2

I am reading from Daphne Koller's book and I don't understand how she find this probability:

She provides this table (on page 105):

http://i.imgur.com/Kg53aga.png

and claims that "For example, by summing out A, C, and D, we obtain P($b^{1}$) ≈ 0.732 and P($b^{0}$) ≈ 0.268" which is understandable to me. Then she says that: "On the other hand, if we now observe that Charles does not have the misconception ($c^{0}$), we obtain P($b^{1} | c^{0}$) ≈ 0.06." This is on the end of page 105.

I don't get it how this 0.06 comes up. Any help much appreciated.

the book: Probabilistic Graphical Models: Principles and Techniques, by D. Koller and N. Friedman, MIT Press, 2009.

Md. Poe
  • 23
  • 4
  • @Creator yeah sure, i did that , but it doesnt come up as 0.06. – Md. Poe Dec 04 '16 at 21:54
  • 1. Please give a full reference to the book (author, year, title publisher) 2. What answer do you get? Can you outline your calculations? I just got an answer that is approximately 0.06. – Glen_b Dec 04 '16 at 22:10
  • @Glen_b I basically add the following "Assignments" from the given table: $(a^{1}b^{1}c^{0}d^{1})+(a^{1}b^{1}c^{0}d^{0})+(a^{0}b^{1}c^{0}d^{1})+(a^{0}b^{1}c^{0}d^{0})$ (added the ref to the book btw). – Md. Poe Dec 04 '16 at 22:25
  • 1
    Well you're correct that those four don't add to 0.06 ... but how does that produce a conditional probability? – Glen_b Dec 04 '16 at 22:29
  • @Glen_b I don't know really. I just thought that I have to add up all instances where $b^{1} \text{ and } c^{0}$ happened at the same time. Do I have to use bayes rule or something? – Md. Poe Dec 04 '16 at 22:40
  • To add to Glen_b's reply to your previous comment: For an assignment from the table that includes $c^0$, will it ever NOT include $b^1$? (If so, these are not in your sum.) – GeoMatt22 Dec 04 '16 at 22:45

1 Answers1

2

Consider this conditional probability fact (/definition):

$$P(A|B) = P(AB)/P(B)$$

So in the context of the current problem

$\quad P(b^{1}|c^{0})= P(b^{1}c^{0})/P(c^{0})$ where

$\quad P(b^{1}c^{0})=6.9\times 10^{-5}+6.9\times 10^{-5}+... $

$\quad P(c^{0}) = 0.04 + ....$

(I'm leaving some details for you to fill in for yourself. The numerator terms are a subset of the denominator terms)

So $ P(b^{1}c^{0})/P(c^{0}) \approx 0.06$

Glen_b
  • 257,508
  • 32
  • 553
  • 939
  • If it helps, you can think of the [venn diagram](http://stats.stackexchange.com/a/239042/127790) illustration of conditional probability. (Here the set $B$ could be thought of as a "bucket holding all assignments where $c=c^0$".) – GeoMatt22 Dec 04 '16 at 22:54