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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
= 1-p(X<1)=1-P(X=0)=1-b(0;n=3, p=\frac{1}{25})=1-\left(\!<br />
\begin{array}{c}<br />
3 \\<br />
0<br />
\end{array}<br />
\!\right))
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.
=b(3;n=3,p=0.88)=\left(\!<br />
\begin{array}{c}<br />
3 \\<br />
3<br />
\end{array}<br />
\!\right))
Instead of simply cubing the answer I got. Sorry for the long post.
