Expected value, pairwise incompatibility and independency of events $A, B, C$ are three probability events, with probability $1/2, 1/3, 1/6$ respectively.
a) Indicating with X the number of how many of these events will occur, compute E(X).
b) What is the probability that at least one of them will occur $(X\geqslant{1})$ if the three events are pairwise incompatible?
c) What is the probability that at least one of them will occur $(X\geqslant{1})$ if the three events are completely independent?
My attempt:
a) $X=[0,1,2,3]$ so $E[X] = 0*p(0) + 1*p(1) + 2*p(2) + 3*p(3)$
= 0P({None of the events occurs}) + 1P({Just 1 event occur}) + 2P({Just 2 events occurs}) + 3P({All the events occurs})
= 0*$P({\bar{A}\cap\bar{B}\cap\bar{C}}) + 1*P({(A\cap\overline{{B\cup{C}}}})\cup({B\cap\overline{{A\cup{C}}}})\cup({C\cap\overline{{A\cup{B}}}})) + 2*P(({(A\cap{B})\cap\bar{C}})\cup({(A\cap{C})\cap\bar{B}})\cup({(B\cap{C})\cap\bar{A}})) + 3*P(A\cap{B}\cap{C})$
Now, using De Morgan $(\overline{{X\cup{Y}}}=\bar{X}\cap\bar{Y})$ to simplify the second and third addend:
= 0*$P({\bar{A}\cap\bar{B}\cap\bar{C}}) + 1*P({(A\cap\bar{B}\cap\bar{C}})\cup(B\cap\bar{A}\cap\bar{C})\cup(C\cap\bar{A}\cap\bar{B})) + 2*P(({(A\cap{B})\cap\bar{C}})\cup({(A\cap{C})\cap\bar{B}})\cup({(B\cap{C})\cap\bar{A}})) + 3*P({A\cap{B}\cap{C}})$
These events are clearly pairwise disjoint, so:
= 0*$P(\bar{A}\cap\bar{B}\cap\bar{C}) + P(A\cap\bar{B}\cap\bar{C})+P(B\cap\bar{A}\cap\bar{C})+P(C\cap\bar{A}\cap\bar{B}) + 2*P(A\cap{B}\cap\bar{C})+2*P(A\cap{C}\cap\bar{B})+2*P(B\cap{C}\cap\bar{A}) + 3*P({A\cap{B}\cap{C}})$
How can I continue from here if I don't know if there is dependence or not in this latest events?
b) We know from hypothesis that $A\cap{B}=∅, A\cap{C}=∅, B\cap{C}=∅$ so I can say with confidence that the probability of the union of the events is the sum of the probabilities:
P({At least one event occur}) = $P(A\cup B \cup C) = P(A) + P(B) + P(C) = 1/2 + 1/3 + 1/6 = 1$
c) We know from hypothesis that $A∩B∩C=∅$ so I can say with confidence that $P(A∩B∩C)=P(A)*P(B)*P(C).$ It's easy to see that if A, B, C are mutually independent, the complements are independent.
P({At least one event occur}) = $P(A\cup{B}\cup{C})$ = 1 − P({None of the events will occur}) = 1 − $P({\bar{A}\cap\bar{B}\cap\bar{C}}) = 1-[(1-1/2)(1-1/3)(1-1/6)] = 13/18 = 0.72$
Is this correct?

I think the problem is much simpler than initially assumed.
a) Let say that X is an arbitrary variable that counts how many of these events will occur and $\mathbb{1}_{A}$ the characteristic function of A, $\mathbb{1}_{B}$ the characteristic function of B and $\mathbb{1}_{C}$ the characteristic function of C. We even know that [$\mathbb{1}_{X}$]=P().
So $E[X] = E[\mathbb{1}_{A}+\mathbb{1}_{B}+\mathbb{1}_{C}] = E[\mathbb{1}_{A}]+E[\mathbb{1}_{B}]+E[\mathbb{1}_{C}]=P(A)+P(B)+P(C)=1.$
b) We know from hypothesis that $A\cap{B}=∅, A\cap{C}=∅, B\cap{C}=∅$ so I can say with confidence that the probability of the union of the events is the sum of the probabilities:
P({At least one event occur}) = $P(A\cup B \cup C) = P(A) + P(B) + P(C) = 1/2 + 1/3 + 1/6 = 1$
c) We know from hypothesis that $A∩B∩C=∅$ so I can say with confidence that $P(A∩B∩C)=P(A)*P(B)*P(C).$ It's easy to see that if A, B, C are mutually independent, the complements are independent.
P({At least one event occur}) = $P(A\cup{B}\cup{C})$ = 1 − P({None of the events will occur}) = 1 − $P({\bar{A}\cap\bar{B}\cap\bar{C}}) = 1-[(1-1/2)(1-1/3)(1-1/6)] = 13/18 = 0.72$
That's all? Could anyone confirm?
