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A bag contains equal numbers of red, green, and yellow marbles. If Gee
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Updated on: 17 Feb 2019, 21:31
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A bag contains equal numbers of red, green, and yellow marbles. If Geeta pulls three marbles out of the bag, replacing each marble after she picks it, what is the probability that at least one will be red? A. 8/27 B. 19/27 C. 20/27 D. 7/9 E. 22/27 I have a problem with the practice question from Manhattan Prep and need your help with it: The suggested solution is below: The quick way to answer this question is to calculate the probability that none of the marbles are red. For each of the three picks, there is a probability that the marble will not be red. The probability that all three marbles will not be red is 2/3*2/3*2/3=8/27. If the probability that none of the marbles is red is 8/27, then the probability that at least one marble is red is 18/27=19/27. I understand the idea but I can't also get rid of thinking that if there is a 1/3 chance of picking red marble, then the chance of picking 1 red marble after three picks must be 1/3+1/3+1/3=1. And that is even without probability of picking 2 and 3 red marbles. Where am I wrong? Please, help to clarify.
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Originally posted by 720isenough on 05 Feb 2018, 07:15.
Last edited by Bunuel on 17 Feb 2019, 21:31, edited 2 times in total.
Renamed the topic and edited the question.



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Re: A bag contains equal numbers of red, green, and yellow marbles. If Gee
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05 Feb 2018, 08:19
Here's my 2 cents Since there are equal numbers of each colour I guess the probability for each colour to be picked will be 1/3
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Re: A bag contains equal numbers of red, green, and yellow marbles. If Gee
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05 Feb 2018, 09:08
720isenough wrote: A bag contains equal numbers of red, green, and yellow marbles. If Geeta pulls three marbles out of the bag, replacing each marble after she picks it, what is the probability that at least one will be red? I have a problem with the practice question from Manhattan Prep and need your help with it: The suggested solution is below: The quick way to answer this question is to calculate the probability that none of the marbles are red. For each of the three picks, there is a probability that the marble will not be red. The probability that all three marbles will not be red is 2/3*2/3*2/3=8/27. If the probability that none of the marbles is red is 8/27, then the probability that at least one marble is red is 18/27=19/27. I understand the idea but I can't also get rid of thinking that if there is a 1/3 chance of picking red marble, then the chance of picking 1 red marble after three picks must be 1/3+1/3+1/3=1. And that is even without probability of picking 2 and 3 red marbles. Where am I wrong? Please, help to clarify. First of all, the question asks about picking atleast 1 red marble. In case of 'atleast' questions, it is easier to calculate the probability of picking the other colored marbles and then subtracting from 1, to get the required probability. In your calculation, the highlighted portion is incorrect, as the 3 probabilities shall be multiplied and not added. Secondly the probability will be 1/3 incase to pick up red marbles in all 3 picks whereas the question needs us to pick atleast 1 red marble. Hope its clear.
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Re: A bag contains equal numbers of red, green, and yellow marbles. If Gee
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05 Feb 2018, 11:23
souvonik2k wrote: 720isenough wrote: A bag contains equal numbers of red, green, and yellow marbles. If Geeta pulls three marbles out of the bag, replacing each marble after she picks it, what is the probability that at least one will be red? I have a problem with the practice question from Manhattan Prep and need your help with it: The suggested solution is below: The quick way to answer this question is to calculate the probability that none of the marbles are red. For each of the three picks, there is a probability that the marble will not be red. The probability that all three marbles will not be red is 2/3*2/3*2/3=8/27. If the probability that none of the marbles is red is 8/27, then the probability that at least one marble is red is 18/27=19/27. I understand the idea but I can't also get rid of thinking that if there is a 1/3 chance of picking red marble, then the chance of picking 1 red marble after three picks must be 1/3+1/3+1/3=1. And that is even without probability of picking 2 and 3 red marbles. Where am I wrong? Please, help to clarify. First of all, the question asks about picking atleast 1 red marble. In case of 'atleast' questions, it is easier to calculate the probability of picking the other colored marbles and then subtracting from 1, to get the required probability. In your calculation, the highlighted portion is incorrect, as the 3 probabilities shall be multiplied and not added. Secondly the probability will be 1/3 incase to pick up red marbles in all 3 picks whereas the question needs us to pick atleast 1 red marble. Hope its clear. Hi, souvonik2k, Thanks for your input. Unfortunately, it is still not clearer. We would multiply probabilities if we needed to get red AND red AND red. But if we want just one red then the combined probability of 3 picks should be red OR red OR red, therefore sum of 1/3's, shouldn't it? Once again, I know it cannot be true, but I am curious why.



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Re: A bag contains equal numbers of red, green, and yellow marbles. If Gee
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13 Feb 2018, 11:50
720isenough wrote: souvonik2k wrote: 720isenough wrote: A bag contains equal numbers of red, green, and yellow marbles. If Geeta pulls three marbles out of the bag, replacing each marble after she picks it, what is the probability that at least one will be red? I have a problem with the practice question from Manhattan Prep and need your help with it: The suggested solution is below: The quick way to answer this question is to calculate the probability that none of the marbles are red. For each of the three picks, there is a probability that the marble will not be red. The probability that all three marbles will not be red is 2/3*2/3*2/3=8/27. If the probability that none of the marbles is red is 8/27, then the probability that at least one marble is red is 18/27=19/27. I understand the idea but I can't also get rid of thinking that if there is a 1/3 chance of picking red marble, then the chance of picking 1 red marble after three picks must be 1/3+1/3+1/3=1. And that is even without probability of picking 2 and 3 red marbles. Where am I wrong? Please, help to clarify. First of all, the question asks about picking atleast 1 red marble. In case of 'atleast' questions, it is easier to calculate the probability of picking the other colored marbles and then subtracting from 1, to get the required probability. In your calculation, the highlighted portion is incorrect, as the 3 probabilities shall be multiplied and not added. Secondly the probability will be 1/3 incase to pick up red marbles in all 3 picks whereas the question needs us to pick atleast 1 red marble. Hope its clear. Hi, souvonik2k, Thanks for your input. Unfortunately, it is still not clearer. We would multiply probabilities if we needed to get red AND red AND red. But if we want just one red then the combined probability of 3 picks should be red OR red OR red, therefore sum of 1/3's, shouldn't it? Once again, I know it cannot be true, but I am curious why. Hi, i'll clarify the answer with another way of tackle this question: The question need us to calculate the probability of getting at least 1 red marble, it's imply that 3 of these scenarios will happen: EXACTLY 1 red marbles drawn OR EXACTLY 2 red marbles drawn OR EXACTLY 3 red marbles drawn P(At least 1 red marble) = P(exactly 1 red marble) + P(exactly 2 red marble) + P(exactly 3 red marble) You will need to find the sum of these 3 to find the answer: For the first scenario, you only want 1 red marble, therefore the probability will be: 3 * (1/3) * (2/3) * (2/3) = 12/27 1/3 is the probability of getting a red marble 2/3 is the probability of getting a marble of another color (If probability of getting red is 1/3 => probability of not getting red is 2/3) you have to multiply it by 3, because the red marble can appear in your 1st, 2nd or 3rd pick For the second scenario, you wanted exactly 2 red marble, therefore the probability will be: 3!/2! * (1/3) * (1/3) * (2/3) = 6/27 you have to multiply it by 3!/2! because 2 of the red marble pick can be appear in 3!/2! = 3 ways For the last scenario, you wanted all 3 are red marbles, therefore the probability will be: (1/3) * (1/3) * (1/3) = 1/27 Noted that for this scenario, the red marbles can only appear in 1 way (all the marbles are considered identical) Sum of all 3 will be: 12/27 + 6/27 + 1/27 = 19/27, the same answer as OA If it's still not clear, please ask!



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Re: A bag contains equal numbers of red, green, and yellow marbles. If Gee
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17 Feb 2019, 10:18
Hi,
Can this question be answered with a probability tree?
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Re: A bag contains equal numbers of red, green, and yellow marbles. If Gee
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31 Mar 2019, 04:50
I solved the question using the negative outcome technique. Probability of any colour being picked is 1/3 therefore the probability of red not being picked is 2/3 Since we pick up marbles 3 times, probability of no red 3 times is 2/3*2/3*2/3 = 8/27 Hence probability of at least one red will be = 1 8/27 = 19/27
Please let me know if my approach is correct?



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Re: A bag contains equal numbers of red, green, and yellow marbles. If Gee
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03 Apr 2019, 18:07
I have also solved with this method  The question need us to calculate the probability of getting at least 1 red marble, it's imply that 3 of these scenarios will happen: EXACTLY 1 red marbles drawn OR EXACTLY 2 red marbles drawn OR EXACTLY 3 red marbles drawn P(At least 1 red marble) = P(exactly 1 red marble) + P(exactly 2 red marble) + P(exactly 3 red marble) You will need to find the sum of these 3 to find the answer: For the first scenario, you only want 1 red marble, therefore the probability will be: 3 * (1/3) * (2/3) * (2/3) = 12/27 1/3 is the probability of getting a red marble 2/3 is the probability of getting a marble of another color (If probability of getting red is 1/3 => probability of not getting red is 2/3) you have to multiply it by 3, because the red marble can appear in your 1st, 2nd or 3rd pick For the second scenario, you wanted exactly 2 red marble, therefore the probability will be: 3!/2! * (1/3) * (1/3) * (2/3) = 6/27 you have to multiply it by 3!/2! because 2 of the red marble pick can be appear in 3!/2! = 3 ways For the last scenario, you wanted all 3 are red marbles, therefore the probability will be: (1/3) * (1/3) * (1/3) = 1/27 Noted that for this scenario, the red marbles can only appear in 1 way (all the marbles are considered identical) Sum of all 3 will be: 12/27 + 6/27 + 1/27 = 19/27, the same answer as OA  I have solved with this approach, however, I feel there should be another approach which is less timeconsuming. Bunuel VeritasKarishma Please throw some light.




Re: A bag contains equal numbers of red, green, and yellow marbles. If Gee
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