Or is it just this?
Hi,
I am trying to solve a problem from the book 'Probability Theory, The logic of Science' from chapter 3 page 71. The problem goes like this:
Suppose an urn contains N = balls, N1 of color 1, N2 of color2, . . . , Nk of color k. We draw m balls without replacement; what is the probabilitythat we have at least one of each color? Supposing k = 5, all Ni = 10, how many do we need to draw in order to have at least a 90% probability for getting a full set?
My solution goes something like this:
For a multinomial distribution where sampling is carried out without replacement, the probability of drawing exactly r1, r2, ..., rk balls from N1, N2, ... Nk total balls is given by
To draw m balls which have atleast k distinct colors, m should be greater than k.
So probability for drawing m balls without replacement = (Probability of drawing k distinct balls) x (1 - Probability of drawing m-k distinct balls)
This reduces to
As I am studying myself, I have no means to verify. Is this approach correct?
As to second part of the problem, I took probability of atleast one of each color as 0.9 and tried to solve for m. But m occurs in binomial expression and even by trial and error I could not get a value of m which could satisfy the above equation.
Can someone help me? Hope I was clear enough.
Hi skanur,So probability for drawing m balls without replacement = (Probability of drawing k distinct balls) x (1 - Probability of drawing m-k distinct balls)
I don't think that's correct-- it seems to assume that the events "draw k distinct balls" and "draw m-k distinct balls" are independent, which I don't think is true.
Are you familiar with the Inclusion/Exclusion principle? If so, this might be a good place to apply it. Suppose a drawing has "property i" if there are no balls of color i drawn. I think it should be possible to apply Inclusion/Exclusion to find the probability of a drawing with none of the properties 1, 2, ..., k. Then you are done, since such a drawing has at least one ball of each color.