Calculating Conditional Probabilities with Mutually Exclusive Events

[quantumnano]

How would I evaluate the following:
P(A|(B or C)) where B and C are two mutually exclusive events.

I have scoured a couple texts and the internet but have made no headway, any insight would be greatly appreciated.

QN

 

[mathman]

My guess: P(A|B or C) = P(A|B) + P(A|C) as long as B and C are mutually exclusive.

 

[statdad]

My guess: P(A|B or C) = P(A|B) + P(A|C) as long as B and C are mutually exclusive.

No. Consider drawing one card from a standard deck.

\begin{align*}<br /> A &amp; = \text{ card is an Ace}\\<br /> C &amp; = \text{ card is a Club}\\<br /> H &amp; = \text{ card is a Heart}<br /> \end{align*}<br />

Then C, H are mutually exclusive. Now

<br /> \begin{align*}<br /> P(A \mid C) &amp; = \frac 1 {13} \\<br /> P(A \mid H) &amp; = \frac 1 {13} \\<br /> P(A \mid C) + P(A \mid H) &amp; = \frac{2}{13} \\<br /> P(A \mid C \cup H) &amp; = \frac{2}{26} = \frac 1{13}<br /> \end{align*}<br />

The problem is this (for general events)

<br /> P(A \mid B \cup C) = \frac{P(A \cap (B \cup C))}{P(B \cup C)} = \frac{P(A \cap B)}{P(B) + P(C)} + \frac{P(A \cap C)}{P(B) + P(C)} \tag{1}<br />

and there is no simple general way to split (1) into a sum of different fractions where one has denominator P(B) and the other has denominator P(C).