Probability question?
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
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
557990
This is a binomial problem.  The number of trials is 4 and the probability of success is 1/6.  No more than twice means either 0, 1, or 2 sixes.  So\n\n4!/(0!4!)(1/6)^0(5/6)^4 + 4!/(1!3!)(1/6)^1(5/6)^3 + 4!/(2!2!)(1/6)^2(5/6)^2\n\n= 0.482 + 0.386 + 0.116\n\n= 0.984
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
97799
i don't believe the person above me got the second part right.  he is right about the beginning - the first probability is 2/5:\n\npossible numbers to draw: 2 4 6 11 13 15\n\nso you can choose any of these six out of the fifteen given (6/15 = 2/5).  the next number you draw must be also a "winning" number.  BUT, you have to have an even number less than 7 AND an odd number greater than 10.  this gives us two scenarios:\n\nIF the first number picked was even and <7, then the possible numbers you can pick from are ONLY: 11 13 15, since picking another even number <7 will not meet your requirements.  this gives us a probablity of 3/14, since there are 3 "winning" numbers out of the 14 numbers left.\n\n\nIF the first number picked was odd and >10, then the possible numbers you can pick from are ONLY: 2 4 6, since picking another odd number >10 will not meet your requirements.  this gives us a probablity of 3/14, since there are 3 "winning" numbers out of the 14 numbers left.\n\nSO, finally, suppose you go with the first scenario.  the probability would be: (6/15)*(3/14) = 3/35  going with the second scenario will give you the same thing, 3/35.  thus adding these two, since you can have either scenario occur and still meet your requirements, gives you: 3/35+3/35 = 6/35.\n\nAnd that would be the odds: 6/35.
933079
There are 2 black 2s in a deck out of 52. The odds of selecting a black 2 is 2/52. \nThe odds of NOT selecting a black 2 is therefore 50/52.
366018
try this site\n\n\nhttp://www.futureaccountant.com/probability/
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.)
551202
1) Find the probability of getting a red ball when a bag contains 2 red balls and one blue ball.\n\n2) There is a red ball, a blue ball and a green ball in a bag. Find the probability of getting either a red ball or a blue ball.\n\n3) There are 10 yellow chips and 20 purple chips in a sack. When one chip is selected at random, find the probability of NOT getting yellow chips.
656922
Okay, intially, you pick one marble, out of 12. and there are 3 yellows, then your initial chances are 1/4 of picking a yellow. Now, you've picked one, so there are 11 marbles left in the bag. There are 4 blue marbles in the bag, so you have a 4/11 chance of picking a blue ball at tthis point. \nTo Calculate Probability based  on an event that 1) Has already happened ( the yellow ball was picked first) and that is an event depending on the results of the , you simply multiply the chances. ( 1/4*4/11) or 4/44, or 1/11 chances that that will happen.
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.
906595
C.\nThis is a binomial probability problem.\nThe binomial formula says \nP(r out of n) = (nCr)(p^r)(q^(n-r)), where n is the number of trials (here 6 rolls), r is the number of successes (here 0, 1, 2, and 3 twos), p is the probability of success (here 1/6), and q is the probability of failure (here 5/6)\nYou need to plug in 0, 1, 2, and 3 for "r" into this formula.  If you do, you get:\nP(0 twos) = .3348979767\nP(1 two) = .401877572\nP(2 twos) = .200938786\nP(4 twos) = .05358367627\nYou add up these answers to get your final answer, about .9913
484411
49/99\n\nProof:\n\n1) Probability that first passenger (P1) sits in last passengers (P100) seat (S100) is 1/99\n2) probability that P1 sits in S99 , and P99 sits in S100 is (1/99) * (1/2)\n3.1) probability that P1 sits in S98, and P98 sits in S100 is (1/99) * (1/3)\n3.2) probability that P1 -> S98, P98->S99, P99->S100 is (1/99) * (1/3) * (1/2)\n4.1) P1->S97, P97->S100 = (1/99)*(1/4)\n4.2) P1->S97, P97->S99, P99->S100 = (1/99)*(1/4)*(1/2)\n4.3) P1->S97, P97->S98, P98->S100 =  (1/99)*(1/4)*(1/3)\n4.4) P1->S97, P97->S98, P98->S99, P99->S100 = (1/99)*(1/4)*(1/3)*(1/2)\n5.1) P1->S96, P96->S100 = (1/99)*(1/5)\n5.2) P1->S96, P96->S99, P99->S100 = (1/99)*(1/5)*(1/2)\n5.3) P1->S96, P96->S98, P98->S100 =  (1/99)*(1/5)*(1/3)\n5.4) P1->S96, P96->S98, P98->S99, P99->S100 = (1/99)*(1/5)*(1/3)*(1/2)\n5.5) P1->S96, P96->S97, P97->S100 = (1/99)*(1/5)*(1/4)\n5.6) P1->S96, P96->S97, P97->S99, P99->S100 = (1/99)*(1/5)*(1/4)*(1/2)\n5.7) P1->S96, P96->S97, P97->S98, P98->S100 =  (1/99)*(1/5)*(1/4)*(1/3)\n5.8) P1->S96, P96->S97, P97->S98, P98->S99, P99->S100 = (1/99)*(1/5)*(1/4)*(1/3)*(1/2)\n\n1) (1/99)\n2) (1/99)(1/2)\n3) (1/99)(1/3)(1 + ½)\n4) (1/99)(1/4)(1 + 1/3)(1 + ½)\n5) (1/99)(1/5)(1 + ¼)(1 + 1/3)(1 + ½)\n\ni) (1/99)(1/i)product(j=2 to (i-1),(1+1/j))\n\n99) (1/99)(1/99)product(j=2 to 98,(1+1/j))\n\nGrand Sum = (1/99)*(1 + ½ +sum(i=3 to 99,(1/i)*product(j=2 to (i-1),(1+1/j))))\n=(1/99)*(1 + ½ + sum(i=3 to 99,(1/2)))\n=(1/99)*(1 + 98/2)\n= 50/99\n\nProbability that last passenger gets bumped = 50/99\n\nHence Probability that last passenger does NOT get bumped = 1 – 50/99 = 49/99\nA bit less than 50%\n\nWould be 1/2 like others suggest if first passenger was not constrained to sit in the wrong seat as given in the problem statement.
791422
Each individual toss is 1 / 2, a 50% probability.\n\nThe possibilities of all outcomes is 1 / (2 x 2 x 2), a 12.5% probability.\n\nOnly 2 possible outcomes do not meet the desired outcome (all heads, all tails).\n\n2 x 12.5% is 25% (but that's the opposite of what you want)\n\n100% - 25% = 75%
395430
Assuming that the chance of having a girl is 50%, the chance would be 3/8.  This answer can be most simply reached a couple different ways.  First, you could list the possibilities:\n\nBBB\nBBG\nBGB\nGBB\nGGB\nGBG\nBGG\nGGG\n\n3 of the 8 possibilities fit the condition (exactly one girl), so the probability is 3/8.\n\nAnother way, better for larger numbers where the above method is impractical, is to use permutations.  The above works fine for this problem, though.
898818
There are sixty-six possible outcomes in the first two scenarios, and fifty-five possible outcomes in the third (Mary does not win):\n\nWhat is the probability a married couple wins the prizes? \n9.09% (6/66 x 100)\n\nWhat is the probability one prize goes to a man and one to a woman? \n54.5% (36/66 x 100)\n\nIf Mary does not win a prize, then what is the probability that John will? \n18.2% (10/55 x 100)\n\n(or 0% if John never made it to the party!)
220111
Greetings:\n\nHaving selected one of the four, the probability that person #2 will have a different date of birth is 364/365 (if we seek no repeats, #2 can have any 364 of 365 possible b.days).  Similarly, having selected two, the probability that #3 will have a unique d.o.b. is 363/365 (any 363 of 365 possible dates okay).  Finally, for #4, the probability is 362/365.  Hence by the multiplication property, the probability that all four will have different b.days is given by, (364/365)(363/365)(362/365) = (364)(363)(362)/(365)^3 = 0.984 to three places.\n\nRegards,\n\nNikkor180