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GRE Math Challenge #107-Two beads are to be independently
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Updated on: 03 Dec 2017, 02:21
Updated on: 03 Dec 2017, 02:21
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Question Stats:
85% (00:42) correct
15% (00:36) wrong based on 40 sessions
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Two beads are to be independently and randomly selected, one from each of tw o bags. If \(\frac{2}{7}\) of the beads in one bag and \(\frac{3}{7}\) of the beads in the other bag are yellow, what is the probability that both beads selected will be yellow?
(A) \(\frac{2}{3}\)
(B) \(\frac{5}{7}\)
(C) \(\frac{6}{7}\)
(D) \(\frac{4}{49}\)
(E) \(\frac{6}{49}\)
(A) \(\frac{2}{3}\)
(B) \(\frac{5}{7}\)
(C) \(\frac{6}{7}\)
(D) \(\frac{4}{49}\)
(E) \(\frac{6}{49}\)
Re: GRE Math Challenge #107-Two beads are to be independently
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16 Dec 2015, 17:53
16 Dec 2015, 17:53
sandy wrote:
Two beads are to be independently and randomly selected, one from each of tw o bags. If 2/7 of the beads in one bag and 3/7 of the beads in the other bag are yellow, what is the probability that both beads selected will be yellow?
(A) 2/3
(B) 5/7
(C) 6/7
(D) 4/49
(E) 6/49
(A) 2/3
(B) 5/7
(C) 6/7
(D) 4/49
(E) 6/49
answer please
Re: GRE Math Challenge #107-Two beads are to be independently
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17 Dec 2015, 02:48
17 Dec 2015, 02:48
Expert Reply
Whenever you see a probability question, notice if you have the word AND/OR: and means *, OR means +
In this case you are concerned only of the yellow beads. Just multiply the \(\frac{2}{7} * \frac{3}{7}=\)\(\frac{6}{49}\)
The answer is \(E\)
In this case you are concerned only of the yellow beads. Just multiply the \(\frac{2}{7} * \frac{3}{7}=\)\(\frac{6}{49}\)
The answer is \(E\)
