The original question:
In acceptance sampling, a purchaser samples items from a batch of items and rejects the batch if more than a certain number of the items in the sample are defective. For example, suppose you are a chef inspecting a delivery of 100 avocados from your produce vendor. You randomly select 4 of the avocados to inspect for quality and freshness, and will send the delivery back to the vendor if at least 1 of the 4 avocados you inspect is defective. Find the probability that you send the avocado delivery back, if 10 out of the 100 avocados are defective.
What I'm thinking:
The probability experiment is picking 4 avocados out of the 100. The number of ways you can pick 4 avocados is ${{100}\choose{4}} = 3921225$.
There are four cases in which we will reject the delivery: When there are 1 or 2 or 3 or 4 defective avocados out of the 4 we have picked out for sampling. So that would be ${{4}\choose{1}} + {{4}\choose{2}} + {{4}\choose{3}} + {{4}\choose{4}} = 15$ ways that the delivery can be rejected.
The avocados are non-distinct.
I also notice that the sample space, the 100 avocados, can be partitioned between the defective and the acceptable avocados.
I'm honestly not really sure how to use this information. Is there a formula I'm forgetting about? Any guidance would be a great help. Thanks!
Doing it the additive way will be cumbersome, moreover in your attempt the number of defectives in the full $100$ isn't at all coming into the picture.
This is sampling without replacement, and the distribution is known as the hypergeometric distribution.
The smart way to do it is to use the complement, viz
P(reject lot) $= 1 - \dfrac{\binom{96}4}{\binom{100}4} \approx 0.1528$
And if you want to do it the additive way,
P(reject lot) $$= \dfrac{\binom41\binom{96}3+\binom42\binom{96}2+\binom43\binom{96}1+\binom44\binom{96}0}{\binom{100}4} \approx 0.1528$$
Note that you need to account for the defectives you have found in the lot as well as those not in the lot in the numerator of the formula.