1. ## Help with probability

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)

2. 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...

3. oops i just realized this is super easy. sorry y'all.

4. Originally Posted by dissonance
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.

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