375%\n50 - 70.3125%\n45 - 69.68125%\n\nSo, comparing the two tables, the critical points are 55 and 60.  If the second player has 60, he should stop - the last player's chance of winning improves if the second player spins.  At 55, however, the second player should spin, improving his odds by about 0.2%.\n\n\nMoving on to the three player game... now it gets really messy.  The player to go first needs a score that can beat the other two players.  Does he stop on 55 again, hoping the other two players both bust?  He'll need a better score than that to have a 50% chance of winning, but, as was noted in the two-player analysis, stopping with a 30% chance can be better than making a second spin.  As for how to do a complete analysis, uh, beats me.  This may require a computer to do all the math.\n\n* * * * * *\n\nThinking about the three-player game, maybe it isn't as messy as I initially thought - the only additional consideration for the first player, as opposed to the second player, is that either player will stop if the first player's score is beaten.  This can be calculated from the first table.  For this table, listed is the chance of the first player losing (i.e. either player beats him) if he stands on the score given:\n\n100 - 9.51% \n95 - 26.68%\n90 - 41.29%\n85 - 53.59%\n80 - 63.85%\n75 - 72.31%\n70 - 79.18%\n65 - 84.69%\n\nSimilar to the two-player game, I can now calculate the odds if the first player decides to respin.  Since I’m getting a little tired crunching these numbers, for this last table I’ll only list the critical ones:\n\n70 - 83.4%\n65 - 82.3%\n\nIn other words, the first contestant should respin on 65, stop on 70.\n\nIn conclusion, I think the optimum strategies are.....\n\nFirst player: spin on 65, stop on 70\n\nSecond player: spin on 55, stop on 60 (assuming his score at least matches the first player)
409482
Number of questions you expect to get correct = 20/4 = 5\n\nOdds of getting a question wrong = 3/4\nOdds of getting a question right = 1/4\nDifferent ways of getting R questions right = 20 C R\nOdds of getting R questions right = 20CR * (1/4)^R * (3/4)^(20-R)\nOdds of passing = 1 - odds of getting less than 10\n\nOdds of getting 0-9 right (using excel) \n= POWER(1/4, X) * POWER(3/4, 20-X) *FACT(20)/FACT(X)/FACT(20-X) \n0.003171212\n0.021141413\n0.066947808\n0.133895615\n0.189685455\n0.202331152\n0.168609293\n0.112406195\n0.060886689\n0.027060751\n= 0.986135583 chance of failing\n\nChance of passing = 1 - 0.986135583 \n= 0.013864\nNot very high.
136914
You are dealing with probabilities here and your odds are going to be exponential. But here is what I have for you.\n\nOdds in rolling a:\n4 = 1/12\n5 = 1/9\n6 = 5/36\n7 or 11 = 2/9\n8 = 5/36\n9 = 1/9\n10 = 1/12\n\nNow for each roll after that you would take the odds and times it to the "n" power, where "n" is the number of the roll. So to roll any of these numbers 12 times in a row would mean taking the fraction to the 12th power. Below are those answers. (Hope you understand scientific notation.)\n\n(1/12)^12 = 1.13876838470415E+55\n(1/9)^12  = 1.13771030934265E+55\n(5/36)^12 = 5.1524055910299E-11\n(2/9)^12  = 1.14868734219891E+55\n\nIn other words you would be more likely to win the lottery several times before this would happen. Hope this helps.
399784
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
650826
I'm assuming each person rolls a 20-side die.\n\nFor two to win, two must roll 15, 16, 17, 18 or 19 points more than one. (20+8) - (10+4) = 14\n\n20-1=19\n20-2,19-1=18\n1+2+3+4+5=15 --> # of ways for two to win\n\nTotal combinations: 20*20 = 400\n15/400 * 100 = 3.75%
272863
since it is "at least", use 1-(probability of no aces or no kings), so:\n\n1-((44/52)*(43/51))\n\nfor the second problem, I would use the idea of (probability of no orange + probability of one orange), so:\n\nprob of no orange:\n((5/8)*(4/7)*(3/6)*(2/5))\n\nprob of one orange, assuming you've learned combinations and permutations:\n(3C1*5C3)/8C4\n\nadd the two prob from above.\n\nhope this helps...probabilty can be quite ambiguous.
52786
I got it....easy solution is to list out all possible satisfying combinations and you come out with 21 out of 36 (7/12).\n\nThe OR in your problem is an inclusive-OR (meaning satisfying either condition satisfies the goal).  However, what I forgot to include was the subtraction where both outcomes occur.\n\n1/2 = just getting even on first roll\n1/6 = getting 1 on second roll\n3/36 = getting even on 1st and 1 on 2nd\n\n1/2 + 1/6 - 3/36 = 7/12\n\nThink of it as a Venn diagram.  If we do not subtract out 3/36, then we are double counting some results (namely 2-1, 4-1, 6-1).\n\n\n>>>>old post\n1/2 + 1/6 if it's an inclusive OR\nneed to get even or get the 1 on the die.\n\n\n1/2 * 5/6 + 1/2 * 1/6 if exclusive OR\n50-50 getting even but must have 2-6 on next die.  Then 50-50 on getting odd but must get the 1 on next die.\n\nOr maybe I'm just really wrong.
216122
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.
316750
a) the number of possible tickets is C(42,6) = 5245786\n1/5245786 = .000000190629\nb) the number of tickets with exactly 4 is \nC(6,4)C(36,2)\n= 15*630 = 9450\n9450/5245786 = .00180\nc) You don't win if you have less than 3 winning numbers, so either you can have 0, 1, or 2 winning numbers and not win\nP(0 numbers) = C(36,6)/5245786 = .37130\nP(1 number) = C(6,1)C(36,5)/5245786 = .43119\nP(2 numbers) = C(6,2)C(36,4)/5245786 = .16843\nP(no prize) = .37130 + .43119 + .16843\n= .97092