What is the probability that they are not all chocolates?
205726
Chances of getting a non chocolate on:\n1 draw: 27/47 or 57.4%\n2 draw: 26/46 or 56.5%\n3 draw: 25/45 or 55.6%\n\n57.4% * 56.5% * 55.6% = 18%\n\nThanks Bogusfrog :)
673510
Christians ...Chocolate.....yeah right.
531478
never chocolate is good no matter where it is.\n\nyour welocme\n\nwhat r u waitting 4\n\ngrab it and eat it\n\n\nthanks
672998
What? Yes it would be about half.  Otherwise...if what you said applies, you could eat one, or a whole box and it would be the same.
264538
http://en.wikipedia.org/wiki/Chocolate\nhttp://www.exploratorium.edu/exploring/exploring_chocolate
222830
Chocolate cake\nHa! Ha!
247098
33% is equal to approximately 3 out of 10(3.3)\nTherefore, if six out of ten were to be selected, that would equal 66% of ethnics - more than twice the probability.\nIt follows then that in iself, the odds would be just over two-to-one against this happening.\nSo the probability of more than of no more than six would be two tone ON.\nDoes that make sense?
226451
I'm not sure what you mean by "keys"\n\nIn any event, this is how I would do this problem.\n\n1)  We want exactly one blue ball which means we want exactly three balls of other colors.  Since there are 7 blue balls and 11 total balls, the probability of a blue ball is 7/11.  Also, with 4 other balls, the probability of a non-blue ball is 4/11.  Replacement means these probabilities remain constant.  Thus, one blue ball and three non-blue balls equals: 7/11 * 4/11 * 4/11 * 4/11.  However, since we don't care when we choose the blue ball, it could happen either first, second, third, or fourth.  Thus, we must also multiply the above product by four to take this into account.  This gives us 1792/14641.  This is equivalent to the binomial distribution with n=4, x=1, and p=7/11.  4C1*(7/11)^1*(4/11)^(4-1) = 4 * 7/11 * 4/11 * 4/11 * 4/11 = 1792/14641.\n\n2)  For the second part, we want at least two blue balls.  So, we can have 2, 3, or 4 blue balls.  It will be quicker to calculate the probability of 0 or 1 blue balls and subtract this probability from 1.  We already know the probability of 1 blue ball is 1792/14641.  The probability of 0 blue balls means that every ball is non-blue.  This probability is (4/11)^4 = 256/14641.  Subtracting these probabilities from 1, we get that the probability of at least two blue balls is: 12593/14641.\n\nGood luck!
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
471764
Note: they aren't asking just for the chance that at least 2 children have blue eyes.  They are asking for the probability, given that you know at least one already does.\n\nThis is conditional probability.\n\nP(B | A) = P(A and B) / P(A)\n\nSo the probability that you have at least 2, given you have at least 1 is the probability you have at least 2 divided by the probability you have at least 1.\n\nFor that part though, you would use binomal distribution.\n\nFirst, what is the probability that at least one child has blue eyes?\nThis would be 1 minus the probability that nobody has blue eyes.  \nP ( blue >= 1 ) = 1 - (0.75)^5 \nP ( blue >= 1 ) = 0.762695312 \n\nNow the probability that at least *two* children have blue eyes:\nP( blue = 2 ) + P ( blue = 3 ) + P ( blue = 4 ) + P ( blue = 5 ).  Again you can figure the negative statement and subtract from 1.  So this would be equivalent to:\n1 - P ( blue = 0 ) - P ( blue = 1 ).\n\nThis would be:\nP ( blue >= 2 ) = 1 - (0.75)^5 - (0.25)(0.75)^4 * C(5,1)\nThe C(5,1) means the ways to choose 1 person from 5, which is just 5.\nP ( blue >= 2) = 1 - (0.75)^5 - (0.25)(0.75)^4 * 5 \nP ( blue >= 2) = 0.3671875\n\nSo putting it all together:\nP ( blue >= 2 | blue >= 1 ) = P ( blue >= 2 ) / P ( blue >= 1 )\n= 0.3671875 / 0.762695312\n= 0.481434059\n≈ 48.1%
518092
This is the opposite of the probability of getting all no's.  The probability of getting all No's is (9/10)^7.  The probability of getting at least one yes is 1-(9/10)^7.\n approximately 0.522
185901
No.\n\nSome risk factors of Diabetes:\nObesity, family history, African American\nHigh BP, High cholesterol, inactive lifestyle.\n\nJust eating chocolate would probably not lead to diabetes.
80979
chocolate anything with chocolate
539588
The softer of the two chocolates would melt faster (there is more water content).  I believe that would be milk chocolate.\n\nMake it an experiment.  Get a regular Hershey Bar and a Dark Hersey Bar.  See which one would melt the fastest in the same heat.  Then make s'mores!
50739
Chocolate...for sure!
272392
be happy its not CHOCOLATE
378637
The literature says NO. My experience says YES.\n\nEverytime I eat chocolate I get pimples myself.\n\nThe literature must not be right all the time.