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We are given the following situation:

- The probability that a diskette is defective is \( 0.01 \).
- A pack contains 10 diskettes, and the company offers a money-back guarantee if **more than one** of the 10 diskettes is defective.
- The task is to find the probability that, after buying 3 packs, **at most one** pack will be returned (i.e., either 0 or 1 pack is returned).

### Step 1: Model the number of defective diskettes in one pack

The number of defective diskettes in a pack of 10 diskettes follows a **binomial distribution** because each diskette is independently defective with probability \( p = 0.01 \). Let \( X \) represent the number of defective diskettes in a pack. The probability mass function of \( X \) is:

\[
P(X = k) = \binom{10}{k} (0.01)^k (0.99)^{10-k}, \quad \text{for } k = 0, 1, 2, \dots, 10
\]

The condition for returning a pack is that **more than one** diskette is defective, i.e., \( X > 1 \). Therefore, we need to calculate the probability that a pack is returned, which is \( P(X > 1) \).

### Step 2: Calculate the probability that a pack is returned

To find \( P(X > 1) \), we first calculate the complementary probability, \( P(X \leq 1) \), which is the sum of the probabilities that either 0 or 1 diskettes are defective:

\[
P(X \leq 1) = P(X = 0) + P(X = 1)
\]

Using the binomial distribution:

\[
P(X = 0) = \binom{10}{0} (0.01)^0 (0.99)^{10} = (0.99)^{10}
\]
\[
P(X = 1) = \binom{10}{1} (0.01)^1 (0.99)^9 = 10 \times (0.01) \times (0.99)^9
\]

Now we can compute these values.

### Step 3: Use binomial distribution for 3 packs

Once we have the probability of returning one pack, we treat the three packs as independent Bernoulli trials (where each trial results in a pack either being returned or not). The number of returned packs will follow a binomial distribution with \( n = 3 \) and the calculated probability of a return.

Finally, we calculate the probability that **at most one pack is returned** (i.e., either 0 or 1 packs are returned).

Let me calculate this step-by-step for you.Here are the results:

1. The probability that a single pack will be returned (i.e., more than 1 defective diskette) is approximately \( 0.00427 \).

2. The probability that, after buying 3 packs, **at most one** pack will be returned is approximately **0.99995**.

This means there's a very high chance (nearly 100%) that at most one pack will be returned if you purchase 3 packs.

Would you like a detailed explanation of any part? Here are 5 related questions:

1. What is a binomial distribution, and how is it applied in this scenario?
2. How do you calculate the probability of multiple independent events (like buying 3 packs)?
3. Why is the assumption of independence important in this problem?
4. How would the result change if the defect probability was higher, say 5%?
5. What if the packs contained more than 10 diskettes? How would that affect the calculation?

**Tip:** When working with binomial probabilities, it's often helpful to first compute complementary probabilities for simpler calculations.

The diskettes produced by a certain company are defective with probability 0.01, independently of each other. The company sells the diskettes in packs of size 10 and offers a money-back guarantee if more than one of the 10 diskettes in the pack is found to be defective. If you buy 3 packs, what is the probability that at most one pack will be returned?

Calculate the probability that at most one of three packs of diskettes will be returned under a money-back guarantee. Explore binomial distribution and complementary probabilities based on defective diskette probability of 0.01. Perfect for advanced high school or undergraduate students.

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