Probability of Independent Events: showing abstractly that two events are independent

I haven't had a proofs class, so proofs always seem so complex to me. Here's the question: If A,B, and C are mutually independent, show that the following pairs of events are independent: A' and (B and C') (I understood the other ones). Also show that A', B', and C' are mutually independent.

Here's as far as I could get for the first part:

P(A') and P(B and C') = P(A') P(B) P(C')

1-p(a) and p(b) and (1-p(c))= p(a' and c') and p(b)

p(b)-p(a)p(b)(1-p(c))=(paUc)' and p(b)=

p(b)-p(b)p(c)-p(a)p(b)+p(a)p(b)p(c)=

For the second part I proved that each pair of a,b,c was independent, but I can't get very far with all three mutually independent. I know that p(a' and b' and c')= p(a')p(b')p(c') and then I multiplied out the first part taking the compliment of each as (1-p(x)) and multiplying the three.

I would really appreciate and help with this topic a lot! Thanks!

Re: Probability of Independent Events: showing abstractly that two events are indepen

Quote:

Originally Posted by

**lilyindecember** I haven't had a proofs class, so proofs always seem so complex to me. Here's the question: If A,B, and C are mutually independent, show that the following pairs of events are independent: A' and (B and C') (I understood the other ones). Also show that A', B', and C' are mutually independent.

There are major problems with definitions in this area of probability.

Here is the standard quote:

"If $\displaystyle A_1,\cdots,A_n$ are *mutually independent* events then every event determined by a subcollection of these events is independent of every event determined by a subcollection of the remaining events, Jim Pitman.

If we use that definition, it seems to solve this question.

Have you proved that: If $\displaystyle A~\&~B$ are independent events then $\displaystyle A~\&~B'$ are independent events?

Here is the prove of that.

We know that $\displaystyle \mathcal{P}(A)=\mathcal{P}(A\cap B)+\mathcal{P}(A\cap B')$.

So

$\displaystyle \begin{align*} \mathcal{P}(A) \mathcal{P}(B^{\,\prime}) &= \mathcal{P}(A)[1- \mathcal{P}(B)]\\ &= \mathcal{P}(A)- \mathcal{P}(A) \mathcal{P}(B) \\ &=\mathcal{P}(A)- \mathcal{P}(A\cap B) \\ &=\mathcal{P}(A\cap B^{\,\prime}) \end{align*}$.

Now you should be able to finish.

3 Attachment(s)

Re: Probability of Independent Events: showing abstractly that two events are indepen

Plato's response basically answers your question, but I thought I'd expand on it a "little". Here's my expansion:

Attachment 26938

Attachment 26939

Attachment 26942

If you bother to read the 3rd page, you'll see that Plato really completely answered the question.

Re: Probability of Independent Events: showing abstractly that two events are indepen

I'm sorry, I'm not sure I understand what you're talking about. The proof you gave makes sense @Plato, but I don't understand how it fits into the proof I've been given to work. Thanks for the input, though. I really appreciate your quick response!

-Lily

Re: Probability of Independent Events: showing abstractly that two events are indepen

To prove A', B' and C' are mutually independent, you have to show:

1. The probability of the intersection of any two of them is the product of the probabilities.

2. The probability of the intersection of all three is the the product of the probabilities.

1 is covered in 3 b) of the 1st page of my previous response

2 is covered in 4 c) of the 2nd page.

Finally, the theorem on page 3 says in particular, if A, B, C, D, etc are mutually independent, then A', B', C', D' are mutually independent.