Accepting a batch after checking two components

A batch of \(20\) components contains \(d\) defective components. An inspector chooses \(2\) components from the batch at random, without replacement, and accepts the batch only if neither is defective.

1. Show that the probability that the batch is accepted is \(\dfrac{(20-d)(19-d)}{380}\).[2]
2. The company wants any batch with \(d\) or more defective components to have a probability of acceptance less than \(\dfrac{1}{2}\). Find the least value of \(d\) for which a batch has a probability of acceptance less than \(\dfrac{1}{2}\).[3]
3. For a batch with this number of defective components, the inspector proposes to check \(3\) components instead, accepting only if none of the three is defective. Calculate the new probability of acceptance and explain whether the change helps the company.[3]
4. State one assumption about how the components are chosen that this model needs.[1]