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

A is correct if you want to get a Red ball First and then a White one.
If the order doesn't matter, then the answer is D.

(2/27)*2 = 4/27

Here is how I solved it

First of all we have
Probability of drawing a Red ball is 3/9
Probability of drawing a White ball is 3/9

There are two ways in which the balls can be drawn


Case 1: Red ball in the first draw and white in the second draw
Hence the combined Probability is 3/9*2/9=6/81

Case 2: White ball in the first draw and red in the second draw
Hence the combined Probability is 2/9*3/9=6/81

both these cases satisfy our requirement
Hence either of them will do i.e OR
Hence the final probability comes to be
Case 1 OR Case 2 = 6/81 + 6/81 (OR means addition)
Hence the Ans is 12/81=4/27

Responding to a pm:

The status of "replacement" has nothing to do with the "sequence". It only changes the probability.

Say we have 2 red and 3 white balls in a bag. We pull out two one after another with replacement. What is the probability that one is red and the other is white.
Now note that there are 4 ways in which you can pull out two balls from the bag:
1. You pull a Red and then a Red again RR - (2/5)*(2/5) (Note that it is with replacement)
2. You pull a Red and then a White RW - (2/5)*(3/5)
3. You pull a White and then a Red WR - (3/5)*(2/5)
4. You pull a White and then a White WW - (3/5)*(3/5)

Total probability = (2/5)*(2/5) + (2/5)*(3/5) + (3/5)*(2/5) + (3/5)*(3/5) = 1

In how many cases do we have a red and a white ball? In case 2 and case 3.
Probability of picking a red and a white with replacement = (2/5)*(3/5) + (3/5)*(2/5) = (3/5)*(2/5) * 2
Since the probability of picking a red and then a white is same as probability of picking a white and then a red, you simply write down one case and multiply it by 2. You do the same in case of 'without replacement' too. The only thing that changes is the probability.

Without Replacement:
1. You pull a Red and then a Red again RR - (2/5)*(1/4)
2. You pull a Red and then a White RW - (2/5)*(3/4)
3. You pull a White and then a Red WR - (3/5)*(2/4)
4. You pull a White and then a White WW - (3/5)*(2/4)
Probability of picking a red and a white WITHOUT replacement = (2/5)*(3/4) + (3/5)*(2/4) = (3/5)*(2/4) * 2

As for the link you have mentioned, this is exactly what is done there too. Check it out - I will show you how it is done there.
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3R, 4B, 2W balls.

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

Note that you cannot use 9C2 here because you are drawing balls with replacement.
9C2 means draw 2 balls out of 9 or draw 1 out of 9 and then 1 out of 8.
But here, we need to draw 1 out of 9 and then again 1 out of 9 (since the first ball is put back)

Also, why only 4? Wouldn't we have other cases too such as a Red and a Black?

Look at the solutions above to see how to solve it.
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Hi All,

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:

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