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:

$\displaystyle p(X=0)=f(0;N=25, n=1, k=3)=

\frac{\left(\!

\begin{array}{c}

3 \\

0

\end{array}

\!\right) \left(\!

\begin{array}{c}

22 \\

1

\end{array}

\!\right)}{\left(\!

\begin{array}{c}

25 \\

1

\end{array}

\!\right)} = 0.88$

Because, this would happen for each inspector I cubed the answer I got and got a final answer of$\displaystyle 0.6815$

for (b) I did:

$\displaystyle p(X=1)=f(1;N=25,n=1, n=1)=\frac{\left(\!

\begin{array}{c}

1 \\

1

\end{array}

\!\right) \left(\!

\begin{array}{c}

24 \\

0

\end{array}

\!\right)}{\left(\!

\begin{array}{c}

25 \\

1

\end{array}

\!\right)} = \frac{1}{25}$

Then I cubed the answer and got $\displaystyle \frac {1}{15,625}$

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)

$\displaystyle p(X=1)=f(1;N=25,n=1, n=1)=\frac{\left(\!

\begin{array}{c}

1 \\

1

\end{array}

\!\right) \left(\!

\begin{array}{c}

24 \\

0

\end{array}

\!\right)}{\left(\!

\begin{array}{c}

25 \\

1

\end{array}

\!\right)} = \frac{1}{25}$

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

$\displaystyle p(X\geq 1)= 1-p(X<1)=1-P(X=0)=1-b(0;n=3, p=\frac{1}{25})=1-\left(\!

\begin{array}{c}

3 \\

0

\end{array}

\!\right)$$\displaystyle *(\frac{1}{25})^0*(\frac{24}{25})^3=1-0.8847=0.1153$

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

$\displaystyle p(X=0)=f(0;N=25, n=1, k=3)=

\frac{\left(\!

\begin{array}{c}

3 \\

0

\end{array}

\!\right) \left(\!

\begin{array}{c}

22 \\

1

\end{array}

\!\right)}{\left(\!

\begin{array}{c}

25 \\

1

\end{array}

\!\right)} = 0.88$

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

$\displaystyle p(X=3)=b(3;n=3,p=0.88)=\left(\!

\begin{array}{c}

3 \\

3

\end{array}

\!\right)$$\displaystyle *(0.88)^3*(0.12)^0=0.6815$

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