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**mikau****Member**- Registered: 2005-08-22
- Posts: 1,504

This problem is driving me crazy. I got part(a) and I really think I'm doing (b) correctly but I'm not getting the same answer as my book.

**Componants 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% 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 componants 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**

Here's how i solved (a)

Let Bk be the event that k defective components are in a batch, and let Tk be the event that k of the two tested components are defective.

We seek P(B0 | T0), P(B1 | T0), and P(B2 | T0)

First, observing that Bk are mutually exclusive and exhaustive, we may use Bayes' theorem and compute these as:

P(Bi | T0)=P(T0 | Bi) P(Bi) /[ P(T0 | B0)P(B0) + P(T0 | B1)P(B1) + P(T0 | B2)P(B2)]

look at P(T0 | B0), P(T0 | B1) and P(T0 | B2).

P(T0 | B0) = 1, for if our batch contains no defective components, the probability of testing 0 defective components is 1.

P(T0 | B1) = (9 choose 2)/(10 choose 2), because we may test 2 out of all but the single defective component, and there are 10 choose 2 ways to test them total.

P(T0 | B2) = (8 choose 2)(10 choose 2), because we may test 2 out of all but the two defective components

Using these results in the formula for Bayes' theorem, gives me the correct answers for (a).

Now for (b), We seek P(B0 | T1), P(B1 | T1), and P(B2 | T1).

Using the same method as before, we now have to find the probabilities of P(T1 | B0), P(T1 | B1) and P(T1 | B2)

P(T1 | B0) = 0, for if there are no defective components, we cannot test 1 defective one

P(T1 | B1) = (9 choose 1)/(10 choose 2), as there are (9 choose 1) ways to select the remaining non-defective component to be tested out of (10 choose 2) ways total

P(T1 | B2) = (8 choose 1)/(10 choose 2), as there are (8 choose 1) ways to select the remaining non-defective component.

Plugging all this in to bayes' theorem gives me:

P(B0 | T1) = 0

P(B1 | T1) = 0.628

P(B2 | T1) = 0.372

Seems absolutely sound to me. However, my book claims the answers are:

0

0.457

0.543

I am particularly frustrated since my reasoning seemed to work for (a)

A logarithm is just a misspelled algorithm.

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**George,Y****Member**- Registered: 2006-03-12
- Posts: 1,316

P(T1 | B2) = (8 choose 1)/(10 choose 2), as there are (8 choose 1) ways to select the remaining non-defective component.

--Mistake here

--Just Check out HyperGeometric Distribution

**X'(y-Xβ)=0**

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**Gannicus****Guest**

You have to meet both requirements, that 1 be defective and the other be working. So for the B1 (batch with 1 defective component) scenario, you need to multiply 9c1 [ out of 9 working components choose 1 working component) and 1c1 (out of 1 defective component choose 1). So the correct P (C1|B1) it should be (9c1) * (1c1) / 10c2.

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