Of 120 auto ignition circuits, there are 18 defects. If 2 circuits are randomly selected, find the probability that they are both defective in each case

We choose the two defects from the 18 defects. We choose 0 good ones from the 102 good ones. Overall, we choose two from the 120.


\frac{C(18,2)C(102,0)}{C(120,2)}

or

\frac{18}{120}\cdot\frac{17}{119}=

\frac{3}{20}\cdot\frac{1}{7}