Question

The probabilities of three events $A$, $B$ and $C$ are given by $P\left(A\right)=0.6$, $P\left(B\right)=0.4$ and $P\left(C\right)=0.5$. If $P(A\cup B)=0.8$, $P(A\cap C)=0.3$, $P(A\cap B\cap C)=0.2$, $P(B\cap C)=\beta$ and $P(A\cup B\cup C)=\alpha$, where $0.85\le \alpha \le 0.95$, then $\beta$ lies in the interval

• A. $[0.20,0.25]$
• B. $[0.25,0.35]$
• C. $[0.35,0.36]$
• D. $[0.36,0.40]$

Answer 1

The correct option is B $[0.25,0.35]$

$P(A\cup B\cup C)=P\left(A\right)+P\left(B\right)+P\left(C\right)-P(A\cap B)-P(B\cap C)-P(C\cap A)+P(A\cap B\cap C)$
$\alpha =0.6+0.4+0.5-P(A\cap B)-\beta -0.3+0.2$
$\alpha =1.4-P(A\cap B)-\beta \Rightarrow \alpha +\beta =1.4-P(A\cap B)\cdots \left(i\right)$
again
$P(A\cup B)=P\left(A\right)+P\left(B\right)-P(A\cap B)$
$0.8=0.6+0.4-P(A\cap B)$
$P(A\cap B)=0.2\cdots \left(ii\right)$
Put the value $P(A\cap B)$ in equation $\left(i\right)$
$\alpha +\beta =1.2$
$\alpha =1.2-\beta$
$0.85\le \alpha \le 0.95\Rightarrow 0.85\le 1.2-\beta \le 0.95$
$\beta \in [0.25,0.35]$