Q. A bag contains 3 red, 4 black and 2 white balls. What is the probability of drawing a red and a white ball in two successive draws, each ball being put back after it is drawn?
(A) 2/27
(B) 1/9
(C) 1/3
(D) 4/27
(E) 2/9
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Hi Joy Shaha,
This question can be solved in a couple of different ways. I'm going to approach this with permutation/probability math.
We're asked for the probability of pulling a red and a white (in either order) from a group of 3 red, 4 black and 2 white balls (with replacement). There are two ways to accomplish this:
(White first)(Red second) = (2/9)(3/9) = 6/81
(Red first)(White second) = (3/9)(2/9) = 6/81
Total = 6/81 + 6/81 = 12/81 = 4/27
Final Answer: D
GMAT assassins aren't born, they're made,
Rich
This question can be solved in a couple of different ways. I'm going to approach this with permutation/probability math.
We're asked for the probability of pulling a red and a white (in either order) from a group of 3 red, 4 black and 2 white balls (with replacement). There are two ways to accomplish this:
(White first)(Red second) = (2/9)(3/9) = 6/81
(Red first)(White second) = (3/9)(2/9) = 6/81
Total = 6/81 + 6/81 = 12/81 = 4/27
Final Answer: D
GMAT assassins aren't born, they're made,
Rich
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OptimusPrep
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3R, 4B, 2W balls.Joy Shaha wrote:Q. A bag contains 3 red, 4 black and 2 white balls. What is the probability of drawing a red and a white ball in two successive draws, each ball being put back after it is drawn?
(A) 2/27
(B) 1/9
(C) 1/3
(D) 4/27
(E) 2/9
P(1R, 1W) = (3/9)*(2/9) = 1/3*2/9 = 2/27
Now the Red and the While balls can be drawn in any order
Hence probability = 2*2/27 = 4/27
Correct Option: D
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Matt@VeritasPrep
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You could also do
1 - P(Both Red) - P(Both White) - P(Either Black)
or
1 - (3/9)*(2/9) - (2/9)*(1/9) - (1 - (5/9)*(4/9))
or
1 - 6/81 - 2/81 - 61/81
or
12/81 => 4/27
1 - P(Both Red) - P(Both White) - P(Either Black)
or
1 - (3/9)*(2/9) - (2/9)*(1/9) - (1 - (5/9)*(4/9))
or
1 - 6/81 - 2/81 - 61/81
or
12/81 => 4/27
