# Is there a multivariate hypergeometric distribution?

I'm attempting a problem that is as follows:

In a bag full of fruits, $$13\%$$ are apples, $$37\%$$ are oranges, $$20\%$$ are bananas, $$10\%$$ are pears, and $$20\%$$ are avocados. Find the probability that $$12$$ randomly selected fruits are $$3$$ apples, $$2$$ oranges, $$5$$ bananas, $$1$$ pear, and $$1$$ avocado.

At first glance this looks like a hypergeometric problem if we pretend there are $$100$$ fruits in total inside the bag, where $$p_{\text{fruit}}(x)=\frac{\binom{b}{x}\binom{r}{k-x}}{\binom{b+r}{k}}$$. Where $$p$$ is the number of fruits in the bag, $$x$$ the number of a specific fruit we want, $$r$$ is the number of other fruits, and $$k$$ is the total number of fruits we are selects. for example $$p_{\text{apples}}(3)=\frac{\binom{13}{3}\binom{100-13}{12-3}}{\binom{100}{12}}$$. Then the desired probability would just be $$p_{\text{apples}}(3)\ \ \ p_{\text{oranges}}(2)\ \ \ p_{\text{banana}}(5)\ \ \ p_{\text{pear}}(1)\ \ \ p_{\text{avocado}}(1)$$. However I'm not sure if this is a valid move, so I'm wondering if there is such a thing as a multivariate hypergeometric that considers $$k$$ types of objects instead of just two.

The question as stated implies a multinomial distribution, because the total number of fruits in the bag is not stated; therefore, we presume that its value is sufficiently large that the sampling, although technically without replacement, is very nearly equivalent to sampling with replacement.

However, your point is valid: if we had exactly $$n = 100$$ fruits in the bag and the distribution of the types of fruits were not percentages but actual numbers, then we would have something akin to a multivariate hypergeometric distribution. The probability of the desired outcome is

$$\frac{\binom{13}{3}\binom{37}{2}\binom{20}{5}\binom{10}{1}\binom{20}{1}}{\binom{100}{12}} = \frac{7970688}{14175722687}.$$

Note we do not need multinomial coefficients, just binomial.

If we use the multinomial model, the probability is

$$\frac{12!}{3! 2! 5! 1! 1!} (0.13)^3 (0.37)^2 (0.20)^5 (0.10)^1 (0.20)^1 = \frac{6252993747}{9765625000000}.$$

To show that the first type of calculation tends toward the second, we can assume $$n = 100k$$ for some positive integer $$k$$, and compute $$P(k) = \frac{\binom{13k}{3}\binom{37k}{2}\binom{20k}{5}\binom{10k}{1}\binom{20k}{1}}{\binom{100k}{12}}$$ for successive integer values. This gives us the following plot: and this suggests that the resulting probability does tend toward the multinomial probability we found above.