To solve these problems, we will use the binomial probability formula because each employee’s result (pass/fail) is independent, and there are a fixed number of trials (10 employees) with a consistent probability of success (0.7).

The binomial probability formula is: $P(X = k) = \binom{n}{k} p^k (1 - p)^{n - k}$ where:

• $n$ is the number of trials (10 employees),
• $k$ is the number of successful outcomes we are interested in (e.g., 6 employees passing),
• $p$ is the probability of success (0.7),
• $\binom{n}{k}$ is the number of ways to choose $k$ successes from $n$ trials, calculated as $\binom{n}{k} = \frac{n!}{k!(n - k)!}$.

a) Probability that exactly 6 employees pass the test

We are looking for $P(X = 6)$.

$P(X = 6) = \binom{10}{6} \times 0.7^6 \times (1 - 0.7)^{10 - 6}$ Calculating each part:

• $\binom{10}{6} = \frac{10!}{6! \cdot 4!} = 210$
• $0.7^6 \approx 0.1176$
• $(1 - 0.7)^4 = 0.3^4 \approx 0.0081$

Then, $P(X = 6) \approx 210 \times 0.1176 \times 0.0081 \approx 0.2001$

So, the probability that exactly 6 employees pass the test is approximately 0.2001.

b) Probability that at most 6 employees pass the test

"At most 6" means that 0 to 6 employees pass the test. We need to calculate $P(X \leq 6) = P(X = 0) + P(X = 1) + \dots + P(X = 6)$.

Using the binomial formula, each probability can be calculated similarly to part (a), but it’s often more efficient to use cumulative binomial tables or a calculator for this.

After calculating or using a binomial calculator: $P(X \leq 6) \approx 0.3828$

Thus, the probability that at most 6 employees pass the test is approximately 0.3828.

c) Probability that more than 6 employees pass the test

"More than 6" means 7, 8, 9, or 10 employees passing. We need $P(X > 6) = P(X = 7) + P(X = 8) + P(X = 9) + P(X = 10)$.

Alternatively, we can use the complement: $P(X > 6) = 1 - P(X \leq 6)$ Using the result from part (b): $P(X > 6) = 1 - 0.3828 = 0.6172$

Thus, the probability that more than 6 employees pass the test is approximately 0.6172.

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Would you like more details on any of these steps?

Here are some related questions for deeper understanding:

1. What is the probability that exactly 7 employees pass the test?
2. How would the probability change if the probability of passing the test were 0.8 instead of 0.7?
3. What is the probability that fewer than 4 employees pass the test?
4. How would we find the expected number of employees to pass the test?
5. What is the variance and standard deviation of the number of employees passing?

Tip: Using the complement rule ($P(A') = 1 - P(A)$) is often helpful for finding probabilities of events like "at most" or "more than" in binomial problems.