Law of total probability for conditional probability

Unparseable latex formula:

\( P(A \mathbin{\vert} C) = \Sigma P(A \mathbin{\vert} Bi \cap C)P(Bi \mathbin{\vert} C) \)

. Can someone give me a tip on proving this? I get how the proof for just P(A) works but I'm unsure on what I should do with the C. Thanks.

Original post by Roger The Doger

Unparseable latex formula:

\( P(A \mathbin{\vert} C) = \Sigma P(A \mathbin{\vert} Bi \cap C)P(Bi \mathbin{\vert} C) \)

. Can someone give me a tip on proving this? I get how the proof for just P(A) works but I'm unsure on what I should do with the C. Thanks.

You could start from

\displaystyle P(A|C)=\frac{P(A\cap C)}{P(C)}

, if that's acceptable.

Then split the AnC over the Bi's - I assume they're mutually exclusive and exhaustive.

Original post by ghostwalker

You could start from

\displaystyle P(A|C)=\frac{P(A\cap C)}{P(C)}

, if that's acceptable.

Then split the AnC over the Bi's - I assume they're mutually exclusive and exhaustive.

yeh we can do that but I don't see how it will help? Were meant to adapt theorem 3.11 (attached).

Original post by Roger The Doger

yeh we can do that but I don't see how it will help? Were meant to adapt theorem 3.11 (attached).

The significant step to 3.11 is splitting the event A over the Bi's. If you do that with AnC we have:

\displaystyle P(A|C)=\sum_{i\in I}\frac{P(A\cap C\cap B_i)}{P(C)}

Now start splitting up the numerator - keep an eye on what you're aiming for.

Original post by ghostwalker

The significant step to 3.11 is splitting the event A over the Bi's. If you do that with AnC we have:

\displaystyle P(A|C)=\sum_{i\in I}\frac{P(A\cap C\cap B_i)}{P(C)}

Now start splitting up the numerator - keep an eye on what you're aiming for.

Thanks.

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