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View Full Version : Probability if both are defective in each case, 120 auto ignition circuits, 18 defect


myra1078
Jun 30, 2010, 07:58 AM
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

galactus
Jun 30, 2010, 08:45 AM
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}