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# A bag contains 3 white, 4 black, and 2 red marbles. Two marb

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A bag contains 3 white, 4 black, and 2 red marbles. Two marb [#permalink]  15 Sep 2017, 07:47
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59% (01:34) correct 40% (01:40) wrong based on 42 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. $$\frac{7}{9}$$
[Reveal] Spoiler: OA

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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

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Re: A bag contains 3 white, 4 black, and 2 red marbles. Two marb [#permalink]  13 May 2018, 16:45
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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.

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Re: A bag contains 3 white, 4 black, and 2 red marbles. Two marb [#permalink]  19 Nov 2020, 10:57
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Expert's post
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. $$\frac{7}{9}$$

This question reminds me of my childhood, when my friends and I would sometimes "draw straws" to randomly select one person to do something awful.
Someone would hold up n pieces of grass (for n people), and one of those pieces was very short. The person who selected the shortest piece was the one who had to perform the task.

There was always one guy who wanted to choose his piece last. His reasoning was that his chances of drawing the shortest piece would be minimized, since every person before him had a chance of drawing the short piece before it got to his turn.

The truth of the matter is that each of the n people had a 1/n chance of selecting the shortest piece, regardless of the order in which they selected a straw.

The same applies to the original question here.

2 of the 9 marbles are red, so P(red ball selected SECOND) = P(red ball selected FIRST) = 2/9

Cheers,
Brent
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Brent Hanneson – Creator of greenlighttestprep.com

Re: A bag contains 3 white, 4 black, and 2 red marbles. Two marb   [#permalink] 19 Nov 2020, 10:57
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