Question

The probabilities of three events $A,B$ and $C$ are $P\left(A\right)=0.6,P\left(B\right)=0.4$ and $P\left(C\right)=0.5$. If $P(A\cup B)=0.8,P(A\cap C)=0.3,P(A\cap B\cap C)=0.2$ and $P(A\cup B\cup C)\ge 0.85$, $0.04k\le P(B\cap C)\le 0.07k$. Find the value of $k$.

Answer 1

$P(A\cap B)=P\left(A\right)+P\left(B\right)-P(A\cup B)=0.6+0.4-0.8=0.2$
$P(A\cup B\cup C)=P\left(A\right)+P\left(B\right)+P\left(C\right)-P(A\cap B)+P(A\cap B\cap C)-P(A\cap B)-P(B\cap C)\Rightarrow P(B\cap C)=1.2-P(A\cup B\cup C)\because 0.85\le P(A\cup B\cup C)\le 1\therefore \left(1\right)\Rightarrow P(B\cap C)\le 1.2-1-0.85$
and $P(B\cap C)\ge 1.2-1\Rightarrow 0.2\le P(B\cap C)\le \mathrm{0.35.}$