Solve this following
Question:

The probabilities of three events A, B and C are given by $P(A)=0.6, P(B)=0.4$ and $P(C)=0.5$. If $\mathrm{P}(\mathrm{A} \cup \mathrm{B})=0.8, \mathrm{P}(\mathrm{A} \cap \mathrm{C})=0.3, \mathrm{P}(\mathrm{A} \cap \mathrm{B} \cap$ 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:

1. $[0.36,0.40]$

2. $[0.35,0.36]$

3. $[0.25,0.35]$

4. $[0.20,0.25]$

Correct Option:

Solution:

$\mathrm{P}(\mathrm{A} \cup \mathrm{B})=\mathrm{P}(\mathrm{A})+\mathrm{P}(\mathrm{B})-\mathrm{P}(\mathrm{A} \cap \mathrm{B})$

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

$P(A \cap B)=0.2$

$\mathrm{P}(\mathrm{A} \cup \mathrm{B} \cup \mathrm{C})=\Sigma \mathrm{P}(\mathrm{A})-\Sigma \mathrm{P}(\mathrm{A} \cap \mathrm{B})+\mathrm{P}(\mathrm{A} \cap \mathrm{B} \cap \mathrm{C})$

$\alpha=1.5-(0.2+0.3+\beta)+0.2$

$\alpha=1.2-\beta \in[0.85,0.95]$

(where $\alpha \in[0.85,0.95]$ )

$\beta \in[0.25,0.35]$