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Let A, B and C are independent events. How am I supposed to prove that:

  1. A′, B′ and C′ are independent.

  2. A, B′ and C′ , are independent.

  3. A, B and C' are independend.

This is my approach:

for Nr 3.

P(ABC') = P(A)P(B)P(C'). But P(AB)=P(ABC)+P(ABC') and using independence P(A)P(B) = P(A)P(B)P(C)+ P(ABC'), therefore P(A)P(B)(1-P(C))=P(ABC'), P(ABC') = P(A)P(B)P(C').

for Nr 2.

P(AB'C') = P(A)P(B')P(C'). But P(AC')=P(ABC')+P(ABC') and using independence P(A)P(C') = P(A)P(B)P(C')+ P(A B' C'), therefore P(A)P(C')(1-P(B))=P(AB' C'), P(AB'C') = P(A)P(B')P(C').

And for Nr 1.

P(A'B'C') = P(A')P(B')P(C'). But P(A'B')=P(A'B' C )+P(A'B'C') and using independence P(A')P(B') = P(A')P(B')P(C)+ P(A'B'C'), therefore P(A')P(B')(1-P(C))=P(A'B'C'), P(ABC') = P(A')P(B')P(C').

What do you think people? is this way of proving right?

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    I have finally sovled it !! Thanks all for your help, i figured out! You may close this thread ;)2012-03-30

2 Answers 2

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You have shown this, but for clarity, using independence you have $\Pr(ABC)=\Pr(A)\Pr(B)\Pr(C)$ and $\Pr(AB)=\Pr(A)\Pr(B)$ so \Pr(ABC')=\Pr(AB)-\Pr(ABC) $=\Pr(A)\Pr(B)- \Pr(A)\Pr(B)\Pr(C) $ $= \Pr(A)\Pr(B)(1-\Pr(C)) $ =\Pr(A)\Pr(B)\Pr(C')

so $A$, $B$ and C' are independent events.

Since this is all commutative, as martini says, you can then derive (2) and (1).

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    Actually, depending on what one assumes, proving that $\Pr(ABC^\prime)=\Pr(A)\Pr(B)\Pr(C^\prime)$ is not quite sufficient to prove that $A$, $B$, $C^\prime$ are independent. It is also necessary to assert (or remind the reader) that the other conditons $\Pr(AB)=\Pr(A)\Pr(B)$, $\Pr(AC^\prime)=\Pr(A)\Pr(C^\prime)$, $\Pr(BC^\prime)=\Pr(B)\Pr(C^\prime)$ follow from the results on pairwise independence of $A, B, C$.2012-03-29
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Your answer/proof is incomplete. In order to assert that three events $D$, $E$, $F$ are mutually independent, you have to verify that four equations hold: $\begin{align*} P(DEF) &= P(D)P(E)P(F)\\ P(DE) &= P(D)P(E)\\ P(DF) &= P(D)P(F)\\ P(EF) &= P(E)P(F)\\ \end{align*}$ Taking $D=A$, $E=B$, $F=C^\prime$, you have verified the first of the four equations above. Now you need to say that $P(AB)=P(A)P(B)$ follows from the independence of $A$ and $B$, and either prove that $P(AC^\prime) = P(A)P(C^\prime)$ and $P(BC^\prime) = P(B)P(C^\prime)$, or assert that these follow from the independence of $A$ and $C$, and $B$ and $C$ respectively if you have done these kinds of calculations previously.

Another definition of independence of $n$ events $A_i$ is that all $2^n$ equations $P(A_1^*A_2^*\cdots A_n^*) = P(A_1^*)P(A_2^*)\cdots P(A_n^*)$ must hold, where each $A_i^*$ stands for either $A_i$ or $A_i^\prime$, the same on both sides of the equation. With this definition, the statements to be proved in the OP's problem are true by definition and there is nothing to prove.

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    ^ quite enlightening post there, this is what i missed, finally , thanks, it seems that i totally forgot pairwise independence. Thats why, i had a weird feeling that my proof was just not right... Thanks!2012-03-29