Independence and Events.

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?

Solutions Collecting From Web of "Independence and Events."

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

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

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

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:
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)\\
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.