Probability (1 Viewer)

[YannY]

a bag contains two red balls, one black ball, and one white ball. andrew selects one ball from the bag and keeps it hidden. he then selects a seconda ball, also keeping it hidden.
andrew drops one of the selected balls and we can see that it is red. what is the probability that the ball that is still hidden is red?

I would have said 1/3 but the answer is 1/5...

I have already thought about the answer being wrong but i want to consider if keeping the ball hidden had something to do with it lolz.

 

[lyounamu]

a bag contains two red balls, one black ball, and one white ball. andrew selects one ball from the bag and keeps it hidden. he then selects a seconda ball, also keeping it hidden.
andrew drops one of the selected balls and we can see that it is red. what is the probability that the ball that is still hidden is red?

I would have said 1/3 but the answer is 1/5...

I have already thought about the answer being wrong but i want to consider if keeping the ball hidden had something to do with it lolz.

I reckon it is 1/3 as well.

 

[vds700]

1/3 seems correct to me

 

[bored of sc]

1/3 must be the correct answer - the answers are incorrect...

unless all four balls are involved in the second selection - which in that case the answer would be 1/2... hmm?

 

[lolokay]

a bag contains two red balls, one black ball, and one white ball. andrew selects one ball from the bag and keeps it hidden. he then selects a seconda ball, also keeping it hidden.
andrew drops one of the selected balls and we can see that it is red. what is the probability that the ball that is still hidden is red?

I would have said 1/3 but the answer is 1/5...

I have already thought about the answer being wrong but i want to consider if keeping the ball hidden had something to do with it lolz.

there are 12 ways of selecting two balls. there are 2 ways of not selecting red, ie 10 ways of selecting red. 2 of these ways include two red balls. so, probability that they were both red is 2/10 = 1/5

btw: the question doesn't really make sense, since you have to assume that the guy has chosen to show a red ball - that if he had a red and black ball for example, that he wouldn't show the black one. really it should indeed be 1/3. (reminds me of the monty hall paradox)

 

[milton]

assuming the question is interpreted as: he randomly selects 2 balls initially, then given that he randomly chooses one of those 2 balls and reveals it to be red, then the probability of the other ball being red is indeed 1/3

if instead, he randomly selects 2 balls, and reveals that at least one of those 2 balls is red, then we would conclude the probability of of the "other ball" being red is 1/5. the soln is-
the sample space consists of
{ R1, R2 }
{ R1, B }
{ R1, W }
{ R2, B }
{ R2, W }
{ B, W}
All 6 of these are equally likely, but we're given that at least one of them is red, so all of the above except the last one are equally likely. Then clearly, the prob of the "other ball" being red is 1/5.

the question might be worded slightly ambigiously
see http://en.wikipedia.org/wiki/Boy_or_Girl_paradox for something similar

 

[YannY]

Ah i get it, thanks guys but honestly i think this Q is stupid.

 

[morning storm]

i like how 2 pius guys have posted here... im guessing vds is andy and namu is well... namu...