Calculating Probabilities from a Venn Diagram A city that has 150,000 inhabitants sells three different newspapers (A, B, and C). It's known that 10% of the population read A; 30% read B; 5% read C; 8% read A and B; 2% read A and C; 4% read B and C; 1% read A, B, and C. 
It's also known that A and B are "daily" newspapers, while C is a "nightly" newspaper. 
If 1 person is randomly selected, what's the probability of him belonging to the group that:
a) read only one of the three newspapers. 
b) read a maximum of 2 newspapers. 
c) don't read any of the 3 newspapers. 
d) read at least one daily newspaper and the nightly newspaper.
e) read only one daily newspaper and the nightly newspaper.
First, I constructed a Venn diagram to organize the information given (D stands for daily and N stands for nightly): 

For a), I added up the sections of the diagram which don't include any intersection: 
28500+1500+0 = 30,000; then I divided that result by omega to get my first answer: 
P(only read one of three newspapers) = 30,000/150,000 = 0.2
For b), I added up everything inside the diagram except the intersection between all three sets: 
P(read maximum 2) = 28500+1500+10500+1500+4500+0 = 46.500/150.000 = 0.31
For c), I simply divided the number of people outside the three sets by omega: 
P(read none) = 102,000/150,000 = 0.68
How can I solve d) and e)?


If a person is randomly selected, what is the probability that he or she belongs to a group that reads at least one daily newspaper and the nightly newspaper?

The two daily newspapers are $A$ and $B$; the nightly newspaper is $C$.    
Method 1:  We want $p((A \cap C) \cup (B \cap C))$. Since
$$p((A \cap C) \cup (B \cap C)) = p(A \cap C) + p(B \cap C) - p(A \cap B \cap C)$$
we obtain 
$$p((A \cap C) \cup (B \cap C)) = 0.02 + 0.04 - 0.01 = 0.05$$
Method 2: If we instead work with the Venn diagram, we can find the number of people who read at least one daily newspaper and the nightly newspaper by finding 
$$|A \cap B^C \cap C| + |A^C \cap B \cap C| + |A \cap B \cap C|$$
Dividing that number by the city's population gives the desired probability.
$$p((A \cap B^C \cap C) \cup (A^C \cap B \cap C) \cup (A \cap B \cap C)) 
= \frac{1500 + 4500 + 1500}{150000} = \frac{7500}{150000} = 0.05$$ 

If a person is randomly selected, what is the probability that he or she belongs to a group that reads only one daily newspaper and a nightly newspaper.

Method 1: We want $p(((A \cap C) \cup (B \cap C)) \setminus (A \cap B \cap C))$.  Since 
$$p((A \cap C) \cup (B \cap C)) = p(A \cap C) + p(B \cap C) - p(A \cap B \cap C)$$
we obtain
\begin{align*}
p(((A \cap C) \cup (B \cap C)) \setminus (A \cap B \cap C)) & = p(A \cap C) + p(B \cap C) - 2p(A \cap B \cap C)\\ 
& = 0.02 + 0.04 - 2 \cdot 0.01\\
& = 0.04
\end{align*}
Method 2: If we instead work with the Venn diagram, we can find the number of people who read only one daily newspaper and the nightly newspaper by finding
$$|A \cap B^C \cap C| + |A^C \cap B \cap C|$$
Dividing that number by the city's population gives the desired probability.
$$p(A \cap B^C \cap C) \cup (A^C \cap B \cap C)) = \frac{1500 + 4500}{15000} = 0.04$$
