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Re: A bag contains 12 marbles: 5 of the marbles are red, 3 are g [#permalink]
qty A = (5/12 * 4/11 * 3/10) => Without replacement
qty B = 5/12* 4/12* 3/12) => with Replacement

Therefore B>A
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Re: A bag contains 12 marbles: 5 of the marbles are red, 3 are g [#permalink]
1
Answer: B
12 marbles:
Red: 5
Green: 3
Blue: 4
A: The probability of consecutively choosing two red marbles and a green marble without replacement.
The probability of choosing first red marble is 5/12, because there is not any replacement now 4 red marbles are remained and totally 11 marbles. Thus the probability of choosing the second red marble is 4/11. The probability of choosing a green marble in these 10 remaining marbles is 3/10. So P(A) = 5/12 * 4/11 * 3/10

B: The probability of consecutively choosing a red and two blue marbles with replacement
As there are 5 red marbles, the probability of choosing a red marble between 12 marbles is 5/12. The probability of choosing first blue marble between 12 remained marbles(because there is the replacement, we put back drawn marble in last step) is 4/12. The probability of choosing second blue marble between 12 remaining marbles and 4 remaining blue marbles is 4/12. So P(B) = 5/12 * 4/12* 4/12
B is a little bigger.
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Re: A bag contains 12 marbles: 5 of the marbles are red, 3 are g [#permalink]
Arpine wrote:
qty A = (5/12 * 4/11 * 3/10) => Without replacement
qty B = 5/12* 4/12* 3/12) => with Replacement

Therefore B>A


I think there is a problem here. In quantity B it should be: 5/12 * 4/12 * 4/12 since its with replacement. Please correct me if I am wrong.
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Re: A bag contains 12 marbles: 5 of the marbles are red, 3 are g [#permalink]
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