To solve this problem, we can model it using the binomial probability formula. The binomial probability formula is:

$P(X = k) = \binom{n}{k} p^k (1 - p)^{n - k}$

Where:

• $P(X = k)$ is the probability of getting exactly $k$ successes in $n$ trials.
• $n$ is the number of trials (in this case, 7 students).
• $k$ is the number of successes (students passing the polygraph test).
• $p$ is the probability of success on a single trial (in this case, 85%, or 0.85).
• $\binom{n}{k}$ is the number of combinations of $n$ trials taken $k$ at a time.

(a) All the students are able to pass the polygraph examination.

This corresponds to the probability that all 7 students pass the test, i.e., $k = 7$ successes out of $n = 7$ trials. Using the binomial formula:

$P(X = 7) = \binom{7}{7} (0.85)^7 (1 - 0.85)^{7 - 7}$

$P(X = 7) = 1 \cdot (0.85)^7 \cdot (0.15)^0 = (0.85)^7$

Now, calculate $(0.85)^7$:

$P(X = 7) \approx 0.85^7 = 0.323$

Thus, the probability that all 7 students pass the polygraph test is approximately 0.323.

Would you like further clarification or details?

Here are 5 follow-up questions to expand on this problem:

1. What is the probability that exactly 6 students pass the polygraph test?
2. How would the probability change if only 5 students were able to pass the polygraph?
3. Can you calculate the probability that fewer than 5 students pass the polygraph?
4. What are the assumptions behind using a binomial model in this scenario?
5. How might increasing the sample size from 7 to 10 students affect the probability?

Tip: Binomial probability calculations are useful when dealing with independent trials where each trial has the same probability of success.