Multinomial distribution problem help

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 *= $\displaystyle \sum{Ni}$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

$\displaystyle h(r_{1}, r_{2}, ..., r_{k}|N_{1}, N_{2}, ... N_{k}) \equiv \frac{\binom{N_{1}}{r_{1}} ... \binom{N_{k}}{r_{k}}} {\binom{\sum{N_{i}}}{\sum{r_{i}}}}$

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

$\displaystyle \equiv \frac{\binom{N_{1}}{1} ... \binom{N_{k}}{1}} {\binom{\sum_{i=0}^{k}{N_{i}}}{\sum_{i=0}^{k}{r_{i }}}}. (1 - \frac{\binom{N_{1} - 1}{1} ... \binom{N_{k}-1}{1}} {\binom{\sum_{i=k+1}^{m}{N_{i}}}{\sum_{i=k+1}^{m}{ r_{i}}}})$

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.(Worried)

Re: Multinomial distribution problem help

Or is it just this?

$\displaystyle \frac{\binom{N_{1}}{1} ... \binom{N_{k}}{1}}{\binom{\sum{N_{i}}}{m}}$

Re: Multinomial distribution problem help

Quote:

So probability for drawing m balls without replacement = (Probability of drawing k distinct balls) x (1 - Probability of drawing m-k distinct balls)

Hi skanur,

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.