# Math Help - hypergeometric distribution problem

1. ## hypergeometric distribution problem

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:

$p(X=0)=f(0;N=25, n=1, k=3)=
\frac{\left(\!
\begin{array}{c}
3 \\
0
\end{array}
\!\right) \left(\!
\begin{array}{c}
22 \\
1
\end{array}
\!\right)}{\left(\!
\begin{array}{c}
25 \\
1
\end{array}
\!\right)} = 0.88$

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

for (b) I did:

$p(X=1)=f(1;N=25,n=1, n=1)=\frac{\left(\!
\begin{array}{c}
1 \\
1
\end{array}
\!\right) \left(\!
\begin{array}{c}
24 \\
0
\end{array}
\!\right)}{\left(\!
\begin{array}{c}
25 \\
1
\end{array}
\!\right)} = \frac{1}{25}$

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

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)
$p(X=1)=f(1;N=25,n=1, n=1)=\frac{\left(\!
\begin{array}{c}
1 \\
1
\end{array}
\!\right) \left(\!
\begin{array}{c}
24 \\
0
\end{array}
\!\right)}{\left(\!
\begin{array}{c}
25 \\
1
\end{array}
\!\right)} = \frac{1}{25}$

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

$p(X\geq 1)= 1-p(X<1)=1-P(X=0)=1-b(0;n=3, p=\frac{1}{25})=1-\left(\!
\begin{array}{c}
3 \\
0
\end{array}
\!\right)$
$*(\frac{1}{25})^0*(\frac{24}{25})^3=1-0.8847=0.1153$

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

$p(X=0)=f(0;N=25, n=1, k=3)=
\frac{\left(\!
\begin{array}{c}
3 \\
0
\end{array}
\!\right) \left(\!
\begin{array}{c}
22 \\
1
\end{array}
\!\right)}{\left(\!
\begin{array}{c}
25 \\
1
\end{array}
\!\right)} = 0.88$

Let X= the number of inspectors that do not find a defective production item.
$p(X=3)=b(3;n=3,p=0.88)=\left(\!
\begin{array}{c}
3 \\
3
\end{array}
\!\right)$
$*(0.88)^3*(0.12)^0=0.6815$

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

2. Hello, downthesun01!

You've made the problem much harder . . .

Since the items are replaced, the probabilities are independent.

A company uses three-step testing scheme on items before they are shipped.
Boxes of 25 are readied for shipment.
An inspector takes one item at random, inspects it, then replaces it in the box.
A second inspector does the same. Finally, a third inspector does likewise.

If any defectives are found, the entire box is sent back.
If no defectives are found, the box is shipped.

(a) What is the probability that a box containing 3 defectives will be shipped?
There are 3 Defective and 22 Good: . $\begin{Bmatrix}P(D) &=& \frac{3}{25} \\ \\[-3mm] P(G) &=& \frac{22}{25}\end{Bmatrix}$

The box will be shipped if each inspector draws a Good.

. . $P(\text{3 Good}) \;\;=\;\;\left(\frac{22}{25}\right)^3 \;\;=\;\;\frac{10,\!648}{15,\!625}$

(b) What is the probability that a box containing only one defective will be sent back?
There is 1 Defective and 24 Good: . $\begin{Bmatrix}P(D) &=& \frac{1}{25} \\ \\[-3mm] P(G) &=& \frac{24}{25} \end{Bmatrix}$

The box is sent back if at least one inspector finds a Defective.

$P(\text{no D}) \:=\:\left(\frac{24}{25}\right)^3 \;=\;\frac{13,\!824}{15,\!625}$

. . $P(\text{at least 1 D}) \;\;=\;\;1 - \frac{13,\!824}{15,\!625} \;\;=\;\;\frac{1,\!801}{15,\!625}$