Probability question - balls in urn (hypergeometric?)

[dizzle1518]

Hi all,

I need help with the following problem:

The urn contains 5 black and 8 red balls. You close your eyes and
remove balls from the urn one by one without replacement. What is
the probability that the last ball is black?

This looks to me like it is a hypergeometric distribution problem. I set it up in the following way: ((5 choose 4)*(8 choose 8))/(13 choose 12). In order for the last ball to be black we have to remove 12 balls such that the remaining 13th one is black. so we have 5 choose 4 orderings of the black balls and 8 choose 8 orderings of the red balls. Since we are taking out a total of 12 balls then there are 13 choose 12 possible orderings or the red and black balls.

Is this correct?

Thanks,
--David

 

[Dick]

It's correct, but it's the complicated way to compute the answer. What do you get when you do it that way? Now think about it a different way. What's the probability the first ball is black? What's the probability the second ball is black? What's the probability that ANY ball is black?

 

[dizzle1518]

It's correct, but it's the complicated way to compute the answer. What do you get when you do it that way? Now think about it a different way. What's the probability the first ball is black? What's the probability the second ball is black? What's the probability that ANY ball is black?

would the answer to a modified version of this problem be 5/12?

The urn contains 5 black and 8 red balls. You close
your eyes and remove balls from the urn one by one without replacement.
What is the probability that the last ball is black given that the 1st ball
is red?

 

[Dick]

would the answer to a modified version of this problem be 5/12?

The urn contains 5 black and 8 red balls. You close
your eyes and remove balls from the urn one by one without replacement.
What is the probability that the last ball is black given that the 1st ball
is red?

Sure it is.