Aldrich Ames is a convicted traitor who leaked American...

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 backward by 7 are countermeasures that are difficult to detect but can change the results of a polygraph examination (Source: Lies' Liesi't Lies\%' The Psychology of Deceit, by C. V. Ford, professor of psychiatry, University of Alabama). 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 nine 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 is the probability that (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 four of the students are able to pass the polygraph examination? (d) all the students fail the polygraph examination?

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 backward by 7 are countermeasures that are difficult to detect but can change the results of a polygraph examination (Source: Lies' Liesi't Lies\%' The Psychology of Deceit, by C. V. Ford, professor of psychiatry, University of Alabama). 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 nine 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 is the probability that
(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 four of the students are able to pass the polygraph examination?
(d) all the students fail the polygraph examination?

Key Concepts

Cumulative Probabilities

Cumulative probabilities involve summing the probability mass function over a range of values to determine the likelihood of an event occurring within a specified interval. This concept is crucial when the problem asks for compound probabilities like 'more than half' or 'no more than four' successes.

Independent Trials

An essential assumption underlying the binomial model is that each trial is independent, meaning the outcome of one trial does not affect another. This assumption allows for the multiplication of individual probabilities across trials.

Binomial Distribution

The binomial distribution describes the probability of obtaining a fixed number of successes in a specific number of independent trials, where each trial has the same probability of success. It is fundamental in probability theory for modeling binary outcomes like pass/fail scenarios.

Probability Mass Function

The probability mass function specific to the binomial distribution calculates the probability of exactly k successes in n trials. It uses the combination formula C(n, k) multiplied by the success probability raised to the k power, and the failure probability raised to the (n-k) power.

Transcript

00:01 In question 21, we have a situation in which we are looking at a binomial setting.

00:07 It says, it is reported in professor ford's book that only after 20 minutes of instruction by buzz, fay, 85 % of those trained were able to pass the polygraph examination, even when guilty of the crime.

00:23 Suppose that a random sample of nine students, we're just going to look at a sample of nine students.

00:30 They're told a secret and then given instructions on how to pass the polygraph exam without revealing their knowledge of the secret.

00:38 What is their probability that, and we have a few situations here, so the probability of passing, so let's just do the probability of passing is 0 .85.

00:53 And part a says, what's the probability that all students are able to pass the polygraph exam? so we're going to say the probability of x being equal to nine.

01:11 So this is a binomial setting in which we have n to be nine and the probability is 0 .85.

01:20 So we can use our ti -80 for calculator, which has a cool binomial pdf feature in which we plug in the number of trials to be 9.

01:30 The probability of 0 .85.

01:33 And we want to set our x value of what we're solving to be none.

01:38 So when we do this, we get 0 .232.

01:50 We could also do this by hand, where we figure out the number of ways to get non -successes out of non -possible trials, where we have the probability of success to be none, and failure to be zero.

02:08 The same model just changed the exponents depending on how many successes you're looking for for any of these problems.

02:17 And we get the same answer, which is ultimately just 0 .85 to the ninth power in this case, in this particular part.

02:25 All right, in part b, we want to know what's the probability that more than half the students were able to pass the polygraph exam? so the probability that x is more than half the students, which is 4 .5, so greater than or equal to five in this case because we can't have a half of a student so five six seven eight so this is also a binomial setting same parameters in as nine and the probability is point eight five so we can also use our t a eighty four calculator in this case looking at a binomial cdf cdf basically takes a value and adds together that probability and all the probabilities to the left of it um so in this case we have 0 all the way up to 9 possibilities of a success.

03:22 What we're looking for here is 5, 6, 7, 8, or 9, but our calculator is left -sided, meaning our calculator wants to take a certain value we plug in and calculate that value in every single value to the left of it.

03:36 So the greatest value that i can plug in is 4.

03:42 And why i'm going to do that, where n is 9, p is 0 .85, and i'm going to plug in four here because this is going to tell me this red box of probabilities the probability of 0 1, 2, 3, or 4...

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