Probability Question: What percentage of the bag of balls is marked? In a bag of reds and black balls, $30\%$ were red, and $90\%$ of the black balls and $80\%$ of the red balls are marked balls. What percentage of the bag of balls is marked?
I thought I would have to use: Let 
$A = \text{marked red balls},$
$B = \text{marked black balls}.$ 
$P(A \cup B) = P(A) + P(B) - P(A \cap B)$
But the answer is simply $0.8\times0.3 + 0.7\times0.9$. How come we don't have to take $P(A \cap B)$ into consideration?
 A: Consider the following:
$30\%$ of the balls are red and $80\%$ of the red balls are marked, so we have $P(A) = 30\% \times 80\%$.
Now $70\%$ of the balls are black and $90\%$ of the black balls are marked, so we have $P(B) = 70\% \times 90\%$.
Now we can evaluate the formula $P(A \cup B) = P(A) + P(B) - P(A \cap B) = 30\% \times 80\% + 70\% \times 90\%$ because $P(A \cap B) = 0$. This is the percentage of all the balls that are marked.
The reason we don't consider the intersection mentioned in your question is because there is no intersection. Indeed, your reasoning with the equation $P(A \cup B) = P(A) + P(B) - P(A \cap B)$ is correct, but $P(A \cap B) = 0$ because a ball cannot be both black and red at the same time.
A: Let in bag be $x$ balls.
Number of red balls: $30\%x$
Number of black balls: $(100-30)\%x=70\%x$
Number of marked red balls: $80\%\cdot30\%x=24\%x$
Number of marked black balls: $90\%\cdot70\%x=63\%x$
So, answer will be: $24\%x+63\%x=87\%x$
A: Hint:
$P\left(\text{marked}\right)=P\left(\text{marked and black}\right)+P\left(\text{marked and red}\right)$
$P\left(\text{marked}\right)=P\left(\text{marked}\mid\text{black}\right)P\left(\text{black}\right)+P\left(\text{marked}\mid\text{red}\right)P\left(\text{red}\right)$

By definition: $$P\left(\text{marked}\mid\text{black}\right)P\left(\text{black}\right)=P\left(\text{marked and black}\right)$$
and likewise: $$P\left(\text{marked}\mid\text{red}\right)P\left(\text{red}\right)=P\left(\text{marked and red}\right)$$
