Ask Me Help Desk - Probability if both are defective in each case, 120 auto ignition circuits, 18 defect

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probability if both are defective in each case, 120 auto ignition circuits, 18 defect

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

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

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

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