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.


1 Answer 1


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:

enter image description here

and this suggests that the resulting probability does tend toward the multinomial probability we found above.


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