A company receives a very large shipment of components. A random sample of 20 of these
components will be checked, and the shipment will be accepted if fewer than 2 of these components
are defective. What is the probability of accepting a shipment containing 5% defectives?

This problem can be modeled using the **binomial distribution** because we have a fixed number of trials (20 components), two possible outcomes for each trial (defective or not defective), and a constant probability of a defective component (5% or 0.05).

Let \( X \) be the random variable representing the number of defective components in the sample of 20 components. \( X \) follows a binomial distribution:
\[
X \sim \text{Binomial}(n = 20, p = 0.05)
\]
Where:
- \( n = 20 \) (sample size),
- \( p = 0.05 \) (probability of a defective component).

The shipment is accepted if fewer than 2 components are defective, meaning we want to find the probability that \( X < 2 \).

This can be expressed as:
\[
P(X < 2) = P(X = 0) + P(X = 1)
\]

Using the binomial probability formula:
\[
P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}
\]
Where:
- \( \binom{n}{k} \) is the binomial coefficient,
- \( p^k \) is the probability of having exactly \( k \) defective components,
- \( (1 - p)^{n - k} \) is the probability of the remaining \( n - k \) components being non-defective.

Let's calculate \( P(X = 0) \) and \( P(X = 1) \).

1. **For \( X = 0 \):**
\[
P(X = 0) = \binom{20}{0} (0.05)^0 (0.95)^{20} = 1 \times 1 \times (0.95)^{20}
\]

2. **For \( X = 1 \):**
\[
P(X = 1) = \binom{20}{1} (0.05)^1 (0.95)^{19} = 20 \times 0.05 \times (0.95)^{19}
\]

Let's now compute the final probability.The probability of accepting a shipment containing 5% defectives, by finding fewer than 2 defective components in the random sample of 20, is approximately **0.736** or **73.6%**.

Would you like further details or have any other questions? Here are five related questions you might find helpful:

1. How would the probability change if the sample size increased to 30 components?
2. What happens to the probability if the defective rate increases to 10%?
3. How does the binomial distribution change if we allow up to 3 defective components for acceptance?
4. Can this problem be solved using a Poisson approximation instead of a binomial distribution?
5. How would the company optimize the sample size to minimize the risk of accepting a bad shipment?

**Tip:** When working with small probabilities and large sample sizes, consider using approximations like the Poisson distribution for efficiency.

A company receives a very large shipment of components. A random sample of 20 of these components will be checked, and the shipment will be accepted if fewer than 2 of these components are defective. What is the probability of accepting a shipment containing 5% defectives?

Find the probability of accepting a shipment with 5% defectives using a binomial distribution. This problem covers probability concepts, the binomial theorem, and key formulas to compute acceptance probability based on sample size and defect rates. Suitable for advanced high school or college students.

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