# Expected value for a sequence of group samplings from a fixed population

Suppose I've 100 apples where 25 of them are bad and the remaining 75 are good.

I draw apples 20-by-20 from this group of apples. That is, I draw the first set of 20 apples from 100 apples, second set of 20 apples from the remaining 80 apples and so on.

I call a sampling of 20 apples good if it has 15 or more good apples in it. I'd like to calculate the expected number of good samplings.

I know that, for the very first sampling, I can compute the probability of it being good by using the hypergeometric distribution as follows,

$$\Pr\big[Good Sampling \big] = \Pr\big[X \geq 15 \big] = \sum_{x=15}^{20} \frac{\binom{75}{x}\binom{25}{20-x}}{\binom{100}{20}}$$ where X is the random variable which represents the number of good apples in the sampling.

However, the probability of 2nd sampling being good is not exactly this as it depends on the first sampling.

I tried to approximate the expected number of good samplings as $$\Pr\big[Good Sampling \big] \cdot \frac{100}{20}$$ but I'm not sure how to quantify the approximation error in this so I can't tell if it's a good approximation or not.

So, what's the proper way to compute the expected number of good samplings in such a scenario?

• The expected number of good samples is exactly $5\Pr\big[Good Sampling \big]$ by linearity of expectation, but the distribution is not a binomial distribution with $n=5$ (for example you cannot get $0$ good samples) – Henry Sep 9 at 20:53
• Thanks for the answer Henry. How does linearity of expectation apply to here, could you please elaborate a bit on that? – SpiderRico Sep 9 at 21:08
• Each of the five samples has the same probability of being "good" (they are not independent but do not need to be) – Henry Sep 9 at 21:10
• So this is what I didn't understand. How do each sampling has the same probability of being good? Doesn't the probability of a sampling being good is conditioned on all the samplings been done before it? – SpiderRico Sep 9 at 21:28
• Line up the $100$ apples in a random order, and then split into five groups of $20$: the unconditional distribution for say the fourth group is the same as the distribution for the first group (both are hypergeometric) – Henry Sep 9 at 23:23