there are three newspapers A, B and C in a certain city. below are the percentages how the people read the newspapers:
A : 50
B: 25
C: 35
A & B: 10
A & C: 8
B & C: 5
A & B & C: 3
a) probability to read at least on of the newspapers?
$$P(A \cup B\ \cup C)=P(A)+P(B)+P(C)-P(A \cap B)-P(A \cap C)-P(B \cap C)+P(A \cap B \cap C)$$
$$P(A \cup B\ \cup C)=.5+.25+.35-.1-.08-.05+.03=.9$$
b)probability that a person reads B or C but not A.
$$P(B \cap C \cap A^c ) = ?$$
I draw a Venn diagram and I got $0.02$ as result, but I do not khow to show this using set properties.
c)probability that the person reads B OR C given he reads A.
$$P(B \cup C | A )=\frac{P(B \cup C \cap A)}{P(A)}$$
again, using Venn diagran I got $\frac{15}{50}$ but I do not know how to get it using set properties.
I would appreciate help with this.
Probability for two sets is given by following rule
$$ P(A \cup B) = P(A) + P(B) - P(B \cap C)$$
Now since, we know that sets $$ (B \cap C) \cap A^C $$ and $$ (B \cap C) \cap A $$ are mutually disjoint and further their union is equal to $$ (B \cap C) $$ i.e. $$ [(B \cap C) \cap A^C] \cup [(B \cap C) \cap A^C] = (B \cap C) $$ we can get
$$ P(B \cap C) = P([(B \cap C) \cap A^C]) + P([(B \cap C) \cap A])$$
which can be solved to $$ P(B \cap C \cap A^C) = P(B \cup C) - P(B \cap C \cap A) $$ and thus solving your first problem.
Now your second problem can be solved by distributive law, which states that
$$ (B \cup C) \cap A = (B \cap A) \cup (C \cap A)$$ but since $$ (B \cap A) \cap (C \cap A) = A \cap B \cap C $$, we can get $$ P[(B \cap A) \cup (C \cap A)] = P((B \cap A) + P(C \cap A) - P[(B \cap A) \cap (C \cap A)]$$ which can solve your second problem.