tirumalai Posted September 6, 2008 Share Posted September 6, 2008 A bag has six red marbles and six blue marbles. If two marbles are drawn randomly from the bag, what is the probability that they will both be red? i want to know which process is right (6C1*5C1) / (12C1*11C1) OR (6P1*5P1)/12P2 Pls do let me know. Quote Link to comment Share on other sites More sharing options...
prem.muthuswamy Posted September 6, 2008 Share Posted September 6, 2008 its 6c1*5c1/12c2 Quote Link to comment Share on other sites More sharing options...
tirumalai Posted September 6, 2008 Author Share Posted September 6, 2008 i asked between the alternatives i had written. thanks. Quote Link to comment Share on other sites More sharing options...
gmatissimple Posted September 6, 2008 Share Posted September 6, 2008 Permutation : Permutation means arrangement of things. The word arrangement is used, if the order of things is considered. nPr = n!/(n-r)! Combination: Combination means selection of things. The word selection is used, when the order of things has no importance. nCr= n!/r!(n-r)! The following link may help in this case: http://tutors4you.com/permutationcombinationtutorial.htm Randomly drawn means selection without order. So, combination is required. Quote Link to comment Share on other sites More sharing options...
sid_j Posted September 8, 2008 Share Posted September 8, 2008 2 balls are to be selected, arrangement is not mentioned..so it will be C and not P.. it shud be 6C2/12C2... it doesnt say one after the another, had that been the case..i wud have gone with 6C1X5C1.. please correct,if i am wrong...plzz..plzzz... Quote Link to comment Share on other sites More sharing options...
Jolly Goodfellow Posted September 8, 2008 Share Posted September 8, 2008 6 marbles taken from 12 * 5 marbles left taken from 11 remaining marbles = 6/12 * 5/11 = 1/2 * 5/11 = 5/22 Quote Link to comment Share on other sites More sharing options...
Shooter Posted September 9, 2008 Share Posted September 9, 2008 I too think 6C2/12C2 (15/66) is the answer. Could you please confirm the OA. Quote Link to comment Share on other sites More sharing options...
Retake_GMAT Posted September 9, 2008 Share Posted September 9, 2008 IMO it is 6C2/12C2 Quote Link to comment Share on other sites More sharing options...
Tristar Posted May 19, 2015 Share Posted May 19, 2015 I will try to explain this concept in an easy way. When you are given more than one spot where you need to select objects from a wider population (in our case the population of marbles), first find out the required probability for the first spot. In our example, for the first marble to be drawn, the probability that it is going to be red is 6 (favorable number of cases) / 12 (total number of cases) = 1/2. The next step, is to observe that for the second marble the probability of being selected (without replacement) will be different from the probability found in step 1. This is so, because we have one less marble to choose from the remaining population, meaning that the chances of choosing the second marble in the red color is LESS than, say, choosing the second marble in the blue color (5 red marbles left vis-a-vis 6 blue marbles). So, the probability of choosing the second marble in the red color is 5 (remaining red marbles) / 11 (remaining total number of all marbles). So, our required probability is: (1/2)*(5/11) = 5/22. Hope the confusion related to this question is resolved for good. Quote Link to comment Share on other sites More sharing options...
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