Computer keyboard failures can be attributed to electrical defects or mechanical defects. A repair facility currently has 25 failed keyboard, 7 or which have electrical defects and 18 of which have mechanical defects. If a sample of 7 keyboards is randomly selected, what is the probability that at least 6 of these will have a mechanical defect? Round your answer to four decimal places

To solve this problem, we can model the situation using the **hypergeometric distribution** because we are sampling without replacement from a finite population.

### Hypergeometric Distribution Formula:
The probability of selecting \( k \) items of a certain type in a sample of size \( n \) from a population of \( N \) items, where \( K \) are of the desired type, is given by:

\[
P(X = k) = \frac{\binom{K}{k} \binom{N - K}{n - k}}{\binom{N}{n}}
\]

Where:
- \( N \) = Total population size (25 keyboards)
- \( K \) = Total number of mechanical defect keyboards (18 keyboards)
- \( n \) = Sample size (7 keyboards)
- \( X \) = Number of mechanical defect keyboards in the sample
- We want to find the probability that at least 6 of the 7 keyboards have mechanical defects.

### Probability Calculation:
We need to calculate \( P(X \geq 6) \), which is the sum of the probabilities for \( X = 6 \) and \( X = 7 \):

\[
P(X \geq 6) = P(X = 6) + P(X = 7)
\]

#### Step 1: Calculate \( P(X = 6) \)

\[
P(X = 6) = \frac{\binom{18}{6} \binom{7}{1}}{\binom{25}{7}}
\]

#### Step 2: Calculate \( P(X = 7) \)

\[
P(X = 7) = \frac{\binom{18}{7} \binom{7}{0}}{\binom{25}{7}}
\]

Now, let's compute these probabilities.The probability that at least 6 out of the 7 selected keyboards will have a mechanical defect is approximately \( 0.3365 \) (rounded to four decimal places).

Would you like more details or have any questions about this solution?

Here are 5 related questions to further explore the topic:
1. What is the probability of selecting exactly 5 keyboards with mechanical defects?
2. How would the probability change if the sample size were 5 keyboards instead of 7?
3. Can we calculate the expected number of mechanical defects in a random sample of 7 keyboards?
4. What if 10 keyboards had electrical defects instead of 7? How would that impact the probability?
5. How does the hypergeometric distribution differ from the binomial distribution in this context?

**Tip:** The hypergeometric distribution is useful when sampling without replacement, while the binomial distribution is better for sampling with replacement.

Computer keyboard failures can be attributed to electrical defects or mechanical defects. A repair facility currently has 25 failed keyboards, 7 of which have electrical defects and 18 of which have mechanical defects. If a sample of 7 keyboards is randomly selected, what is the probability that at least 6 of these will have a mechanical defect? Round your answer to four decimal places.

This problem explores the probability of selecting at least 6 mechanical defect keyboards from a sample of 7, using hypergeometric distribution. The dataset consists of 25 keyboards, with 18 having mechanical defects. Suitable for advanced high school or undergraduate students.

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