Expert: Paul Klarreich - 12/25/2009

Question
Setswana Nosks estimates that the probability its lathe tool is properly adjusted is 0.8. When the lathe is properly adjusted, there is a 0.9 probability that the parts produced pass inspection. If the lathe is out of adjustment, howeever, the probability of a good part being produced is only 0.2. A part randomly chosen is inspected and found to be acceptable. What is the posterior probability that lathe tool is properly adjusted?

This is what I have so far:
P(adjusted)= .50     P(out of adjustment)=.50
P (.08|adjusted)       P(.08|out of adjustment)


Answer
Questioner: Liza
Country: Puerto Rico
Category: Advanced Math
Private: No
Subject: Probability
Question: Setswana Nosks estimates that the probability its lathe tool is properly adjusted is 0.8. When the lathe is properly adjusted, there is a 0.9 probability that the parts produced pass inspection. If the lathe is out of adjustment, howeever, the probability of a good part being produced is only 0.2.
A part randomly chosen is inspected and found to be acceptable. What is the posterior probability that lathe tool is properly adjusted?

This is what I have so far:
P(adjusted)= .50     P(out of adjustment)=.50
P (.08|adjusted)       P(.08|out of adjustment)
..........................................

Notation:  If I write AB, that is intersection, A+B is union.


The basic rule is that  P(A / B) = P(AB)/P(B)

So if  A = 'lathe is adjusted'
and    G = 'part is good'  (passes inspection)

you have:

the probability its lathe tool is properly adjusted is 0.8 means:

P(A) = 0.8

When the lathe is properly adjusted, there is a 0.9 probability that the parts produced pass inspection means:

P(G / A) = 0.9

If the lathe is out of adjustment, howeever, the probability of a good part being produced is only 0.2.  means:

P(G / A') = 0.2
......................
Now if:

P(A) = 0.8,  and
P(G / A) = 0.9

then

P(G / A) = P(AG) / P(A)

So:  0.9 = P(AG) / 0.8

and  P(AG) = 0.72
.......................
And if:
P(A) = 0.8,  then  P(A') = 0.2

and if P(G / A') = 0.2

then

P(G / A') = P(A'G) / P(A')

0.2 = P(A'G) / 0.2

P(A'G) = 0.04
..................................
Now:

A part randomly chosen is inspected and found to be acceptable.

G is known.


What is the posterior probability that lathe tool is properly adjusted?

PLEASE!   That is 'a posteriori'!  The posterior is the part of you that touches the toilet seat.

(Actually, it's "a priori", but who's counting?)


Now  event G = AG + A'G

So: p(G) = P(AG) + P(A'G) = 0.72 + 0.04 = 0.76

Does that make sense?