probably a super easy question
 

Hey - I have an interview tomorrow to which my possible employer told me there will be 'basic probability questions.' In surfing the web to prepare, I stumbled upon one that I am having trouble wrapping my mind around.

There are two bags of balls in front of you. You look inside and see that in 1 there are 10 red balls, and in one there are 5 red balls and 5 green balls. You close your eyes, and someone spins the bags around after closing them. You reach inside and select a red ball. You close your eyes again, he spins the bags, and you reach in and select another red ball. What are the chances that both balls you picked were from the bag with 10 red balls?


Seems like it should just be 10/15*9/14...but I am not sure. Sorry because I'm sure this is super basic.

James

Re: probably a super easy question
 

P(10R)/ P(10R) + P(1 each) + P(5/5)
.25(1)/.25(1) + .5(.5) + .25(.5)(.4) = 5/11 = .45

Re: probably a super easy question
 

I suck at those but here is my try :

You could draw 2 red balls in (15 2) = 105 ways (every pair with same probabiilty).
2 red balls from first bag in (10 2) = 45 ways

So its 45/105 = 9/21.

Explanation :
I think its like that : there is 190 possibilites of drawing (including green balls). We are only interested in draws which has 2 red balls , they are 105 of them (also there are 75 draws with one green and on red and 10 with both green).
So now we calculate probability of drawing 2 red balls from 1st bag assuming one OUT OF THOSE 105 draws happened because we know drawing 2 red balls already happened (bayes theorem).
So its 45/105.

EDIT : Sry for my english I dont have time to correct it now I hope reasoning is clear; would love to have somebody smart to comment on it.

Re: probably a super easy question
 

perfect guys, thanks so much.

James

Re: probably a super easy question
 

[ QUOTE ]
I suck at those but here is my try :

You could draw 2 red balls in (15 2) = 105 ways (every pair with same probabiilty).
2 red balls from first bag in (10 2) = 45 ways

So its 45/105 = 9/21.

Explanation :
I think its like that : there is 190 possibilites of drawing (including green balls). We are only interested in draws which has 2 red balls , they are 105 of them (also there are 75 draws with one green and on red and 10 with both green).
So now we calculate probability of drawing 2 red balls from 1st bag assuming one OUT OF THOSE 105 draws happened because we know drawing 2 red balls already happened (bayes theorem).
So its 45/105.

EDIT : Sry for my english I dont have time to correct it now I hope reasoning is clear; would love to have somebody smart to comment on it.

[/ QUOTE ]

This solution is invalid because the 105 ways to draw 2 red balls are not equally likely. Huskerfan has the right answer assuming that the 1st red ball is not replaced before the 2nd spin. The easiest way to solve this type of Bayes' theorem problem is to compute the probability of each case as follows:

The probability of selecting the bag with all reds both times is (1/2)*(1/2) = 1/4, and if we select this bag both times, then the probability of both balls being red is 1, so the probability of picking this bag both times and getting both reds is (1/4)*1 = 1/4. Now the probability of selecting the bag with both red and green both times is also (1/2)*(1/2) = 1/4, and if we select this bag both times, then the probability of both balls being red is (5/10)*(4/9) = 2/9, assuming that the 1st ball is not replaced before the 2nd spin, so the probability of picking this bag both times and getting both reds is (1/4)*(2/9) = 1/18. The remaining possibility is that we pick one of each bag, and this has probability 1/2. In this case, the probability of both balls being red is just the probability that we draw a red ball from the bag with half reds, or 1/2, so the probability of picking one of each bag and getting both reds is (1/2)*(1/2) = 1/4. So the probability of NOT picking the bag with all reds both times and picking 2 red balls is 1/18 + 1/4 = 11/36, while the probability of picking the bag with all reds both times and picking both reds is 1/4 = 9/36, so the odds are 11:9 against having picked the bag with all reds both times, or a probability of 9/20 = 0.45.

Note that if the 1st ball is replaced before the 2nd spin, this would change the probability of the 2nd case from (1/4)*(2/9) to (1/4)*(5/10)*(5/10) = 1/16, and this would change the probability of NOT picking the bag with all reds both times and picking 2 red balls to 1/16 + 1/4 = 5/16. Then the odds against having picked the bag with all reds both times would be (5/16):(1/4) = 5:4 or a probability of 4/9.

Re: probably a super easy question
 

Thanks much BruceZ.
I forgot that its more probable to pick a pair of balls from one bag than from two other bags.

Re: probably a super easy question
 

I know this comes a little late for your interview, but this truly is a super easy question which everyone tried to make too complicate. I think you didn't really bother to read the question which was chances of taking both balls from the bag with ten red balls. This is obviously 1/4 or 25%. [img]/images/graemlins/grin.gif[/img]

[ QUOTE ]
Hey - I have an interview tomorrow to which my possible employer told me there will be 'basic probability questions.' In surfing the web to prepare, I stumbled upon one that I am having trouble wrapping my mind around.

There are two bags of balls in front of you. You look inside and see that in 1 there are 10 red balls, and in one there are 5 red balls and 5 green balls. You close your eyes, and someone spins the bags around after closing them. You reach inside and select a red ball. You close your eyes again, he spins the bags, and you reach in and select another red ball. What are the chances that both balls you picked were from the bag with 10 red balls?


Seems like it should just be 10/15*9/14...but I am not sure. Sorry because I'm sure this is super basic.

James

[/ QUOTE ]

Re: probably a super easy question
 

[ QUOTE ]
I think you didn't really bother to read the question which was chances of taking both balls from the bag with ten red balls.

[/ QUOTE ]
trickster, that isn't correct.I think you should be careful when making such statements as the above...
BruceZ's answer is correct. You are confusing two questions. The question is NOT: "What is the probability of picking a ball from the bag with all red balls both times?". The question is: "GIVEN that I have picked 2 red balls, what is the probability that I picked from the bag with all red balls both times?"
These are absolutely not the same thing but its a mistake that many people who don't use Bayesian statistics on a regular basis often make.
You are working out P(all red bag both times)=1/4
The question asks for P(all red bag both times|both balls picked were red) where P(A|B) means probability that A is true given that B is true.
This is computed using P(A|B)=P(B|A)P(A)/P(B) which is what BruceZ did without explicitly stating it (I actually thought it was a very clear explanation).
If this isn't clear (and if you care!), google "Bayes Theorem" and think about it for a while.

Re: probably a super easy question
 

I apologize. I didn't read the question carefully. At first I calculated what are the chances of picking up two red balls. Then I read the question again and got all mixed up. [img]/images/graemlins/blush.gif[/img] Now I understand what this problem is all about and even understand the solution. I do know my math but my english ain't that good [img]/images/graemlins/smirk.gif[/img].

Re: probably a super easy question
 

Hehe, np. Everyone mis-reads stuff from time to time :-)