1

Suppose there are 30 balls in an urn, 2 red, 2 black and the other 26 white.

After selecting at random 4 balls without replacing, what is the probability of having at least one red and at least one black ball? Also, what if I were to continue picking balls one by one after the initial 4, how would I go about calculating that.

I know how to use the hypergeometric distribution to know the probability of getting at least one red ball, but I don't know how to compute for one red and one black.

Jürgen
  • 121

2 Answers2

1

I would say (analogous to hypergeometric probability distribution):

When four draft:

$\displaystyle P(red = r, black = b)=\frac{{2 \choose r}{2 \choose b}{26 \choose 4-r-b}}{{30 \choose 4}},\quad r,b\in\langle 0,2 \rangle$

$\displaystyle P(red >0, black >0)=\sum_{r=1}^2 \left(\sum_{b=1}^2 \frac{{2 \choose r}{2 \choose b}{26 \choose 4-r-b}}{{30 \choose 4}}\right)=\frac{281}{5481}$

When 'n > 4' draft

$\displaystyle P(red >0, black >0)=\sum_{r=1}^2 \left(\sum_{b=1}^2 \frac{{2 \choose r}{2 \choose b}{26 \choose n-r-b}}{{30 \choose n}}\right)$

E.g.: n = 28, P = 0.995; n = 29, P = 1; n = 30, P = 1

georg
  • 2,749
  • I am trying to recreate the result of your second example:

    P(red>0,black>0)=∑r=12(∑b=12(2r)(2b)(264−r−b)(304))=2815481

    So, the result of both sigmas is 1+2= 3, so 3 * (3 * the result of the division, which for the numerator would be 44325 divided by the denominator 27405). How you come to 281/5481 eludes me completely. Could you perhaps enlighten me? Much appreciated!

     – Jürgen Oct 05 '15 at 16:41
  • Sorry, I made a typo, It should have been 22325 and not 44325. – Jürgen Oct 05 '15 at 17:23
  • I posted the question here: http://math.stackexchange.com/questions/1465861/unable-to-recreate-the-result – Jürgen Oct 05 '15 at 18:05
  • @Jürgen I saw that you understand it (-: I had little time, so I let the machine count: http://www.wolframalpha.com/input/?i=%5Csum_%7Br%3D1%7D%5E2+%5Cleft%28%5Csum_%7Bb%3D1%7D%5E2+%5Cfrac%7B%7B2+%5Cchoose+r%7D%7B2+%5Cchoose+b%7D%7B26+%5Cchoose+4-r-b%7D%7D%7B%7B30+%5Cchoose+4%7D%7D%5Cright%29 – georg Oct 06 '15 at 08:17
0

I'm afraid I don't know of a clever way to solve this problem, you could enumerate all the possibilities of getting at least one red and one black (e.g. [b,r,white,white] and all permutations, [b,b,r,white] and all perms, [r,r,b,white] and all perms and [r,r,b,b] and all perms), calculate the probabilities for each via the multinomial distribution and then sum them together.

David
  • 66