To solve this problem, we can use the binomial probability formula because we are dealing with a fixed number of trials (6 parts) and two possible outcomes for each trial (defective or not defective).

Given:

• Probability of a defective part, $p = 0.13$
• Probability of a non-defective part, $q = 1 - p = 0.87$
• Number of trials (parts), $n = 6$
• We need to find the probability that at most 1 part is defective.

This is the probability that 0 or 1 part is defective, so we need to calculate: $P(X \leq 1) = P(X = 0) + P(X = 1)$

Step 1: Calculate $P(X = 0)$

The binomial probability formula is: $P(X = k) = \binom{n}{k} p^k q^{n-k}$

For $X = 0$: $P(X = 0) = \binom{6}{0} (0.13)^0 (0.87)^6$ $P(X = 0) = 1 \times 1 \times (0.87)^6$ $P(X = 0) \approx 0.4870$

Step 2: Calculate $P(X = 1)$

For $X = 1$: $P(X = 1) = \binom{6}{1} (0.13)^1 (0.87)^5$ $P(X = 1) = 6 \times 0.13 \times (0.87)^5$ $P(X = 1) \approx 6 \times 0.13 \times 0.4984$ $P(X = 1) \approx 0.3885$

Step 3: Sum the probabilities

Now, add the two probabilities: $P(X \leq 1) = P(X = 0) + P(X = 1)$ $P(X \leq 1) \approx 0.4870 + 0.3885 = 0.8755$

Final Answer:

The probability that at most 1 of the 6 parts are defective is approximately 0.88 (rounded to two decimal places).

Would you like more details or have any further questions?

Here are some related questions:

1. How do you interpret the result of a binomial probability in a real-world context?
2. What is the formula for the binomial coefficient $\binom{n}{k}$?
3. How would the probability change if the defective rate increased to 20%?
4. What is the difference between binomial and normal distributions?
5. How can you use a cumulative binomial probability table?
6. What is the expected number of defective parts in a batch of 6?
7. How does the variance of a binomial distribution relate to the probability of success?
8. What are the conditions required for a binomial distribution to be a good model?

Tip: When dealing with binomial probabilities, it's useful to remember that the binomial coefficient $\binom{n}{k}$ represents the number of ways to choose $k$ successes out of $n$ trials.