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A bag contains 3 green marbles, three red marbles and 2 blue marbles.

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A bag contains 3 green marbles, three red marbles and 2 blue marbles.  [#permalink]

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New post 14 Jun 2017, 03:34
00:00
A
B
C
D
E

Difficulty:

  35% (medium)

Question Stats:

66% (01:44) correct 34% (01:45) wrong based on 132 sessions

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Re: A bag contains 3 green marbles, three red marbles and 2 blue marbles.  [#permalink]

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New post 14 Jun 2017, 04:05
2
Bunuel wrote:
A bag contains 3 green marbles, three red marbles and 2 blue marbles. If 3 marbles are randomly drawn from the bag, what is the probability of selecting 2 red marbles and 1 green marble?

A. 1/336
B. 1/56
C. 3/56
D. 9/56
E. 9/28


Probability is Favorable outcomes/Total outcomes

Total Outcomes = 8C3 = 8*7*6/3*2 = 56

Favorable outcomes = 3C2 * 3C1 = 3*3 = 9

Hence D 9/56
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A bag contains 3 green marbles, three red marbles and 2 blue marbles.  [#permalink]

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New post 14 Jun 2017, 05:03
3R, 3G, 2B
P(2r, 1g)= \(\frac{no of desired outcome}{total outcome}\)
Total outcome= 8C3 ( Choosing any 3 from total 8 balls)
Desired outcome= 3C2*3C1
3C2= Choosing 2 red balls from 3 red
3C1= Choosin1 green ball from 3 green

P(2r,1g)=\(\frac{3C2*3C1}{8C3}\) =9/56
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Re: A bag contains 3 green marbles, three red marbles and 2 blue marbles.  [#permalink]

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New post 14 Jun 2017, 05:32
Bunuel wrote:
A bag contains 3 green marbles, three red marbles and 2 blue marbles. If 3 marbles are randomly drawn from the bag, what is the probability of selecting 2 red marbles and 1 green marble?

A. 1/336
B. 1/56
C. 3/56
D. 9/56
E. 9/28


Availability= 3G 3R 2B
Requirements: 2R 1G

Favourable cases = 3C2*3C1 = 3*3= 9
Total cases = 8C3 = 56

Probability= 9/56

Answer Option D
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Re: A bag contains 3 green marbles, three red marbles and 2 blue marbles.  [#permalink]

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New post 15 Jun 2017, 16:26
Bunuel wrote:
A bag contains 3 green marbles, three red marbles and 2 blue marbles. If 3 marbles are randomly drawn from the bag, what is the probability of selecting 2 red marbles and 1 green marble?

A. 1/336
B. 1/56
C. 3/56
D. 9/56
E. 9/28


We are given that a bag contains 3 green marbles, 3 red marbles, and 2 blue marbles. We need to determine the probability of selecting 2 red marbles and 1 green marble.

The red marbles can be selected in 3C2 = 3 ways. The green marble can be selected in 3C1 = 3 ways.

The total number of ways to select the three marbles is 8C3 = 8!/[3!(8-3)!](8 x 7 x 6)/3! = 56.

Thus, the probability is (3 x 3)/56 = 9/56.

Answer: D
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Re: A bag contains 3 green marbles, three red marbles and 2 blue marbles.  [#permalink]

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Re: A bag contains 3 green marbles, three red marbles and 2 blue marbles.   [#permalink] 11 Apr 2019, 02:16
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