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? (Round your answers to three decimal places.)
a.) all the students are able to pass the polygraph examination
b.) more than half the students are able to pass the polygraph examination
c.) no more than half of the students are able to pass the polygraph examination
d.) all the students fail the polygraph examination
To calculate the probabilities, we need to know the success rate of passing the polygraph examination after receiving the instructions. The information you provided states that 85% of those trained were able to pass the polygraph examination. We will assume this success rate applies to the students in the psychology laboratory as well.
Let's go through each probability:
a) Probability that all students are able to pass the polygraph examination:
Since each student's success is independent of the others, we can multiply the individual probabilities. The probability of a student passing is 85%, so the probability of all seven students passing is (0.85)^7. Calculating this:
(0.85)^7 = 0.371.
Therefore, the probability that all students are able to pass the polygraph examination is approximately 0.371.
b) Probability that more than half the students are able to pass the polygraph examination:
To calculate this probability, we need to sum the probabilities of each possible scenario where more than three students pass (half of seven is 3.5, so we consider scenarios with four or more passing). We can use the binomial probability formula:
P(X > 3) = P(X = 4) + P(X = 5) + P(X = 6) + P(X = 7),
where X follows a binomial distribution with n = 7 (number of trials) and p = 0.85 (probability of success). Calculating this:
P(X > 3) = C(7, 4) * (0.85)^4 * (1 - 0.85)^3 + C(7, 5) * (0.85)^5 * (1 - 0.85)^2 + C(7, 6) * (0.85)^6 * (1 - 0.85)^1 + C(7, 7) * (0.85)^7 * (1 - 0.85)^0,
where C(n, r) denotes the number of combinations of choosing r elements from a set of size n. Evaluating this expression:
P(X > 3) ≈ 0.912.
Therefore, the probability that more than half the students are able to pass the polygraph examination is approximately 0.912.
c) Probability that no more than half of the students are able to pass the polygraph examination:
This is the complement of the probability in part b. Thus:
P(no more than half) = 1 - P(X > 3) = 1 - 0.912 = 0.088.
Therefore, the probability that no more than half of the students are able to pass the polygraph examination is approximately 0.088.
d) Probability that all students fail the polygraph examination:
Similar to part a, since each student's success is independent of the others, we can multiply the individual probabilities of failing. The probability of a student failing is the complement of passing, which is 1 - 0.85 = 0.15. Calculating this:
(0.15)^7 ≈ 0.000170.
Therefore, the probability that all students fail the polygraph examination is approximately 0.000170.
To calculate the probabilities, we will assume that each student's ability to pass the polygraph examination is independent of the others, and all students receive the same 20 minutes of instruction.
a.) To find the probability that all the students are able to pass the polygraph examination, we can use the probability of success for a single student. From the information given, 85% of those trained were able to pass the examination. Therefore, the probability that a student can pass is 0.85. Since each student's ability is independent, the probability that all seven students pass is (0.85)^7, which is approximately 0.368.
b.) To find the probability that more than half the students are able to pass the polygraph examination, we need to consider the possibilities where 4, 5, 6, or 7 students pass. We can use the binomial distribution formula to calculate this probability. Let's define X as the number of students who pass. Then, the probability is:
P(X > 3) = P(X = 4) + P(X = 5) + P(X = 6) + P(X = 7)
Using the binomial distribution formula, where n is the number of trials (7 students) and p is the probability of success (0.85):
P(X > 3) = C(7, 4)*(0.85)^4*(1-0.85)^(7-4) + C(7, 5)*(0.85)^5*(1-0.85)^(7-5) + C(7, 6)*(0.85)^6*(1-0.85)^(7-6) + C(7, 7)*(0.85)^7*(1-0.85)^(7-7)
Using the binomial coefficient C(n, r) = n! / (r!(n-r)!):
P(X > 3) = 35*(0.85)^4*(0.15)^3 + 21*(0.85)^5*(0.15)^2 + 7*(0.85)^6*(0.15)^1 + 1*(0.85)^7*(0.15)^0
After calculating the expression above, we find that P(X > 3) is approximately 0.923.
c.) To find the probability that no more than half of the students are able to pass the polygraph examination, we need to consider the possibilities where 0, 1, 2, 3, or 4 students pass. Using the same binomial distribution formula, we can calculate this probability:
P(X <= 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4)
P(X <= 3) = C(7, 0)*(0.85)^0*(1-0.85)^(7-0) + C(7, 1)*(0.85)^1*(1-0.85)^(7-1) + C(7, 2)*(0.85)^2*(1-0.85)^(7-2) + C(7, 3)*(0.85)^3*(1-0.85)^(7-3) + C(7, 4)*(0.85)^4*(1-0.85)^(7-4)
After calculating the expression above, we find that P(X <= 3) is approximately 0.071.
d.) To find the probability that all the students fail the polygraph examination, we can use the complement of the probability that at least one student passes. So,
P(all fail) = 1 - P(at least one passes) = 1 - P(X >= 1)
Using the same binomial distribution formula, we can calculate this probability:
P(all fail) = 1 - (P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6) + P(X = 7))
After calculating the expression above, we find that P(all fail) is approximately 0.004.
Note: Please keep in mind that the assumptions and probabilities used in this calculation are based on the information provided in the question and may not represent the true performance of lie detector tests or the accuracy of passing instructions.
this is a binomial probability ... pass/fail for 7 students
... probability pass (p) = .85 ... probability fail (f) = .15
(p + f)^7 = p^7 + 7 p^6 f + 21 p^5 f^2 + 35 p^4 f^3 + 35 p^3 f^4 + ... + f^7
evaluate the term(s) which represent the given situation
a.) p^7 = .85^7
b.) sum of 1st four terms
c.) wording?
d.) last term