Aldrich Ames is a convicted traitor who leaked American secrets to a foreign power. Yet Ames took routine lie detector tests and each time passed them. How can this be done? Recognizing control questions, employing unusual breathing patterns, biting one's tongue at the right time, pressing one's toes hard to the floor, and counting backwards by 7 are countermeasures that are difficult to detect but can change the results of a polygraph examination†. In fact, it is reported in Professor Ford's book that after only 20 minutes of instruction by "Buzz" Fay (a prison inmate), 85% of those trained were able to pass the polygraph examination even when guilty of a crime. Suppose that a random sample of seven students (in a psychology laboratory) are told a "secret" and then given instructions on how to pass the polygraph examination without revealing their knowledge of the secret. What are the following probabilities?

You do not say what the questions are so for example

all seven pass= .85^7 = .32

To calculate the probabilities, we need additional information. Specifically, we need to know the probability that an individual student trained to pass the polygraph examination will successfully do so. Unfortunately, the information provided does not include this probability. Could you please provide the required information?

To calculate the desired probabilities, we need to determine the success rate of passing the polygraph examination given the 20-minute instructions on how to cheat.

Let's assume that the success rate mentioned in Professor Ford's book (85%) is accurate and applicable to the random sample of seven students in the psychology laboratory. We'll denote the success rate as p = 0.85.

Now, let's calculate the probabilities for the following scenarios:

1. Probability that all seven students pass the polygraph examination:
This can be calculated using the binomial probability formula: P(X=k) = (n choose k) * p^k * (1-p)^(n-k)
P(all pass) = (7 choose 7) * 0.85^7 * (1-0.85)^(7-7) = 0.0851 (rounded to four decimal places)

2. Probability that at least five students pass the polygraph examination:
To find this probability, we need to consider the cases where 5, 6, or all 7 students pass.
P(at least five pass) = P(five pass) + P(six pass) + P(all pass)
P(at least five pass) = (7 choose 5) * 0.85^5 * (1-0.85)^(7-5) + (7 choose 6) * 0.85^6 * (1-0.85)^(7-6) + P(all pass)
P(at least five pass) = 0.1332 (rounded to four decimal places)

Therefore, the probabilities for the given scenarios are approximately:
1. Probability that all seven students pass the polygraph examination: 0.0851
2. Probability that at least five students pass the polygraph examination: 0.1332