probability that... will be shipped; will be sent back

[_N_r_chy]

Thank you very much in advance!
here's the prob:
A manufacturing company uses an acceptance scheme on production items before they are shipped. The plan consists of two stages, that is, boxes of 25 are readied for shipment and a sample of 3 are tested for defectives. If no defectives are found the box is shipped.
What is the probability that
a) a box containing 3 defectives will be shipped?
b) a box containing only 1 defective will be sent back for screening?



I tried to solve the problem.
My own wrong solution to letter (a) is this:
N=25
n=3
k=3
N-k=22
x=3
h(x;N,n,k)={(kCOMBINATIONx)*[(N-k)COMBINATION(n-x)]} / [(N)COMBINATION(n)]
h(x;N,n,k)={3C3*22C0} / 25C3

While my own wrong solution to letter (b) is this:
N=25
n=3
k=1
N-k=24
x=1
h(x;N,n,k)={(kCOMBINATIONx)*[(N-k)COMBINATION(n-x)]} / [(N)COMBINATION(n)]
h(x;N,n,k)={1C1*24C2} / 25C3

 

[pka]

Why do you say that your solution to part b is wrong?
It is correct if I understand the problem.

Now your solution to part a is incorrect.
You have it backwards.
\(\displaystyle \frac{{\left( {\begin{array}{c}
{22} \\
3 \\
\end{array}} \right)}}{{\left( {\begin{array}{c}
{25} \\
3 \\
\end{array}} \right)}}.\)

 

[_N_r_chy]

Why do you say that your solution to part b is wrong?
It is correct if I understand the problem.

I just felt that I was incorrect. I suck at math actually.
But thank you very very much for the help.

Now your solution to part a is incorrect.
You have it backwards.
\(\displaystyle \frac{{\left( {\begin{array}{c}
{22} \\
3 \\
\end{array}} \right)}}{{\left( {\begin{array}{c}
{25} \\
3 \\
\end{array}} \right)}}.\)

oh. thank you very much again for the help!