The probabilities of three events $\mathrm{A}, \mathrm{B}$ and $\mathrm{C}$ are given by $P(A)=0.6, P(B)=0.4$ and $P(C)=0.5$. If $P(A \cup B)=0.8$, $\mathrm{P}(\mathrm{A} \cap \mathrm{C})=0.3, \mathrm{P}(\mathrm{A} \cap \mathrm{B} \cap \mathrm{C})=0.2, \mathrm{P}(\mathrm{B} \cap \mathrm{C})=\beta$ and $\mathrm{P}(\mathrm{A} \cup \mathrm{B} \cup \mathrm{C})=\alpha$, where $0.85 \leq \alpha \leq 0.95$, then $\beta$ lies in the interval:
Correct Option: , 2
$\because P(A \cap B)=P(A)+P(B)-P(A \cup B)$
$=1-0.8=0.2$
Now, $\because P(A \cup B \cup C)=P(A)+P(B)+P(C)-P(A \cap B)$
$-P(B \cap C)-P(C \cap A)+P(A \cap B \cap C)$
$\Rightarrow \alpha=0.6+0.4+0.5-0.2-\beta-0.3+0.2$
$\Rightarrow \beta=1.2-\alpha$
$\because \alpha \in[0.85,0.95]$ then $\beta \in[0.25,0.35]$
Click here to get exam-ready with eSaral
For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.