A package contains 10 resistors, 2 of which are defective. if 3 are selected, find the probability of getting 1 defective resistor.

To select 3 resistors of 10 resistors, the possible ways would be:

10C3 = 10! / (10-3) ! 3! = 10*9*8 / 3*2 = 120 ways

There are 2 defective resistors and 8 normal resistors. T he possible ways to take 1 defective resistors and 2 normal resistors would be: 2C1 * 8C2 = 2 * 8*7/2 = 56 possible ways

Then, the probability of getting 1 defective resistor would be: 56ways / 120 ways = 46.7%