Question

The probabilites 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 $P(A\cup B\cup C)\ge 0.85$, then

• A. $0.2\le P(B\cap C)\le 0.35$
• B. $0.5\le P(B\cap C)\le 0.85$
• C. $0.1\le P(B\cap C)\le 0.35$
• D. None of these

Answer 1

Answer: (A)

$0.2\le P(B\cap C)\le 0.35$

Given $P\left(A\right)=0.6,P\left(B\right)=0.4$ 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 $P(A\cup B\cup C)\ge 0.85$

$P(A\cap B)=P\left(A\right)+P\left(B\right)-P(A\cup B)=0.6+0.4-0.8=0.2$

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

$P(A\cup B\cup C)=0.6+0.4+0.5-0.3+0.2-0.2-P(B\cap C)$

$\Rightarrow P(B\cap C)=1.2-P(A\cup B\cup C)$ ...(1)

$\because 0.85\le P(A\cup B\cup C)\le 1$

$\Rightarrow P(B\cap C)\le 1.2-0.85$ [From (1)]

and $P(B\cap C)\ge 1.2-1\Rightarrow 0.2\le P(B\cap C)\le \mathrm{0.35.}$