Probability

[BTGmoderatorRO]

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
OA is d
what is wrong with option D here? can anyone help me out?
Thanks

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Hi Roland2rule,

We're told that a bag contains 3 red, 4 black and 2 white balls. We're asked for the probability of drawing a red and a white ball in two successive draws, while replacing each ball after it is drawn.

The prompt does NOT state that the red ball has to be drawn first, so there are two options that we have to consider: red first, white second and white first, red second.

The probability of pulling a red first and a white second (with replacement) = (3/9)(2/9) = 6/81
The probability of pulling a while first and a red second (with replacement) = (2/9)(3/9) = 6/81
The total probability of pulling a red and white ball is 6/81 + 6/81 = 12/81 = 4/27

Final Answer: D

GMAT assassins aren't born, they're made,
Rich

[Scott@TargetTestPrep]

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
OA is d
what is wrong with option D here? can anyone help me out?
Thanks

There are 9 balls in the bag, so the probability of drawing a red ball is P(Red) = 3/9 = 1/3, and the probability of drawing a white ball is P(White) = 2/9. We will draw two balls, replacing each ball after it is drawn.

The probability of drawing a red ball first and then a white ball is: P(Red) x P(White) = 1/3 x 2/9 = 2/27. But we can also draw a white ball first and then a red ball: P(White) x P(Red) = 2/9 x 1/3= 2/27. Either of these outcomes satisfies our outcome of interest, and so we add the two probabilities: 2/27 + 2/27 = 4/27.

Answer: D