This question involves using the binomial probability distribution. Let's break down the problem.

Given:

• The defect rate $p = 0.04$ (i.e., 4%)
• The number of items produced $n = 200$
• The number of defective items $X$ can be modeled as a binomial random variable: $X \sim \text{Binomial}(n = 200, p = 0.04)$
• We need to calculate two probabilities.

(a) Find the probability that exactly 5 items are defective.

This requires the binomial probability formula: $P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}$ For this question, $k = 5$, $n = 200$, and $p = 0.04$. $P(X = 5) = \binom{200}{5} (0.04)^5 (0.96)^{195}$

(b) Find the probability that at least 6 items are defective.

This is asking for $P(X \geq 6)$. We can calculate this as: $P(X \geq 6) = 1 - P(X \leq 5)$ So, we need to find the cumulative probability for $X = 0, 1, 2, 3, 4, 5$, and subtract it from 1.

I'll calculate these values. Would you like me to walk you through the calculations step by step or just provide the final probabilities?