# Hypergeometric Distrbution issue

So I am trying to figure out some probabilities using hypergeometric distribution, and I am a little confused on one part I am trying to figure out.

I know how to figure out what the probability is of me getting X in sample Y is. However, I have no idea how to figure out what the probability would be of me getting X and Z in sample Y.

Let me give you an example of what I mean.

I have a basket with 5 Nectarines, 5 Apples, 5 Pears, and 5 Oranges. I want to grab, lets say 5 fruit from the basket. Of the 5 fruit I need exactly 1 or more apple, and exactly one or more pears, yet I can't have 0 of either (for example, 1 apple, 3 pears, and 1 orange is acceptable, but 1 apple, 2 oranges, and 2 nectarines is not).

How do I figure this out exactly?

• I should point out that the hypergeometric distribution in probability is (mostly) distinct from the hypergeometric special functions. The 'almost' comes from fact that the cumulative density function can be written in terms of the hypergeometric functions. In the context of your problem, though, I don't see that as being much help. – Semiclassical Sep 3 '14 at 18:39

In total, you're drawing 5 pieces of fruit from the basket containing 20. You need at least one apple and at least one pear. So let's classify the fruits into apples, pears and others. First, we can use the hypergeometric distribution to find a specific outcome: $$P(\text{exactly 1 apples, 2 pears})=\frac{{5 \choose 1} \cdot {5 \choose 2} \cdot {10 \choose 2}}{20 \choose 5}$$ $20\choose 5$ is in the denominator because that's the total number of fruit you're picking, and in the numerator you have subsets of fruits with the specific numbers you desire. Take care that the sums within the binomial coefficients will match - i.e. $5+5+10=20$, $1+2+2=5$.
Now we're interested in all the possible combinations. So we can use the sum notation to go through all the possible cases where you fulfill your conditions. We can sum from 1 to 4 apples (since that's the most apples you can have). Then we see that the maximum number of pears is $5-\text{apples}$, and the minimum number of pears is 1. The sum of all other fruit in your sample must equal $5-\text{apples}-\text{pears}$, so we get the following double sum as your answer: $$P(\text{at least 1 apple, at least 1 pear})=\sum\limits_{i=1}^{4} \sum\limits_{j=1}^{5-i}\frac{{5 \choose i}\cdot{5 \choose j}\cdot{10 \choose {5-i-j}}}{20 \choose 5}$$