1. The problem is to find the probability of having exactly 2 defective parts in a given scenario. 2. We use the binomial probability formula for this type of problem: $$P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}$$ where $n$ is the total number of parts, $k$ is the number of defective parts, and $p$ is the probability of a part being defective. 3. Important rules: - $\binom{n}{k}$ is the number of ways to choose $k$ defective parts from $n$ parts. - $p^k$ is the probability that those $k$ parts are defective. - $(1-p)^{n-k}$ is the probability that the remaining parts are not defective. 4. To proceed, we need the values of $n$ and $p$. Assuming $n$ and $p$ are given or known, substitute them into the formula. 5. Calculate $\binom{n}{2}$, then multiply by $p^2$ and $(1-p)^{n-2}$. 6. Simplify the expression to get the final probability. Without specific values for $n$ and $p$, the general formula for the probability of exactly 2 defective parts is: $$P(X=2) = \binom{n}{2} p^2 (1-p)^{n-2}$$