# Thread: Prove the probability statement

1. ## Prove the probability statement

Let events A, B and C be independent. Show that (A union B) and C are independent.

Attempt at solution:

Setup: If [A intersection B intersection C] = [A][B][C], then show that [(A union B) intersection C] = [A union B][C].

I'm not sure how to go about doing this. Would using conditional probability theorem be a good start?

2. Originally Posted by My Little Pony
Let events A, B and C be independent. Show that (A union B) and C are independent.

Attempt at solution:

Setup: If [A intersection B intersection C] = [A][b][C], then show that [(A union B) intersection C] = [A union B][C].

I'm not sure how to go about doing this. Would using conditional probability theorem be a good start?
Suppose events $A$, $B,$ and $C$ are independent. Then,

$P(A\cap\\C)=P(A)P(C)$, $P(B\cap\\C)=P(B)P(C)$, and

$P(A\cap\\B\cap\\C)=P(A)P(B)P(C)$ .

Now, $(A\cup\\B)\cap\\C=(A\cap\\C)\cup\\(B\cap\\C)$, by the distributive law.

So,

$P((A\cup\\B)\cap\\C)=P((A\cap\\C)\cup\\(B\cap\\C))$

$=P(A\cap\\C)+P(B\cap\\C)-P((A\cap\\C)\cap\\(B\cap\\C))$

$=P(A)P(C)+P(B)P(C)-P(A)P(C)P(B)P(C)$

$=P(C)[P(A)+P(B)-P(A)P(B)]=P(C)P(A\cup\\B)=P(A\cup\\B)P(C)$, as was to be shown.

All that was needed to prove this was the distributive law, multiplicative law of probability, and the additive law of probability.

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### lf A,B andC are indipendent events show that A and BunionC(not) are also independent

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