Question

If $P\left(A\right)=0.7,P\left(B\right)=0.55,P\left(C\right)=0.5,P(A\cap B)=x,P(A\cap C)=0.45,P(B\cap C)=0.3$ and $P(A\cap B\cap C)=0.2$, then

• A. $0.2\le x\le 0.45$
• B. $0.2\le x\le 0.5$
• C. $0.2\le x\le 0.65$
• D. $0.2\le x\le 1$

Answer 1

The correct option is B $0.2\le x\le 0.5$

$P(A\cup B\cup C)=P\left(A\right)+P\left(B\right)+P\left(C\right)-P(A\cap B)-P(A\cap C)-P(B\cap C)+P(A\cap B\cap C)$

$=>P(A\cap B)=P\left(A\right)+P\left(B\right)+P\left(C\right)-P(A\cup B\cup C)-P(A\cap C)-P(B\cap C)+P(A\cap B\cap C)$

$=>P(A\cap B)=0.7+0.55+0.5-P(A\cup B\cup C)-0.45-0.3+0.2=1.2-P(A\cup B\cup C)$

Maximum $P(A\cup B\cup C)$ is 1, hence, minimum $P(A\cap B)$ = $1.2-1=0.2$

Maximum $P(A\cap B)$ could have been the minimum of $P\left(A\right)$ and $P\left(B\right)$ i.e. $0.55$,

however, that would mean that $1.2-P(A\cup B\cup C)=0.55=>P(A\cup B\cup C)=0.65$, i.e. less than A: This is not possible

hence, Maximum $P(A\cap B)$ could be $1.2-0.7=0.5$.

$\therefore 0.2\le P(A\cap B)\le 0.5\Rightarrow 0.2\le x\le 0.5$