Simple Probability Q

[Ncoe]

The reason I post this is not to have it solved, but to discuss and clearify if the to me given answer really is correct.

But just solve it and give your reasoning and I'll write what it's "supposed" to be in a while.

The Q:
There's four balls in a bag. One black, one blue and two red. The bag is shaken and some guy takes two of the balls out. He lookes at the balls and states that one of them is red. What is the probability that the other one is red, aswell?

[rufus]

This is the 'supposed to be' answer:
The probability that both balls will be red is 1/6
The probability that at least one ball is red is 5/6
Therefore the probability that the other ball is red (1/6)/(5/6)=1/5.

This assumes that the balls are picked with equal probability, and that the guy pulling them says that one is red every time that there is at least one red ball.

The answer you probably wanted to see is 1/3, but that's incorrect here. Despite your labeling the thread to the contrary, this isn't really a simple probability question.

[jay_shark]

1)B=black , b=blue , r1=red 1, r2=red2

There are 4c2=6 ways to select two balls .
Bb,(Br1,Br2,br1,br2,r1r2)

Notice that only 1 in 5 is the prob. that both balls are red . Bb is ommitted from our sample space since neither of the balls are red . This means our sample space consists of 5 elements and only r1r2 satisfies the condition stated in the question .

[im a model]

to help you (or whomever youre proving this to) understand it better, think of it not as two red balls, but a red and a green ball. and the guy says, "one of the balls is red or green." then it becomes clear that the only thing that this information changes is the one instance where there is no red or green ball (the one case where there is a black ball and a blue ball). only when you think of the two red balls as being "the same" does the problem seem difficult.

[UtzChips]

16.67%

[HoldenFoldem]

He has removed one red ball already. there are now 3 balls in the bag. the chance of removing the red one is 33.33%

[HoldenFoldem]

oops , misread the question.
Odds are 2/4 times 1/3 = 1/6 = 16 2/3%

[Copernicus]

[ QUOTE ]
oops , misread the question.
Odds are 2/4 times 1/3 = 1/6 = 16 2/3%

[/ QUOTE ]

now read the correct answers above. its 1/5