Question

The first outcome is one of X, Y, and Z, and the second outcome in each case is one of A and B. Use the following probabilities:
Pr[X]=4/7 Pr[Y]=1/7 Pr[Z]= 2/7
Pr[A|X]= 1/5 Pr[B|X]= 4/5 Pr[A|Y]=2/3 Pr[B|Y]=1/3 Pr[A|Z]=3/8 Pr[B|Z]= 5/8
Find the missing probabilities:
1. Pr[B|Y]=___? 2. Pr[Y|B]=___? 3. Pr[(XorY|B]___?

Answer 1

I need savved Mr. Dumbcow

Answer 2

haha ok, hold on for a bit...it might take awhile to sort everything out

Answer 3

No problem

Answer 4

I have a ton of these and they are all related to Bayes probablilites

Answer 5

are you sure #1 is correct?
P(B|Y) = 1/3

Answer 6

Yes it is correct
Sorry I didnt mean to post that question

Answer 7

P(B) = P(B|X)*P(X) +P(B|Y)*P(Y) +P(B|Z)*P(Z) = 41/60

P(Y|B) = [P(B|Y)*P(Y)]/P(B) = 20/287

P(XorY|B) = P(X|B) +P(Y|B)

P(X|B) = [P(B|X)*P(X)]/P(B) = 192/287

P(XorY|B) = 212/287

Answer 8

WOW you are a genius
right on the money

Answer 9

I have quite a few of these

Answer 10

just remember

P(a|b) = P(ab)/P(b)
and
P(a) = P(ax)+P(ay)+P(az)...