Probability - An assortment of 20 parts An  assortment of 20 parts is considered to be good if it has not more than 2
defective parts, it is considered bad if it contains at least 4 defective parts. Buyer and seller of the assortment agree to test 4 randomly picked out parts. Only when all 4 are good, the purchase will take place. In this procedure the seller bears the risk of not selling a good assortment, the buyer to buy a bad assortment. Who carries the greater risk?
So I have two problems here:
1) If at least 1 part is defective, the purchase won't take place (The Seller has the risk of not selling a good assortment) 
2) If all 4 are good, the purchase will take place (But the buyer has the risk of buying a bad assortment)
My Idea for 1)  (at least 1 part is defective)
$$P(X\ge 1)=1-P(X=0)=1-\begin{pmatrix}
        1  \\
        0   
        \end{pmatrix}* \begin{pmatrix}
        19  \\
        4   
        \end{pmatrix}/ \begin{pmatrix}
        20  \\
        4   
        \end{pmatrix}=1-4/5=0,2$$
My idea for 2) (All 4 parts are good)
$$
       P(B)= \begin{pmatrix}
        4  \\
        0   
        \end{pmatrix}* \begin{pmatrix}
        16  \\
        4   
        \end{pmatrix}/ \begin{pmatrix}
        20  \\
        4   
        \end{pmatrix} =364/969=0,3756
$$
So the Buyer has a higher risk than the seller. However I think I made a mistake somewhere. 
Thanks in advance! 
 A: 1. Seller's Risk (worst case): You want the probability of getting at least one defective part from a 'good' batch. That is a batch that contains 3 (or fewer) defective parts. The risk of a wrong decision is greatest if there are three defective parts, so use that. [It seems you are using one defective part to represent a good batch.]
$$P(X \ge 1) = 1 - P(X = 0) = 1 - \frac{{3\choose 0}{17 \choose 4}}{20 \choose 4} = 0.5088.  $$
This is a hypergeometric distribution. The computation in R statistical software is:
1 - dhyper(0, 3, 17, 4)
## 0.5087719
1 - choose(17,4)/choose(20,4)
## 0.5087719

2. Buyer's Risk (worst case): You want the probability of getting no defective parts from a 'bad' batch.
That is a batch that contains 4 (or more) defective parts. The risk is greatest if there are four defective parts, so we use that. 
$$P(Y = 0) = \frac{{4 \choose 0}{16 \choose 4}}{20 \choose 4} = 0.3756.$$
dhyper(0, 4, 16, 4)
## 0.375645
choose(16, 4)/choose(20,4)
## 0.375645

Here is a plot that shows the two hypergeometric distributions.
The Seller's Risk is the sum of the heights of the blue bars to the
right of the vertical dotted line. Buyer's risk the height of the maroon
bar to the left of the vertical line.

