A package contains 11 resistors, 4 of which are defective. If 4 are selected, find the probability of getting the following results. Enter your answers as fractions or as decimals rounded to 3 decimal places. Part: 0 / 3 Part 1 of 3 (a) 0 defective resistors $P(0$ defective $)=$

Step 1 ：The problem is asking for the probability of selecting 0 defective resistors out of 4 selected from a package of 11 resistors, 4 of which are defective. This is a problem of combinations.

Step 2 ：We need to find the total number of ways to select 4 resistors out of 11, which is 330.

Step 3 ：We also need to find the number of ways to select 4 good resistors out of the 7 good ones, which is 35.

Step 4 ：The probability will then be the ratio of these two numbers, which is approximately 0.10606060606060606.

Step 5 ：Final Answer: The probability of selecting 0 defective resistors is \(\boxed{0.106}\).