This type of problem is not quantum theory or differential geometry. It only requires a bit of logic and the ability to understand the semantics of some simple English language sentences. I am reminded of high school classmates who flunked easy courses in mathematics and English. I did not understand what was wrong with them and I do not understand what is missing in the minds of some who posted to this thread. It is not stupidity, although one might jump to that conclusion. I apologize if some posters are hampered due to English being their second language, which would explain some of the nonsense being posted.
Really? Lets think about that. Lets call the the always solid ball A, and the indeterminate ball B. There are four possibilities for the first draw. 1. You draw ball A, and ball B was solid 2. You draw ball A, and ball B was striped 3. You draw ball B, and ball B was solid 4. You draw ball B, and ball B was striped Without your freak coincidence, all 4 possibilities are equally likely. Your coincidence makes possibility 4 never happen, but the other three possibilities are still all equally likely. The results of the second draw would be 1. Solid 2. Striped 3. Solid So, out of three equally likely possibilities, two are solid, and one is striped. The odds of pulling a solid on the second draw with your freak coincidence affecting the first draw is 2/3.
Still, it stretches my imagination how one can possibly read A ball is now selected randomly from the bag and it turns out to be a solid as A ball is now selected randomly from the bag and it turns out to be either a solid or a striped but we ignore the striped ones later on. But just in case, here is a simplified example which illustrates the error some people have been making here: assume two candidates A and B which are to be elected by a group of 100 people. Now, the election is not a normal election but the constraint on it is that all vote for candidate A. So, how many votes does candidate A get then? Obviously 100. But some posters in this thread argue that A and B initially both get 50 votes, then they remember that A should get all the votes and reduce B's votes to 0, leaving A with 50 and B with 0. This is obviously an incorrect result due to not applying the constraint correctly. Thomas
No one has done this, no one is ignoring anything. It is disingenious for you to claim to understand how anyone else in this thread would approach your proposed problem (which is very different from the original) based on the arguments they have given when it is quite clear you couldn't tell a conditional probability from a hole in the ground. You can either admit your ignorance when it comes to conditional probability and take steps to rectify this gap or not. If you choose the latter, I suggest you just give up discussing this problem, you will never understand it or the replies people have given.
You should read about Rejection Sampling, which shows why the two are equivalent: http://en.wikipedia.org/wiki/Rejection_sampling
Ahh, I interpreted that red bit completely differently. If he's refering to this kind of rejection sampling, then it's the same as the original problem.
If some sneaky aditions are allowed then, that changes every thing to Err~ to Err~ :bugeye: Please Register or Log in to view the hidden image! {Days later} I got it! No wait! Some one may have snuck in more balls while I wasnt looking~ Please Register or Log in to view the hidden image!
One of the most funny threads , that I have seen for long : True answer : 1/3 A= first solid ball B= second solid ball C= striped ball Posibilities in the bag - first ball drawn mentioned first: AB BA AC CA = we know that can not be the case , because first ball drawn was solid !! So remaining possibilities : AB BA AC only one of these 3 possibilities gives a striped ball in the second draw !!! Please Register or Log in to view the hidden image!
That is not correct because the remaining possibilities are not of equivalent value. Consider the following: Solid 1 = A (known to be in bag) Solid 2 = B (50% chance pick) Stripe 1 = C (50% chance pick) Possibilities after first pick AB = 50% AC = 50% BA is not possible (solid 1 is clearly given as first) CA is not possible (solid 1 is clearly given as first) Examining the first scenario, AB, we can conclude that the chance of then picking a stripe would be 0%. Even though there are two ways it can happen (picking A or B first), there is a combined 50% chance of a 0% outcome. 50% * 0% = 0% Examining the AC scenario, we are told that a randomly picked ball is solid. Given that this event occured, there is only one possibility left and that is for the remaining ball to be stripe (100%). 50% * 100% = 50% 50% + 0% = 50% chance of picking a stripe under these conditions. If one wants to test this experimentally, one would need to disregard the cases in which the second pick turns out to be a stripe.
Sorry - you are sooooooooooooooooooo wrong !!!! 1/3 is the correct answer - just ask soduko !!! Please Register or Log in to view the hidden image! I think you can even send her a private message !! This is a VERY simple problem - NO doubt 1/3 is correct ....... This thread is getting more and more funny Please Register or Log in to view the hidden image!
so what the answer be if the probability of picking a solid for the second ball was 1/10 instead of 1/2?
