Answer
$P(accepted)=\frac{2380}{4845}\approx0.4912$
Work Step by Step
Combinations of 20 distinct modems (defective or not) taken 4 at a time (the order in which the modems are selected does not matter):
$N(S)=~_{20}C_4=\frac{20!}{4!(20-4)!}=\frac{20!}{4!\times16!}$
But, $20!=20\times19\times18\times17\times(16\times15\times14\times...\times3\times2\times1)=20\times19\times18\times17\times16!$
$_{20}C_4=\frac{20\times19\times18\times17\times16!}{4!\times16!}=\frac{20\times19\times18\times17}{4\times3\times2\times1}=4845$
There are $20-3=17$ non defective modems.
Combinations of 17 distinct non defective modems taken 4 at a time (the order in which the modems are selected does not matter):
$N(non~defective)=~_{17}C_4=\frac{17!}{4!(17-4)!}=\frac{17!}{4!\times13!}$
But, $17!=17\times16\times15\times14\times(13\times12\times11\times10\times...\times3\times2\times1)=17\times16\times15\times14\times13!$
$_{16}C_4=\frac{17\times16\times15\times14\times13!}{4!\times13!}=\frac{17\times16\times15\times14}{4\times3\times2\times1}=2380$
Using the Classical Method (page 259):
$P(accepted)=\frac{N(non~defective)}{N(S)}=\frac{2380}{4845}\approx0.4912$
