Question

The probabilities of three events A, B & C are P(A) = 0.6, P(B) = 0.4 and P(C) = 0.5. If P($A\cup B$) = 0.8, P($A\cap C$) = 0.3, P($A\cap B\cap C$) = 0.2 & P($A\cup B\cup C$)$\ge$0.85. Find the range of P($B\cap C$)

• A. $[0.2,0.35]$
• B. $[0.25,0.35]$
• C. $[0.2,0.3]$
• D. $[0.2,0.4]$

Answer 1

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

Given,
$P\left(A\right)=0.6,P\left(B\right)=0.4,P\left(C\right)=0.5,$

$P(A\cup B)=0.8,P(A\cap C)=0.3,P(A\cap B\cap C)=0.2$

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

now,$P(A\cap B)=P\left(A\right)+P\left(B\right)-P(A\cup B)=0.2$

$P(A\cup B\cup C)=0.6+0.4+0.5-0.2-0.3-p_1,+0.2$

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

$\Rightarrow 0.2\le p_1\le 0.35$