Reply 1
Reply 2
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Suppose you want to work out the usual marginal, joint and conditional probabilities, so the question gives
p(C)=1/2, p(B)=1/6, p(F) = 1/3
p(L|C) = 1/5, p(L|B)=1/10, p(L|F)=1/20
so put the relevant ones on your tree and think which ones you can work out.
p(C)=1/2, p(B)=1/6, p(F) = 1/3
p(L|C) = 1/5, p(L|B)=1/10, p(L|F)=1/20
so put the relevant ones on your tree and think which ones you can work out.
Reply 3
Original post
by mqb2766Suppose you want to work out the usual marginal, joint and conditional probabilities, so the question gives
p(C)=1/2, p(B)=1/6, p(F) = 1/3
p(L&C) = 1/5, p(L&B)=1/10, p(L&F)=1/20
so put the relevant ones on your tree and think which ones you can work out. Id probably start by getting p(L) and p(C|L)...
p(C)=1/2, p(B)=1/6, p(F) = 1/3
p(L&C) = 1/5, p(L&B)=1/10, p(L&F)=1/20
so put the relevant ones on your tree and think which ones you can work out. Id probably start by getting p(L) and p(C|L)...
Hm...I started by using the conditional probability formula to work out P(L∩C), P(L∩B) and P (L∩F), I then feel like it would be a good idea to use the P(A∩B) = P(A) x P(B) formula to find P(L). Is this the right strategy? Can I assume independence?
Reply 4
21
Original post
by TwisterBlade596Hm...I started by using the conditional probability formula to work out P(L∩C), P(L∩B) and P (L∩F), I then feel like it would be a good idea to use the P(A∩B) = P(A) x P(B) formula to find P(L). Is this the right strategy? Can I assume independence?
Edit -
Its generally wrong to assume stuff (like independence), but youre correct in working out the joints which you can mark on as 3 leaves. Can you then work out the marginal probability p(L) from the joints (think about each joint in terms of the other conditionals like p(C|L), ...) and things should start to fall into place.
(edited 1 year ago)
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