I= 10 percent
II= 30% percent
II=5 percent
I&II=8 percent
I&III=2 percent
II&III=4 percent
I&II&III=1 percent
How many people read at least two newspapers? Well it would seem to me that it would 15,000 or 8+2+4+1. It is however 12,000 I have no idea how this comes to be....Pictures, equations would be great
\begin{align}&P(A \cap B \cap C)+ P(A \cap B \cap C^c) + P(A \cap B^c \cap C) + P(A^c \cap B \cap C)\\&=[P(A \cap B \cap C)+ P(A \cap B \cap C^c)] + P(A \cap B^c \cap C) + P(A^c \cap B \cap C) \\&=P(A \cap B) + [P(A \cap C) -P(A \cap B \cap C)]+[P(B \cap C) -P(A \cap B \cap C)] \\&=P(A \cap B)+ P(A \cap C)+P(B \cap C)-2P(A \cap B \cap C) \\&=0.08+0.02+0.04-2(0.01) \\&=0.12\end{align}
When you add the three numbers $8+2+4$ for I&II, I&III, and II&III, you have actually counted the people who read all three papers three times. You should really only count them once, so you need to subtract the two overcounts, not count them again. So the correct count is $8+2+4-2\cdot1=12$.
Another way to get the correct answer is to change I&II=8, I&III=2, and II&III=4 into I&II-only=7, I&III-only=1, and II&III-only=3 (i.e., subtract the I&II&III=1 from each count, to get counts of people who read exactly two papers), then add all four numbers, $7+1+3+1=12$.