A bag contains 3 red, 4 black and 2 white balls. What is the probabili

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

This is with replacement case.

\(P=2*\frac{3}{9}*\frac{2}{9}=\frac{4}{27}\)

We are multiplying by 2 as there are two possible wining scenarios RW and WR.

Answer: D.
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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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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

This is how I came up with 2/27. the probability of getting one red is 3/9 (nine is the total number of balls). The probability of getting a white ball is 2/9 (9 again because the ball is put back after each draw) so 3/9*2/9 + 6/81 = 2/27

according to Dookie (who is right) if they are asking for the balls to draw one of the the other, which they are (successive draws) you have to multiply 2/27 by 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

Hello Bunuel,
I m a bit confused about when to consider and when not to consider. I am having a tough time understanding this concept. I was under the impression that in "with replacement" cases, we dont need to consider the cases, however in without replacement cases, scenarios needs to be considered.

In addition,I found this link, where the question is also testing the same concept, but we didn't consider the multiple cases here. Please clarify.
https://gmatclub.com/forum/rich-has-3-gr ... 55253.html


Can you provide a high level conceptual knowledge as in when to consider cases and when not to?
Pls help.

Posted from my mobile device

Thank You Karishma.
I got a key concept here

But i am more comfortable by Combination method

1C3*1C2/ (1C9 * 1C9)
= 6/81 = 2/27

Now we can get this in two ways (as described by u)
2* 2/27 = 4/27

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

karishma Bunuel - Can you please explain as to where my concept is headed in the wrong direction?

At first i saw that there are 9 balls and 2 balls are to be drawn, i selected the 2 balls out of 9 (9C2) giving me the total outcomes and next when i am supposed to draw the balls in the manner of white and red it clicked to me that there are 4 case,
1 - RR
2 - RW
3 - WR
4 - WW

and i applied the same logic to the above 4 cases, but the answer did not match?

Can you please help me with this.

Thanks in advance.

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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There are 9 balls in the bag, so the probability of drawing a red ball is P(Red) = 3/9 = ⅓, 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) = ⅓ 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 ⅓ = 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
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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

Responding to a pm: Combinatorics approach

You can pick a red ball in 3C1 ways (assuming the 3 red balls are R1, R2 and R3). You can pick a white ball in 2C1 ways (assuming the 2 white balls are W1 and W2)

The total number of ways of picking 2 balls with replacement = 9 * 9

You can pick a red and a white ball in (3C1 * 2C1) + (2C1 * 3C1) ways = 12 ways

Required Probability = 12/81 = 4/27
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