(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:

`4/100 times 3/99 = 1/825 = 1.2121 times 10^-3`

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

`4/100 times 4/100 = 1/625 = 0.0016`

(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:

`4/100times96/99times2 = 64/825=7.7576times10^-2`

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