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705-805 Level|   Probability|      
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Bunuel
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A box contains some Blue balls and 9 Red balls. If the probability tha 04 Jan 2021, 10:01
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63% (02:50) correct 37% (03:14) wrong based on 186 sessions

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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 probability that the third ball randomly removed without replacement of the balls drawn is a blue ball?"

A. 2/9
B. 3/11
C. 3/10
D. 9/55
E. 3/7
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C
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A box contains some Blue balls and 9 Red balls. If the probability tha 04 Jan 2021, 10:39
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Number of blue balls is 3 by using 9C2/(9+x)C2 = 6/11 for the initial condition.
The equation I used for the second condition was (9C2 x 3C1)/((9C2 x 3C1) + 9C3) and I got 9/16.
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A box contains some Blue balls and 9 Red balls. If the probability tha 09 Mar 2022, 04:03
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Bunuel
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 probability that the third ball randomly removed without replacement of the balls drawn is a blue ball?"

A. 2/9
B. 3/11
C. 3/10
D. 9/55
E. 3/7
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Let there be 'x' blue balls. It is given that there are 9 red balls.
Given- first two balls removed without replacement are red, so the probability of this = \(\frac{9}{(9+x)} * \frac{8}{(8+x)} = \frac{6}{11}\). Solving this you'd get \(x = -20 , 3\). Since, \(-20\) is not possible, you \(x = 3\) i.e no. of Blue balls.
Now, the question asks what's probability that the third ball randomly removed without replacement of the balls drawn is a blue ball?
And here's where the confusion arises, if the question solely wants to know the probability of an individual third event, then it would be \(\frac{3}{10}\), which is option (C), not 3/12 as two red balls have been removed from the available set.
But, if the question wants to know the probability of the third ball being blue given that the first two are red, then we would need to multiply \(\frac{3}{10}\)*\(\frac{6}{11}\) = \(\frac{9}{55}\), which is option (D).
So, I am not quite sure what is it that the question really wants to know. Dear IanStewart your thoughts? Is this language gmat-like? Or am I just not reading the question properly?
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A box contains some Blue balls and 9 Red balls. If the probability tha 09 Mar 2022, 05:40
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PyjamaScientist

Given- first two balls removed without replacement are red, so the probability of this = \(\frac{9}{(9+x)} * \frac{8}{(8+x)} = \frac{6}{11}\). Solving this you'd get \(x = -20 , 3\). Since, \(-20\) is not possible, you \(x = 3\) i.e no. of Blue balls.
Now, the question asks what's probability that the third ball randomly removed without replacement of the balls drawn is a blue ball?
And here's where the confusion arises, if the question solely wants to know the probability of an individual third event, then it would be \(\frac{3}{10}\), which is option (C), not 3/12 as two red balls have been removed from the available set.
But, if the question wants to know the probability of the third ball being blue given that the first two are red, then we would need to multiply \(\frac{3}{10}\)*\(\frac{6}{11}\) = \(\frac{9}{55}\), which is option (D).
So, I am not quite sure what is it that the question really wants to know. Dear IanStewart your thoughts? Is this language gmat-like? Or am I just not reading the question properly?
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First, you can avoid the quadratic altogether here. If the probability of picking two red balls in a row is as high as 6/11, most of the balls will need to be red. If there are n balls in total, the probability the first two are red is

9/n * 8/(n-1)

and for that to equal 6/11, we need to get an '11' in the denominator somehow, so if n is not much larger than 9, n can only be 11 or 12. If n were anything else not much bigger than 9, the product above would never cancel down to 6/11. So we can just check either of those two values, or notice that 11 is clearly too small, and find n = 12 that way. So there are 3 blue and 9 red balls.

Now the wording of the question is problematic, and as I read it, I'd think the answer is 3/12, which is not among the choices. The question could mean one of three things:

- knowing there are 3 blue and 9 red balls, if you pick three balls without replacement, what is the probability the third ball is blue? The answer to this question is 3/12, and this is the way I'd interpret the wording. The question never explicitly says that we know in advance we've made two red selections before picking the third ball, which is why I'd never assume that reading a question with this wording.

- knowing that there are 3 blue and 9 red balls, if you pick three balls without replacement, and you know the first two balls are red, what is the probability the third ball is blue? This is not what the question says, but it's what the question means, and the answer to this question is 3/10.

- knowing that there are 3 blue and 9 red balls, if you pick three balls without replacement, what is the probability the first two balls are red and the third ball is blue? This is the question you answered above, where you got 9/55. Of the three possible interpretations, this is the only one I don't think is reasonable; the question only asks about the probability the third selection is a certain colour, not that the first three selections are certain colours, so we shouldn't be multiplying by 6/11.

I understood your meaning, but I've highlighted one phrase in your post that I think you didn't mean to use -- if we say "given the first two balls are red, what is the probability the third ball is blue?" that means "assuming you know the first two selections were red, what is the probability the third selection is blue?" which is what the question means to ask, and the answer to that question is 3/10. You instead meant something like "given the probability the first two are red is 6/11, what is the probability the first two are red and the third is blue?" and the answer to that question is 9/55.

edit: I forgot which colours were which when writing my post, so I had "red" and "blue" backwards, hopefully I fixed everything.
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A box contains some Blue balls and 9 Red balls. If the probability tha 09 Mar 2022, 06:06
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We know the no. of red balls but we don't know no. of blue balls so that let that be x. In the question it is given probability that two balls randomly removed without replacement from the box are Red is 6/11 which means---> 6/11= 9/ (9+x)* 8/(8+x). hence we come to know that x is 3. Which means there were total of 12 balls- 9 red and 3 blue. Now the question asks- what is the probability that the third ball randomly removed without replacement of the balls drawn is a blue ball-- ans is 3/10 (we take 3 because there are 3 blue balls and 10 balls left in the bag after two balls randomly removed without replacement ).

Hope this helps!!
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A box contains some Blue balls and 9 Red balls. If the probability tha 09 Mar 2022, 09:09
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I think this question could have been worded better.

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 probability that the third ball randomly removed without replacement of the balls drawn is a blue ball?"

The sections highlighted in red make the question confusing as it appears a sequential process. I would love to know what would be the question if the first two balls had to be considered in the probability calculation.
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A box contains some Blue balls and 9 Red balls. If the probability tha 02 Oct 2023, 01:33
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First two without replacement being red means

9/x * (8/(x-1)) = 6/11 --> 6(x)(x-1) = 9*8*11 --> x(x-1) = 3*4*11 =12*11 --> x^2 - 1 -12*11 = 0 --> (x-12)(x+11) = 0 x=12. so B = 3
so p = 3/10 OA is C
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A box contains some Blue balls and 9 Red balls. If the probability tha 14 Nov 2025, 11:43
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