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29 Jan 2018, 06:28
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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.
A. 1/9
B. 1/3
C. 2/3
D. 8/9
E. Cannot be determined.
29 Jan 2018, 06:34
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Bunuel
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)
Traditional Approach
Probability = \(\frac{P(Red,White)}{P(Total)} = \frac{r+5r+4}{9r+6} = \frac{6r+4}{9r+6} = \frac{2}{3}\)(Option C)
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
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
