Question

$A,B,C$ are three events for which $P\left(A\right)=0.6,P\left(B\right)=0.4,P\left(C\right)=0.5,P(A\cup B)=0.8,P(A\cap C)=0.3$ and $P(A\cap B\cap C)=0.2$. If $P(A\cup B\cup C)\ge 0.85$ then the interval of values of $P(B\cap C)$ is

• A. $[0.2,0.35]$
• B. $[0.55,0.7]$
• C. $[0.2,0.55]$
• D. none of these

Answer 1

The correct option is A $[0.2,0.35]$

We know that $P(A\cup B)=P\left(A\right)+P\left(B\right)-P(A\cap B)$

Hence, $P(A\cap B)=0.6+0.4-0.8=0.2$

We also know that $P(A\cup B\cup C)=P\left(A\right)+P\left(B\right)+P\left(C\right)-P(A\cap B)-P(A\cap C)-P(B\cap C)+P(A\cap B\cap C)$

Now, $1\ge P(A\cup B\cup C)\ge 0.85$

Thus, $1\ge 0.6+0.4+0.5-0.2-P(B\cap C)-0.3+0.2\ge 0.85$

$\Rightarrow$ $0.35\ge P(B\cap C)\ge 0.2$

Hence, (a) is correct.