) = 0 If P(HinE) ≠0 for all i = 1 , 2 , 3 , ... , n, then P(
Hi
E
) =
P(HinE)
P(Hi)
×
P(Hi)
P(E)
=
P(
E
Hi
)×P(Hi)
P(E)
> P(
E
Hi
)×P(Hi) Hence, Statement-1 may not always be true. Statement-2: Clearly, we can write as K1 ∪ H2 ∪ H3 ∪ ... ∪ Hn = S ⇒ P(H1)+P(H2)+ ... + P(Hn) = 1 Hence, Statement-2 is true.