Suppose three events $A$, $B$ and $C$ are such that $P(A) = P(B) = P(C) = \frac{1}{3}$. Further, we have $P(A\cap B) = P(A\cap C) = P(B\cap C) = \frac{1}{10}$. Is it possible to compute the probability that none of $A$, $B$ or $C$ occurs?
I think the probability would be given by $1 -P(A\cup B \cup C)$ , i.e. one take away the probability that at least one event occurs. And for $P(A\cup B \cup C)$ I could use the inclusion-exclusion formula. But to use this formula, I need to know $P(A \cap B \cap C)$, but I don't understand how to compute this?
\begin{align}a+b+c+x+y+z+p+q&=1\\&=P(A)+P(B)+P(C)\\&=a+b+c+2(x+y+z)+3p\\ q&=(x+y+z+3p)-p\\&=P(A\cap B) +P(A\cap C) +P(B\cap C)-p\\&=\frac3{10}-p\\P\big((A\cup B\cup C)^c\big)&=\frac3{10}-P(A\cap B\cap C).\end{align}
Unfortunately, I don't think the given information is sufficient to derive $P(A\cap B\cap C)$ and thus $P\big((A\cup B\cup C)^c\big)$.