Components of a certain type are shipped to a supplier in...

Components of a certain type are shipped to a supplier in batches of ten. Suppose that 50$\%$ of all such batches contain no defective components, 30$\%$ contain one defective component, and 20$\%$ of contain two defective components. Two components from a batch are randomly selected and tested. What are the probabilities associated with $0,1,$ and 2 defective components being in the batch under each of the following conditions? (a) Neither tested component is defective. (b) One of the two tested components is defective. [Hint: Draw a tree diagram with three first-generation branches for the three different types of batches.]

Components of a certain type are shipped to a supplier in batches of ten. Suppose that 50$\%$ of all
such batches contain no defective components, 30$\%$ contain one defective component, and 20$\%$ of contain two defective components. Two components from a batch are randomly selected and tested. What are the probabilities associated with $0,1,$ and 2 defective components being in the batch under each of the following conditions?
(a) Neither tested component is defective.
(b) One of the two tested components is defective.
[Hint: Draw a tree diagram with three first-generation branches for the three different types of batches.]

Key Concepts

Sampling Without Replacement

This refers to the method of drawing items from a batch where each item selected is not returned before the next selection. Since the two components tested are taken from a batch of ten without replacement, the probabilities change after the first draw, and this must be taken into account when calculating the likelihood of various outcomes.

Tree Diagram

A tree diagram is a graphical representation that shows all possible outcomes of a sequence of events along with their corresponding probabilities. It is particularly useful in this problem for mapping out the probabilities associated with the batch types and the outcomes of the tests, simplifying the calculation of conditional probabilities.

Bayes' Theorem

Bayes' Theorem provides a way to revise existing probabilities (the priors) after obtaining new information (the evidence). Here, it allows the calculation of the posterior probabilities of having 0, 1, or 2 defective components in the batch after observing whether the sampled components are defective or not.

Conditional Probability

This concept involves computing the probability of one event occurring given that another event has already occurred. In the context of the question, it is used to update the likelihood of various numbers of defective components in the batch based on the outcome of the tests conducted on the sampled components.

Law of Total Probability

This principle is used to break down complex probabilities into simpler, mutually exclusive cases. In this problem, it helps in determining overall probabilities by considering separate cases for batches with different numbers of defective components before updating with sample outcomes.

Transcript

00:01 Okay, so we can pick a batch of items, of 10 items, and 50%, so 0 .5 contain zero defective components.

00:16 And 0 .3 contain one defective components.

00:21 So i'm using the hint.

00:22 I'm drawing a tree diagram.

00:24 And 0 .2 contain two defective components.

00:27 That's your percent chance.

00:29 Now, once you've picked a batch, let's figure out how many, if i pick two items from the batch, well, there's, here there's a 100 % chance of neither being defective, 0 % chance of one being defective, and 0 % chance of 2 being defective.

00:53 Here, let's see, there's one defective item.

00:58 So how do i, i can't pick two defective items.

01:02 From here.

01:05 There's a, let's see, nine choose two over 10 choose two chance that i pick no defective items from this batch with one defective item.

01:24 Okay, so that's like nine times eight over 10 times nine because the two factorials would cancel.

01:31 So that's 0 .8.

01:33 0 .8 % chance that i get no.

01:36 Defective items which means there's a 0 .2 % chance that i get the one defective item.

01:43 Okay and over here let's see there's an 8 choose 2 over 10 choose 2 chance that i get no defective items 8 times 7 over 10 times 9 is 28 over 45 208 over 45 chance of getting no defective items.

02:09 What about getting both defective items that's one over 10 choose two right there's only one pair okay so then that's one over 45 um okay so total of 29 over 45 uh which means the remaining 16 over 25 over 45 is my chance of getting one defective item okay now i have all the probabilities now we're asking what is the probability if i know neither tested component is defective.

02:46 So i'm looking here.

02:49 Let's see, my probability of ending up on that whole branch is 0 .5 from the beginning.

02:56 Probability of ending up on that whole branch is 0 .24.

03:01 And probability of ending up on that whole branch is, let's see, 28 over 45 times 5, i guess.

03:16 Is that 225? i believe it is.

03:21 Yeah, 28 over 225.

03:28 Okay.

03:29 So that's my whole world in which i've tested nothing to be defective, neither of the two components to be defective.

03:41 So i just need the part where there's zero components over this whole world.

03:46 Okay, so p of zero.

03:49 Tested being defective is 0 .5 plus 0 .24.

03:56 I'm just going to raise this temporarily, plus 28 over 225.

04:01 And then what i need is 0 .5 divided by that.

04:16 So this is over p of 0.

04:20 This is the probability.

04:22 Okay, so that's 0 .5784 or 57 .84 percent.

04:31 This is.

04:31 This is, the probability of zero defective in batch given zero defective of the two tested...

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