A manufacturing company uses an acceptance scheme on production items before they are shipped. The plan is a two-stage one. Boxes of 25 are readied for shipment and an inspector takes one production item at random and inspects it, then replaces it in the box; a second inspector does likewise. Finall, a third inspector goes through the same procedure. If any defectives are found, the entire box is sent back for 100% screening. If no defectives are found, the box is shipped.

(a) What is the probability that a box containing 3 defectives will be shipped?

(b) What is the probability that a box containing only on defective will be sent back for screening?

X is the number of defective production items found.

For (a) I did:

Because, this would happen for each inspector I cubed the answer I got and got a final answer of

for (b) I did:

Then I cubed the answer and got

This however isn't the correct answer. The correct answer is 0.1153. What am I doing wrong?

Ok, I think I figured it out, but I still would love for someone to comment on/confirm my solution.

For (b)

Then I let X= the number of inspectors that find a defective production item

If what I did for (b) is correct, then I assume that for (a) I should have done:

Let X= the number of inspectors that do not find a defective production item.

Instead of simply cubing the answer I got. Sorry for the long post.