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A bag contains 3 white, 4 black, and 2 red marbles. Two marbles are dr

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A bag contains 3 white, 4 black, and 2 red marbles. Two marbles are dr  [#permalink]

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New post 10 May 2018, 01:19
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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 marbles are dr  [#permalink]

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New post 10 May 2018, 01:33
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1
Two cases -
First ball Non red + Second ball Red
and
First ball Red + Second ball Red

= (7/9)*(2/8) + (2/9)*(1/8) = 2/9

Hence option D.
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A bag contains 3 white, 4 black, and 2 red marbles. Two marbles are dr  [#permalink]

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New post 10 May 2018, 01:47
9 marbles in total (3W, 4B, 2R)
(1) 1. Draw= not red, 2. Draw= red
7/9 * 2/8 = 14/72

(2) 1. Draw=red, 2. Draw=red
2/9 * 1/8 = 2/72

14/72 + 2/72 = 16/72 = 2/9 (Answer: D)
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Re: A bag contains 3 white, 4 black, and 2 red marbles. Two marbles are dr  [#permalink]

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New post 10 May 2018, 01:50
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There are two conditions Red Only in 2nd attempt & Red in both 1st and 2nd attempt.
Probability = (Red only in 2nd attempt) + (Red in both attempt)
= (7/9) * (2/8) + ( 2/9) * (1/8)
= 14/72 + 2/72
=16/72
= 2/9

Ans - 2/9
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Re: A bag contains 3 white, 4 black, and 2 red marbles. Two marbles are dr  [#permalink]

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New post 13 May 2018, 16:58
Bunuel 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 marbles are dr  [#permalink]

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New post 22 May 2018, 08:52
Bunuel 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. \(\frac{7}{9}\)



SOLUTION:

There can be 2 cases -

CASE 1 (First ball) Non red + (Second ball) Red

CASE 2 (First ball) Red + (Second ball) Red

= (7/9)*(2/8) + (2/9)*(1/8) = 2/9

ANS : D :-)
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Re: A bag contains 3 white, 4 black, and 2 red marbles. Two marbles are dr  [#permalink]

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New post 22 May 2018, 10:57
Bunuel 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. \(\frac{7}{9}\)



No of white marbles = 3
No of Black marbles = 4
No of Red marbles = 2

so total no of marbles = 9

Probability that in first draw we have either white and black marbles and in second draw we have red marbles = 7 / 9 *2/8
Probability that in both the draws we have red marbles = 2/9 * 1/8

so the probability of choosing red marbles in second draw = 14 / 72 + 2 / 72 = 16 /72 = 2/9
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Re: A bag contains 3 white, 4 black, and 2 red marbles. Two marbles are dr  [#permalink]

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New post 13 May 2019, 10:24
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Bunuel 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 (often either work, like getting wood for the fire, or dumb, like eating something that shouldn't be eaten).

So, 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.

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,
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Re: A bag contains 3 white, 4 black, and 2 red marbles. Two marbles are dr  [#permalink]

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New post 18 Jan 2020, 06:04
Bunuel 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. \(\frac{7}{9}\)


One of the ways to solve this question is to list the possible combinations. Below are the 4-ways in which the first and the second marbles can be drawn.

In total we have 9 marbles (\(3 W + 4 B + 2 R = 9\))

1. White -> Red \(= \frac{3}{9} * \frac{2}{8}\) (The 8 comes as we are not allowed to replace the marbles after picking up the first) \(= \frac{1}{12}\)
2. Blue -> Red \(= \frac{4}{9} * \frac{2}{8} = \frac{1}{9}\)
3. Red -> Red \(= \frac{2}{9} * \frac{1}{8} = \frac{1}{36}\)
4. White -> Blue | Blue -> White \(= \frac{3}{9} * \frac{4}{8} = 0\) (probability of getting a red is zero)

Adding all 4 possibilities - \(\frac{1}{12} + \frac{1}{9} + \frac{1}{36} = \frac{8}{36} = \frac{2}{9}\)

Ans. D
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Re: A bag contains 3 white, 4 black, and 2 red marbles. Two marbles are dr   [#permalink] 18 Jan 2020, 06:04
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