Apackage 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. The 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%

That’s hard