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by itsmebharat » Tue May 31, 2011 1:35 am
Q. A box contains identical balls in three different colors - Black, white and blue. There are 8-x blue balls and 2X+5 black balls. If a ball is picked at random from the box, what is the probablity that the ball is either blue or black?
1) X=2
2) There are 3X+39 white balls in the box.

OA is B
I am not an Expert, please feel free to suggest if there is an error.
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by Frankenstein » Tue May 31, 2011 1:58 am
Hi,
There are (2x+5)black and(8-x)blue balls. Let y be the number of white balls.
Probability that a ball picked is blue or black is, p(E) = (2x+5)+(8-x)/[(2x+5)+(8-x)=y]
From(1): x=2, p(E) = 15/15+y. We need the value of y.
Not sufficient
From(2): y=3x+39. So, p(E) = 13+x/(13+x+3x+39) = 1/4
Sufficient

Hence, answer B

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by GMATGuruNY » Tue May 31, 2011 2:01 am
itsmebharat wrote:Q. A box contains identical balls in three different colors - Black, white and blue. There are 8-x blue balls and 2X+5 black balls. If a ball is picked at random from the box, what is the probablity that the ball is either blue or black?
1) X=2
2) There are 3X+39 white balls in the box.

OA is B
Statement 1: x=2.
No info about the white balls.
Insufficient.

Statement 2: W = 3x+39.
Total = blue + black + white = (8-x) + (2x+5) + (3x+39) = 4x + 52 = 4(x+13).
Number that are blue or black = (8-x) + (2x+5) = x+13.
P(blue or black) = (x+13) / 4(x+13) = 1/4.
Sufficient.

The correct answer is B.

Another way to evaluate statement 2 would be to plug in different values for x:
Let x=1.
Blue = 8-x = 8-1 = 7.
Black = 2x+5 = 2*1 + 5 = 7.
White = 3x+39 = 3*1 + 39 = 42.
Total = 7+7+42 = 56.
P(blue or black) = (7+7)/56 = 14/56 = 1/4.

Let x=2.
Blue = 8-x = 8-2 = 6.
Black = 2x+5 = 2*2 + 5 = 9.
White = 3x+39 = 3*2 + 39 = 45.
Total = 6+9+45 = 60.
P(blue or black) = (6+9)/60 = 15/60 = 1/4.

Since in each case P(blue or black) = 1/4, sufficient.
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