Original post by mqb2766

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. 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?

Original post by TwisterBlade596

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?

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 3 weeks ago)

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can someone please explain what principle domain is and why the answer is a not c?Maths

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