Question

$A,B$ and $C$ are three events such that $P\left(A\right)=0.3,P\left(B\right)=0.4,P\left(C\right)=0.8$,$P(A\cap B)=0.12,P(A\cap C)=0.28$, $P(A\cap B\cap C)=0.09$ and $P(A\cup B\cup C)\ge 0.75$, then the limits of $P(B\cap C)$ are

• A. $[0.32,0.52]$
• B. $[0.23,0.48]$
• C. $[0.28,0.53]$
• D. $[0.21,0.19]$

Answer 1

The correct option is C $[0.28,0.53]$

Since $1\ge A\cup B\cup C\ge 0.75$, we have from the diagram that: $P(B\cup C)=P(A\cup B\cup C)-0.08$
Hence, $0.92\ge P(B\cup C)\ge 0.67$
Now, $P(B\cap C)=P\left(B\right)+P\left(C\right)-P(B\cup C)=1.2-P(B\cup C)$
Hence, $1.2-0.67\ge P(B\cap C)\ge 1.2-0.92=>0.53\ge P(B\cap C)\ge 0.28$