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

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