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This problem involves calculating the probability for a binomial distribution.

Given:
- \( n = 15 \) (the sample size)
- \( p = 0.02 \) (the probability of a defective item)
- \( x = 1 \) (we want the probability of exactly 1 defective item)

The binomial probability formula is:
\[
P(X = x) = \binom{n}{x} p^x (1 - p)^{n - x}
\]
Where:
- \( \binom{n}{x} = \frac{n!}{x!(n - x)!} \) is the binomial coefficient.

Let's calculate the probability step by step.

1. **Binomial coefficient** for \( \binom{15}{1} \) is simply \( 15 \), since:
   \[
   \binom{15}{1} = \frac{15!}{1!(15-1)!} = 15
   \]

2. **Probability part**:
   - \( p^1 = 0.02 \)
   - \( (1 - p)^{14} = (0.98)^{14} \)

Now, using the formula:
\[
P(X = 1) = 15 \times 0.02 \times (0.98)^{14}
\]

I'll compute this for you.

\[
P(X = 1) \approx 15 \times 0.02 \times 0.7540 \approx 0.2261
\]

Thus, the probability that \( x = 1 \) defective is approximately **0.2261**, which matches option **b**.

Would you like more details, or do you have any questions?

Here are 5 related questions:
1. What is the binomial distribution used for in probability theory?
2. How does changing the value of \( p \) affect the binomial probability?
3. How would you calculate the probability for \( x = 0 \) defectives?
4. What is the mean and variance of a binomial distribution?
5. How would you use a cumulative binomial distribution to find the probability of \( x \leq 1 \)?

**Tip:** Always verify your calculations by checking if the sum of probabilities for all possible outcomes equals 1 in binomial distributions.

In a binomial distribution, a sample of n = 15 and p = 0.02 defective. Determine the probability that x equals 1 defective.

Learn how to calculate the probability of exactly 1 defective item in a sample of 15, with p = 0.02, using the binomial distribution formula. This problem covers key concepts in probability theory and the binomial theorem, suitable for college-level statistics.

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