Past records indicate that the probability of online retail orders that turn out to be fraudulent is 

0.060.06. 

Suppose​ that, on a given​ day, 

2121 

online retail orders are placed. Assume that the number of online retail orders that turn out to be fraudulent is distributed as a binomial random variable. Complete parts​ (a) through​ (d) below.

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Part 1

**a.** What are the mean and standard deviation of the number of online retail orders that turn out to be​ fraudulent?

The mean number of online retail orders that turn out to be fraudulent is 

1.26001.2600.

​(Type an integer or a decimal. Round to four decimal places as​ needed.)

Part 2

The standard deviation of the number of fraudulent retail orders is 

1.0881.088.

​(Type an integer or a decimal. Round to three decimal places as​ needed.)

Part 3

**b.** What is the probability that zero online retail orders will turn out to be​ fraudulent?

answer- 0.2727

​(Type an integer or a decimal. Round to four decimal places as​ needed.)

Part 4

**c.** What is the probability that one online retail order will turn out to be​ fraudulent?

​(Type an integer or a decimal. Round to four decimal places as​ needed.)

Let's break down part (c) of the problem step by step.

### Given:
- Probability of a fraudulent order, \( p = 0.06 \)
- Number of orders, \( n = 21 \)
- We need to find the probability that exactly one online retail order will turn out to be fraudulent.

The number of fraudulent orders \( X \) follows a binomial distribution, \( X \sim \text{Binomial}(n, p) \).

### Part (c): Probability that exactly one order is fraudulent

The probability mass function for a binomial distribution is:
\[
P(X = k) = \binom{n}{k} \cdot p^k \cdot (1 - p)^{n - k}
\]
Where:
- \( n = 21 \) is the number of trials (online orders),
- \( k = 1 \) is the number of successes (fraudulent orders),
- \( p = 0.06 \) is the probability of success on each trial.

Substituting these values:
\[
P(X = 1) = \binom{21}{1} \cdot 0.06^1 \cdot 0.94^{20}
\]
\[
P(X = 1) = 21 \cdot 0.06 \cdot 0.94^{20}
\]

Now, we calculate \( 0.94^{20} \):
\[
0.94^{20} \approx 0.302114
\]

Then, calculate the full probability:
\[
P(X = 1) = 21 \cdot 0.06 \cdot 0.302114 \approx 0.381
\]

So, the probability that exactly one online retail order will turn out to be fraudulent is approximately **0.381** (rounded to three decimal places).

### Summary:

- The mean number of fraudulent orders is **1.2600**.
- The standard deviation of the number of fraudulent orders is **1.088**.
- The probability that zero orders are fraudulent is **0.2727**.
- The probability that exactly one order is fraudulent is **0.381**.

Would you like further details or have any other questions? 

Here are some additional questions you might consider:
1. What is the probability of having exactly two fraudulent orders?
2. What is the cumulative probability of having 3 or fewer fraudulent orders?
3. How does the probability change if the number of orders placed increases to 30?
4. What is the variance of the number of fraudulent orders?
5. How would the mean and standard deviation change if the probability of fraud increases to 0.10?

**Tip**: Always remember to carefully round your results according to the problem's instructions to avoid minor errors in final answers.

Explore how to calculate the mean, standard deviation, and probabilities of fraudulent online retail orders using binomial distribution. Understand the concepts with detailed step-by-step solutions and examples.

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