Question

The probablities of three events $A,B\text{and}C$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 $P(A\cup B\cup C)\ge 0.85.$ Then the range of $P(B\cap C)$is

• A. $[0.15,0.35]$
• B. $[0.15,0.45]$
• C. $[0.20,0.35]$
• D. $[0.20,0.45]$

Answer 1

The correct option is C $[0.20,0.35]$

$P(A\cap B)=P\left(A\right)+P\left(B\right)-P(A\cup B)=0.6+0.4-0.8=0.2$
Also, $P(A\cup B\cup C)=P\left(A\right)+P\left(B\right)+P\left(C\right)-P(A\cap B)-P(B\cap C)-P(A\cap C)+P(A\cap B\cap C)$
$\Rightarrow P(B\cap C)=1.2-P(A\cup B\cup C)\cdots \left(i\right)$
Now, $0.85\le P(A\cup B\cup C)\le 1$
From equation $\left(i\right),$
$0.2\le P(B\cap C)\le 0.35$