Could you please explain to me how to solve this problem?
In a city, $45\%$ of the population read magazine $A$, $55\%$ read magazine $B$ and $40\%$ read magazine $C$. $30\%$ read $A$ and $B$, $15\%$ read $A$ and $C$, $15\%$ read $B$ and $C$ and $10\%$ read all three magazines. How many percent read no magazine at all?
[The answer is apparently $20\%$.]
On
This answer is in the same line as the answer of @Calculus and works with numbers in stead of probabilities.
Let it be that the city has $100$ inhabitants and denote $n\left(A\right)=45$, $n\left(B\right)=55$ et cetera.
As you will understand $n(A)$ stands for the number of persons that read magazine $A$
With inclusion/exclusion we find:
$$n\left(A\cup B\cup C\right)=$$$$n\left(A\right)+n\left(B\right)+n\left(C\right)-n\left(A\cap B\right)-n\left(A\cap C\right)-n\left(B\cap C\right)+n\left(A\cap B\cap C\right)$$
Next to that we have:
$$n\left(A^{c}\cap B^{c}\cap C^{c}\right)=100-n\left(A\cup B\cup C\right)$$
Hint: You can apply the inclusion-exclusion-principle.
$P(A \cup B \cup C) =P(A)+P(B)+P(C)-P(A\cap B)-P(A \cap C)-P(C\cap B)+P(A\cap B\cap C )$
It is asked for
$1-P(A\cup B\cup C )$. But I have a result, which is different from 20%.