A Problem my teacher set me

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.
 
tsmid said:
Apart from the repeat method, there is actually a further way to satisfy the constraint, namely if you simply assume that, due to some freak coincidence, you draw only solid balls with the first draw in the first place. This may be an unlikely scenario, but it is the only scenario that satisfies the constraint, and if you evaluate the second draws in this case, you will find that they are distributed 50/50 between solid and striped.

Thomas
Click to expand...

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.
 
Dinosaur said:
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.
Click to expand...
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
 
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tsmid said:
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.
Click to expand...

No one has done this, no one is ignoring anything.

tsmid said:
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.
Click to expand...

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.
 
tsmid said:
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.
Click to expand...

You should read about Rejection Sampling, which shows why the two are equivalent:

http://en.wikipedia.org/wiki/Rejection_sampling
 
Pete said:
That's right, LaidBack. But there might be another solid ball in the bag, not a striped one.
Click to expand...

If some sneaky aditions are allowed then, that changes every thing to Err~ to Err~ :bugeye: :eek:

{Days later}

I got it!

No wait!

Some one may have snuck in more balls while I wasnt looking~ :D
 
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 !!!

;)
 
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.
 
sleeper555 said:
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.
Click to expand...

Sorry - you are sooooooooooooooooooo wrong !!!!
1/3 is the correct answer - just ask soduko !!! :)
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 :p
 
so what the answer be if the probability of picking a solid for the second ball was 1/10 instead of 1/2?
 
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