This problem involves a binomial distribution, where we are given the following information:

• Probability of an item being faulty (success) $p = 0.06$
• Number of items $n = 12$
• We want to find the probability that less than 4 items are faulty, meaning $X < 4$.

The binomial probability mass function (PMF) is given by:

$P(X = k) = \binom{n}{k} p^k (1 - p)^{n - k}$

Where:

• $X$ is the number of faulty items.
• $k$ is the specific number of faulty items.
• $\binom{n}{k}$ is the binomial coefficient, which represents the number of ways to choose $k$ faulty items out of $n$.

To find the probability that less than 4 items are faulty, we need to sum the probabilities for $X = 0, 1, 2, 3$.

$P(X < 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)$

Let’s calculate this.The probability that less than 4 items will be faulty is approximately $0.996$, or 99.57%.

Would you like any further details on the process or breakdown? Here are some related questions to explore further:

1. How is the binomial distribution applied in quality control?
2. What would the probability be for exactly 4 faulty items?
3. How would the result change if the sample size increased to 20?
4. What assumptions does the binomial model rely on in this scenario?
5. How is this approach different from using a normal approximation?

Tip: In binomial distributions, the larger the sample size, the closer the distribution may resemble a normal distribution due to the Central Limit Theorem.