Let the number of blue balls is x, so we have (9/x+9) * (8/x+8) = 6/11

Solving we get x = -20 or 3, we go by x = 3 as -20 is not possible.

P(Picking a third blue ball) = (x/x + 7) = [spoiler]3/10[/spoiler].

varun289 wrote:296. A box contains some Blue balls and 9 Red balls. If the probability that two balls randomly removed without replacement from the box are Red is 6/11, what is the total probability that the third ball randomly removed without replacement of the balls drawn is a blue ball?

Since P(RR) = 6/11, most of the balls must be red.
When we calculate probability WITHOUT REPLACEMENT, each selection reduces the total number of marbles by 1.
The result is that the denominators of the probabilities are consecutive integers in descending order:
5/12 * 4/11 * 3/10...
In order for P(RR) to yield a denominator of 11, the total number of marbles likely is 12:
P(RR) = 9/12 * 8/11 = 6/11.
This works.
Thus, at the start there are 9 red marbles and 3 blue marbles, for a total of 12 marbles.

The wording of the question stem is unclear.
I believe that the phrase "total probability" intends to ask the following:
What is the probability of selecting two red marbles and then a blue marble?
P(RRB) = 9/12 * 8/11 * 3/10 = 9/55.

It is possible -- but unlikely -- that the question stem intends to ask the following:
What is probability of selecting a blue marble on the third pick, REGARDLESS of what is chosen on the first two picks?
In this case, we should take advantage of the following rule:
The probability of selecting X on the NTH pick is equal to the probability of selecting X on the FIRST pick.
Thus, P(blue on the 3rd pick) = P(blue on the first pick) = 3/12 = 1/4.

Other problems that ask for the probability of selecting X on the Nth pick:

https://www.beatthegmat.com/probablity-ques-t60161.html
https://www.beatthegmat.com/manhattan-pr ... 89481.html (2 posts)

GMATGuruNY wrote
It is possible -- but unlikely -- that the question stem intends to ask the following:
What is probability of selecting a blue marble on the third pick, REGARDLESS of what is chosen on the first two picks?
In this case, we should take advantage of the following rule:
The probability of selecting X on the NTH pick is equal to the probability of selecting X on the FIRST pick.
Thus, P(blue on the 3rd pick) = P(blue on the first pick) = 3/12 = 1/4.

Other problems that ask for the probability of selecting X on the Nth pick:

My question is as the process is without replacement and now the number of balls left after two pick ups is 10(7 Red + 3 Blue), so P(picking a blue ball) should be 3/10 and not 3/12.

Please advise!!

Puneet
[/quote]

puneetkhurana2000 wrote:

GMATGuruNY wrote
It is possible -- but unlikely -- that the question stem intends to ask the following:
What is probability of selecting a blue marble on the third pick, REGARDLESS of what is chosen on the first two picks?
In this case, we should take advantage of the following rule:
The probability of selecting X on the NTH pick is equal to the probability of selecting X on the FIRST pick.
Thus, P(blue on the 3rd pick) = P(blue on the first pick) = 3/12 = 1/4.

Other problems that ask for the probability of selecting X on the Nth pick:

My question is as the process is without replacement and now the number of balls left after two pick ups is 10(7 Red + 3 Blue), so P(picking a blue ball) should be 3/10 and not 3/12.

Please advise!!

Puneet

If the question stem is asking for the probability of selecting a blue marble on the 3rd pick -- REGARDLESS of what is chosen on the 1st two picks -- then there are several ways to get a favorable outcome:

P(BBB) = 3/12 * 2/11 * 1/10 = 1/220.
P(BRB) or P(RBB) = 3/12 * 9/11 * 2/10 * 2 = 18/220.
P(RRB) = 9/12 * 8/11 * 3/10 = 36/220.

Since P(BBB) or P(BRB) or P(RBB) or P(RRB) will yield a blue marble on the 3rd pick, we ADD the probabilities:
1/220 + 18/220 + 36/220 = 55/220 = 1/4.

Note that this result is equal to P(B on the FIRST PICK):
3/12 = 1/4.

Hence, P(B on the 3rd pick) = P(B on the 1st pick) = 1/4.

Hi Mitch,

I am able to deduce that number of blue balls are 3
so total are 12 balls ( 9 RED, 3 blue)

Why probability of third blue ball is

9/12 * 8/11 * 3/10 = 18/110

Please tell where I am wrong. Since question is asking for total probability. I put a AND operation

first two RED balls & then a blue ball.

why you are just taking the probability of blue ball only. Also why 3/12 - why not 3/10

GMATGuruNY wrote:

puneetkhurana2000 wrote:

GMATGuruNY wrote
It is possible -- but unlikely -- that the question stem intends to ask the following:
What is probability of selecting a blue marble on the third pick, REGARDLESS of what is chosen on the first two picks?
In this case, we should take advantage of the following rule:
The probability of selecting X on the NTH pick is equal to the probability of selecting X on the FIRST pick.
Thus, P(blue on the 3rd pick) = P(blue on the first pick) = 3/12 = 1/4.

Other problems that ask for the probability of selecting X on the Nth pick:

My question is as the process is without replacement and now the number of balls left after two pick ups is 10(7 Red + 3 Blue), so P(picking a blue ball) should be 3/10 and not 3/12.

Please advise!!

Puneet

If the question stem is asking for the probability of selecting a blue marble on the 3rd pick -- REGARDLESS of what is chosen on the 1st two picks -- then there are several ways to get a favorable outcome:

P(BBB) = 3/12 * 2/11 * 1/10 = 1/220.
P(BRB) or P(RBB) = 3/12 * 9/11 * 2/10 * 2 = 18/220.
P(RRB) = 9/12 * 8/11 * 3/10 = 36/220.

Since P(BBB) or P(BRB) or P(RBB) or P(RRB) will yield a blue marble on the 3rd pick, we ADD the probabilities:
1/220 + 18/220 + 36/220 = 55/220 = 1/4.

Note that this result is equal to P(B on the FIRST PICK):
3/12 = 1/4.

Hence, P(B on the 3rd pick) = P(B on the 1st pick) = 1/4.

Why you didn't consider P(BBR) = 3/12 * 2/11 * 9/10