PyjamaScientist wrote:
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?
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.