It is known that diskettes produced by a certain company...

It is known that diskettes produced by a certain company will be defective with probability.01, independently of one another. The company sells the diskettes in packages of size 10 and offers a money-back guarantee that at most 1 of the 10 diskettes in the package will be defective. The guarantee is that the customer can return the entire package of diskettes if he or she finds more than 1 defective diskette in it. If someone buys 3 packages, what is the probability that he or she will return exactly 1 of them?

Key Concepts

Combinatorial Methods

Combinatorial methods, such as computing combinations (often expressed as 'n choose k'), are used to determine the number of ways a certain number of successes (or defects) can occur in a sequence of trials. This is essential when applying the binomial probability formula, as it provides the coefficient that accounts for the different arrangements in which the successes can appear.

Binomial Distribution

The binomial distribution is a probability model that describes the number of successes in a fixed number of independent trials with the same probability of success in each trial. It is used to calculate probabilities where each trial has only two outcomes, such as a diskette being either defective or not. This model applies when determining the likelihood of observing a certain number of defective items in a package.

Independent Events

Independence in probability means that the occurrence of one event does not affect the probability of another. In the context of diskettes, each diskette being defective is independent of the others, and the outcome for one package is independent of another. This assumption allows for the multiplication of probabilities when calculating the chance of multiple packages meeting a particular defect criterion.

Threshold-based Probability Calculation

This concept involves calculating the probability that a random variable exceeds or falls below a specific threshold. In quality control problems like this, a guarantee may be based on not exceeding a certain number of defective items. Determining the probability that a package has more than the permitted threshold uses the cumulative probabilities derived from the binomial distribution.

Transcript

00:01 For the first problem, we want to find the probability that a test was tested and it failed.

00:08 So the equation that we're going to use is we're going to find the probability that a cd was tested and we'll do t as tested times the probability, the conditional probability that it fails, but it was tested.

00:31 And so what this is going to be is d is going to be that we failed it, but it was tested.

00:39 And so we already know that the probability that it was that the cd was defective.

00:46 So d is denoted as defective.

00:50 So d not is that the cd was not defective.

00:55 So what p of t is, is we have a sample space of four.

01:01 So it will be 1 divided by 4 times because this is denoted as 1 cd 1 minus we know the prevalence rate of the fail rate already so to find d not all we do is do 1 minus each of the fail rates and that will give us the pass rate so we do 1 minus p of d times d1 times p of d2 times p of d3 times p of d4 so you get it's point 25 times 1 minus and when you put all of these together well when you convert them first 0 .99 because your first one is 0 .01 so if you do 1 minus that you get 0 .99 so that's how we got that number, times .97, times .98 times .99.

02:43 And your final answer is .0 -170.

02:50 So for the next question, we want to find the probability that programs 2 and 3 failed from the example from the city.

03:01 And so what we do is that if we know 2 and 3 failed, we know 1 and 4.

03:06 Did not fail and so if we know the fail rates is 0 .01 for instance for the first cd we know that the pass rate is 0 .99 so we're going to do the conditional probability that two or three passes over all in a world where all the cds passed where there was no deficit i mean defectiveness and so what we'll do for this problem is is we do do the probability that the first one and the probability that the second one they do pass.

04:03 And so we're going to do times one minus the probability of the second and third failing because that we know that two or three fails.

04:20 Sorry, and this is supposed to be a four.

04:33 And so what we get is when we put these, when we acquisition, we're going to use, we're already given our fail rates, so we're going to use our pass rates.

04:42 So we're going to do one minus each of the different data's 1 minus 0 .03...

Please give Ace some feedback