# Math Is Fun Forum

Discussion about math, puzzles, games and fun.   Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫  π  -¹ ² ³ °

You are not logged in.

## #1 2010-10-01 15:36:01

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

### Probability and Statistics - Bayes' Theorem

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.

Offline

## #2 2010-10-04 22:51:12

George,Y
Member
Registered: 2006-03-12
Posts: 1,379

### Re: Probability and Statistics - Bayes' Theorem

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

Offline

## #3 2012-10-15 16:16:03

Gannicus
Guest

### Re: Probability and Statistics - Bayes' Theorem

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.