(a) On the first draw, there are 4 defectives in the box out of the 100 total items.

If we have already chosen one of the defectives on the first draw, then on the second draw, there will be 3 defectives left out of the 99 items in the box. The required probability is:

(b) Both the first draw and the second draw have the same probability of getting a defective, i.e. 4 in 100. Required probability is:

(c) We can either:

1. Get a defective on the first draw (4 chances in 100) then a non-defective on the second (96 non-defectives out of 99 left in the box); OR
2. Get a non-defective first (96 chances in 100) then a defective (4 in the remaining 99).

So the probability is (4/100) × (96/99) + (96/100) × (4/99), which can also be written as:

Note: In probability, the word "OR" in the question usually means we need to add the probabilities.