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# A bag contains r red marbles, 5r + 4 white marbles and 3r + 2 blue mar

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Joined: 02 Sep 2009
Posts: 45213
A bag contains r red marbles, 5r + 4 white marbles and 3r + 2 blue mar [#permalink]

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29 Jan 2018, 06:28
00:00

Difficulty:

35% (medium)

Question Stats:

64% (00:52) correct 36% (01:05) wrong based on 45 sessions

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A bag contains r red marbles, 5r + 4 white marbles and 3r + 2 blue marbles. If there are only red, white and blue marbles in the bag, what is the probability of pulling a red or white marble?

A. 1/9
B. 1/3
C. 2/3
D. 8/9
E. Cannot be determined.
[Reveal] Spoiler: OA

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A bag contains r red marbles, 5r + 4 white marbles and 3r + 2 blue mar [#permalink]

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29 Jan 2018, 06:34
Bunuel wrote:
A bag contains r red marbles, 5r + 4 white marbles and 3r + 2 blue marbles. If there are only red, white and blue marbles in the bag, what is the probability of pulling a red or white marble?

A. 1/9
B. 1/3
C. 2/3
D. 8/9
E. Cannot be determined.

Since there are r red marbles, 5r+4 white marbles, and 3r+2 blue marbles,
there will be a total of 9r+6

Lets assume an arbitrary value for r=1. Now, there is 1 red marble, 9 white marbles, and 5 blue marbles.

Therefore, the probability of pulling a red or white marble in this case will be $$\frac{1+9}{1+9+5} = \frac{10}{15} = \frac{2}{3}$$(Option C)

Probability = $$\frac{P(Red,White)}{P(Total)} = \frac{r+5r+4}{9r+6} = \frac{6r+4}{9r+6} = \frac{2}{3}$$(Option C)
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A bag contains r red marbles, 5r + 4 white marbles and 3r + 2 blue mar [#permalink]

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Updated on: 29 Jan 2018, 07:22
1
KUDOS
C

Total marbles = r + 5r + 4 + 3r + 2 = 9r+6.

Probability of getting red marbles = r/(9r+6)
Probability of getting white marbles = (5r+4)/(9r+6)

Hence, Probability of getting red or white marbles = r/(9r+6) + (5r+4)/(9r+6)
= (6r+4)/(9r+6)
= 2(3r+2)/(3(3r+2)
= 2/3
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Help with kudos if u found the post useful. Thanks

Originally posted by Sasindran on 29 Jan 2018, 06:47.
Last edited by Sasindran on 29 Jan 2018, 07:22, edited 1 time in total.
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Re: A bag contains r red marbles, 5r + 4 white marbles and 3r + 2 blue mar [#permalink]

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29 Jan 2018, 07:16
P(RorW)= r/(9r+6) + (3r+2)/(9r+6) = 2/3
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Re: A bag contains r red marbles, 5r + 4 white marbles and 3r + 2 blue mar [#permalink]

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29 Jan 2018, 10:28
Agree with solutions. Since its or hence net probability should be the sum of finding either red marbles or white marbles.
Hence total probability is
Prob of finding red marbles = r/(9r+6)

Prob of finding white marbles= (5r+4)/(9r+6)

sum = r/(9r+6)+ (5r+6)/(9r+6) =6r+4/9r+6 = 2/3
hence C
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Re: A bag contains r red marbles, 5r + 4 white marbles and 3r + 2 blue mar [#permalink]

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30 Jan 2018, 11:14
Bunuel wrote:
A bag contains r red marbles, 5r + 4 white marbles and 3r + 2 blue marbles. If there are only red, white and blue marbles in the bag, what is the probability of pulling a red or white marble?

A. 1/9
B. 1/3
C. 2/3
D. 8/9
E. Cannot be determined.

The total number of marbles is r + (5r + 4) + (3r + 2) = 9r + 6. The total number of marbles that are either red or white is r + 5r + 4 = 6r + 4. Thus, the probability of pulling a red or a white marble is:

(6r + 4)/(9r + 6)

2(3r + 2)/[3(3r + 2)]

2/3

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Re: A bag contains r red marbles, 5r + 4 white marbles and 3r + 2 blue mar   [#permalink] 30 Jan 2018, 11:14
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