discrete probability distribution - exercise

#1
I'm stuck with the following question:

a product is packaged for transport per 20 pieces. 5 pieces per package are examined, without replacement, to control infections. If more than 1 infected piece is found, the entire package will be rejected.
What is the probability that a package containing 4 contaminated items will be rejected?

Solution would be 0,2487
but i only get different answer....

i have tried the following:

chance of acceptance = chance of 0 infections + chance of 1 infection = 16/20 * 15/19 * 14/18 * 13/17 * 12/16 + 4/20 * 16/19 * 15/18 * 14/17 * 13/16 = 0.375
chance of rejection then becomes 0.624 (= 1 - 0,375).
what's wrong?