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

Answer 1

The correct option is B

$[0.25,0.35]$

Explanation for the correct option:

We know that,

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

So,

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

$\alpha =1.4-P(A\cap B)-\beta$

$\Rightarrow \alpha +\beta =1.4-P(A\cap B)....\left(i\right)$

again,

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

$0.8=0.6+0.4-P(A\cap B)$

$\Rightarrow P(A\cap B)=0.2....\left(ii\right)$

Put the value of $P(A\cap B)$ in equation $\left(i\right)$, we get,

$\alpha +\beta =1.2$

$\alpha =1.2-\beta$

So, clearly $0.85\le \alpha \le 0.95$

$\Rightarrow 0.85\le 1.2-\beta \le 0.95$

$\beta \in [0.25,0.35]$

Hence, the correct answer is option (B), $[0.25,0.35]$