• You are currently viewing our forum as a guest, which gives you limited access to view most discussions and access our other features. By joining our free community, you will have access to additional post topics, communicate privately with other members (PM), view blogs, respond to polls, upload content, and access many other special features. Registration is fast, simple and absolutely free, so please join our community today! Just click here to register. You should turn your Ad Blocker off for this site or certain features may not work properly. If you have any problems with the registration process or your account login, please contact us by clicking here.

Help with probability

redacted

Well-known member
Joined
Nov 28, 2007
Messages
4,223
I know I'm probably missing something really easy, but I can't figure this out right now and I'm burnt out from hours of programming.

Anyone?

(The easy answers seem too easy, but they might be right; I can't even think)
 

nomadic

mountain surfing
Joined
Jul 15, 2008
Messages
1,709
MBTI Type
enfp
i think since...

(0,0) = .08
(1,0) = .01

so (1,1) + (0,1) should = .91

im guessing (1,1) * 8 = (0,1)

(1,1) = .101 and (0,1) = .808

can't help but wonder if bayes method applies more tho... too hung over to look it up for you...
 

ygolo

My termites win
Joined
Aug 6, 2007
Messages
5,986
oops i just realized this is super easy. sorry y'all.

Really? It seems like you are missing information.

If you are assuming Y is independent of X because Joe's grading is not influenced by the professors writing, then P(Y=1|X=1)=P(Y=1)=1-P(Y=0)=1-[P(X=0,Y=0)+P(X=1,Y=0)]=1-[0.08+0.01]=0.91.

However, if we believe that Joe cannot grade a problem set unless the professor is writing...P(Y=1|X=0)=0 which means P(Y=1,X=1)=0.91, which means P(Y=1|X=1)=P(Y=1,X=1)/P(X=1)=0.91/(0.01+0.91)=0.989 approximately.

I'm wondering how you solved the conundrum. Independence seems like a more plausible interpretation.
 
Top