From the given values, we know there are 2 defective resistors and 9 not defective. then, we have
[tex]P(1\text{ defective)=}\frac{C^9_3\times C^2_1}{C^{11}_4}[/tex]
where
[tex]C^9_3[/tex]
denote the combinations of 9 elements taking in 3, that is 3 good resistor from the 9 which are not defective. Similarly,
[tex]C^2_1[/tex]
denotes the combinations of 2 elements taking in 1, that is, 1 defective resistor form the 2 defective resistors.
Then, we have
[tex]P(1\text{ defective)=}\frac{84\times2}{330}[/tex]
then, the answer is
[tex]P(1\text{ defective) =}0.509[/tex]
