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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, 02:19
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Question Stats:

60% (01:23) correct 40% (01:49) wrong based on 78 sessions

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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, 02:33
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, 02:47
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, 02:50
1
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, 17:58
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, 09:52
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, 11:57
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
Re: A bag contains 3 white, 4 black, and 2 red marbles. Two marbles are dr &nbs [#permalink] 22 May 2018, 11:57
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