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A bag contains 3 white, 4 black, and 2 red marbles. Two marb [#permalink]
15 Sep 2017, 07:47
Question Stats:
62% (01:39) correct
37% (01:36) wrong based on 37 sessions
A bag contains 3 white, 4 black, and 2 red marbles. Two marbles are drawn from the bag. If replacement is not allowed, what is the probability that the second marble drawn will be red? A. \(\frac{1}{36}\) B. \(\frac{1}{12}\) C. \(\frac{7}{36}\) D. \(\frac{2}{9}\) E. \(\frac{7}{9}\)
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Re: A bag contains 3 white, 4 black, and 2 red marbles. Two marb [#permalink]
21 Sep 2017, 05:13
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The probability of extracting a red marble in the second draw is equal to the probability of extracting whatever marble in the first draw times the probability of extracting a red marble in the second draw, i.e. 1*2/9 = 2/9. Answer D!



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Re: A bag contains 3 white, 4 black, and 2 red marbles. Two marb [#permalink]
04 Mar 2018, 05:40
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The scenario of getting a red marble in the second draw could be like this: 1) no red in first draw and a red in the second draw 2) a red in the first draw and a red in the second draw
1) (7/9)*(2/8)=7/36 2) (2/9)*(1/8)=1/36
So, the probability that the second marble is drawn will be red = (7/36)+(1/36)=2/9
The answer is D.



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Re: A bag contains 3 white, 4 black, and 2 red marbles. Two marb [#permalink]
13 May 2018, 16:45
Carcass wrote: A bag contains 3 white, 4 black, and 2 red marbles. Two marbles are drawn from the bag. If replacement is not allowed, what is the probability that the second marble drawn will be red? A. \(\frac{1}{36}\) B. \(\frac{1}{12}\) C. \(\frac{7}{36}\) D. \(\frac{2}{9}\) E. \(\frac{7}{9}\) If a red marble is selected first and a red marble is selected second, the probability is: 2/9 x 1/8 = 2/72 If a red marble is not selected first and a red marble is selected second, the probability is: 7/9 x 2/8 = 14/72 So the overall probability is 2/72 + 14/72 = 16/72 = 2/9. Answer: D
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Re: A bag contains 3 white, 4 black, and 2 red marbles. Two marb
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13 May 2018, 16:45





