Can anyone help me with a probability question? It is about conditional probability.

Question

Let $A, B \text{ and } C$ be events in a sample space while $A$ and $B$ are disjoint events. We know $P(A) = 2P(B)$, $P(C|A) = \frac{2}{7}$, $P(C|B) = \frac{4}{7}$. What is the result of $P(C | (A \cup B))$?

I was trying to expand the $P(C | (A \cup B))$ and plug in $P(A)$ and $P(B)$ to find the probability. But I found that this is not a good way to approach this problem.

Answer 1

What was the problem you found while expanding? It seems straightforward

$$ P(C | A \cup B) = \frac{P(C\cap (A \cup B)}{P(A \cup B)} = \frac{P((C \cap A)\cup (C \cap B))}{P(A)+P(B)} \stackrel{*}{=} \frac{\frac 47 P(B) + \frac47 P(B)}{3P(B)} = \frac{8}{21} $$

Note that is this situation you have that:

1. $P(A \cup B) = P(A)+P(B)-P(A \cap B) = P(A)+P(B)$

2. Since $A$ and $B$ are disjoint, $C\cap A$ and $C\cap B$ are also disjoint and so $P((C\cap A) \cup (C\cap B)) = P(C\cap A)+P(C \cap B)$

3. $P(C \cap A)= P(C|A) P(A) = \frac 27 P(A) = \frac 47 P(B)$

4. $P(C\cap B) = P(C|B) P(B) = \frac 47 P(B)$

Answer 2

Sketch the Venn diagram: three linked rings like part of the Olympic rings... I drew them going left to right in the order A, C, B.

Because A and B are disjoint there is no overlap between A and B.

In the intersection of A and C you can put $2a$ and in the rest of A you can put $5a$. Can you see that this gives you the result $P(C|A)=\frac 2 7$?

In the intersection of B and C you can put $4b$ and in the rest of B you can put $3b$. That gives you the result $P(C|B)=\frac 4 7$.

You should see that $P(A)=7a$ and $P(B)=7b$.

Since $P(A)=2P(B)$ we have $7a=14b \Rightarrow a=2b$.

Rewrite the probabilities that were $5a$ and $2a$ as $10b$ and $4b$, respectively.

We now have $P(A \cup B)=10b+4b+4b+3b=21b$

$P(C|A \cup B)=\frac{P(C \cap (A \cup B))}{P(A \cup B)}=\frac{4b+4b}{21b}=\frac{8}{21}$