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 1 of 3
(a) 0 defective resistors
\[
P(0 \text { defective })=0.106
\]
Part: 1 / 3
Part 2 of 3
(b) 1 defective resistor
\[
P(1 \text { defective })=
\]

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

Step 2 ：We can solve this by calculating the number of ways to select 1 defective resistor and 3 non-defective resistors, divided by the total number of ways to select 4 resistors from the package.

Step 3 ：The number of ways to select 1 defective resistor out of 4 is given by the combination formula \(C(n, k) = \frac{n!}{k!(n-k)!}\), where n is the total number of items, k is the number of items to select, and ! denotes factorial.

Step 4 ：Similarly, the number of ways to select 3 non-defective resistors out of 7 (since there are 11 total resistors and 4 are defective) is also given by the combination formula.

Step 5 ：The total number of ways to select 4 resistors out of 11 is also given by the combination formula.

Step 6 ：The probability is then given by the product of the number of ways to select 1 defective and 3 non-defective resistors, divided by the total number of ways to select 4 resistors.

Step 7 ：Final Answer: The probability of selecting 1 defective resistor out of 4 selected from a package of 11 resistors, 4 of which are defective, is approximately \(\boxed{0.424}\).