Quote (feanur @ Jan 17 2017 04:17pm)
When it comes to conditional probabilities, I am used to this formulae :
p(A|B) = p(A ∩ B ) / p(B)
(assuming, of course, that p(B) ≠ 0, but in this case all the results are obvious).
(a)
p ( C | ∪Di ) = p ( C ∩ (∪Di) ) / p (D)
where D = ∪Di
... = p ( ∪ (C∩Di) ) / p (D)
... = Σ p (C∩Di) / p (D)
since all Di are disjoint, then all C∩Di are also disjoint
... = Σ p*p (Di) / p (D)
since, for all i, p (C∩Di) = p (C|Di)*p(Di) = p*p(Di)
... = p * (Σ p (Di)) / p (D)
... = p * p(D) / p(D)
... = p
(b)
p (∪Ci | D) = p ( (∪Ci) ∩ D) / p(D) = p( ∪ (Ci ∩ D)) / p(D)
... = Σ p(Ci ∩ D) / p(D)
since all Ci are disjoint, then all (Ci ∩ D) are also disjoint
... = Σ p(Ci | D)
(c)
Σ p(Ei | D)*p(C | Ei ∩ D) = Σ p(Ei ∩ D) * p(C ∩ Ei ∩ D) / ( p(D)*p(Ei ∩ D) )
... = Σ p(C ∩ Ei ∩ D) / p(D)
just after a cancelletion,
... = p(C ∩ D) / p(D)
because, for every A, p(A) = Σ p(A ∩ Ei)
... = p(C | D)
(d)
p(A | ∪Ci ) =p(A ∩(∪Ci)) / p(∪Ci) = p(∪(A∩Ci)) / p(∪Ci)
... = Σ p(A∩Ci) / Σ p(Ci)
since all Ci are disjoint, then all (A ∩ Ci) are also disjoint
... = Σ p(B∩Ci) / Σ p(Ci)
just because of your hypothesis : p ( A | Ci ) = p ( A ∩ Ci) / p(Ci) = p ( B | Ci ) = p ( B ∩ Ci) / p(Ci) , hence, p ( A ∩ Ci) = p ( B ∩ Ci)
... and just 'walk' back to B (or do the same for B )
... = p(B | ∪Ci )
I hope it helps, good luck !
woww! thx a lot feanur i really appreciate it. sry dont have any fg to give atm
