A box contains some Blue balls and 9 Red balls. If the probability tha

Question

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

A. 2/9
B. 3/11
C. 3/10
D. 9/55
E. 3/7

RookieDaCookie · Answer

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.

PyjamaScientist · Answer

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?

IanStewart · Answer

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.

Kirtika2507 · Answer

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!!

sanjitscorps18 · Answer

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?"

A. 2/9
B. 3/11
C. 3/10
D. 9/55
E. 3/7

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.

ocelot22 · Answer

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

Adit_ · Answer

So basically since there is no replacement you will notice denominator is either 12*11 or 11*10 since the prob is 6/11 and we also know both are red so 9*8 is the numerator for sure so by using these two we can say denominator is 12*11 which means the prob is being carried out of 12 balls or 9 red balls and 3 blue balls.

A box contains some Blue balls and 9 Red balls. If the probability tha

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