1. Given data

We have: $P(A)=0.6, \quad P(B)=0.4, \quad P(C)=0.5$ $P(A\cup B)=0.8, \quad P(A\cap C)=0.3, \quad P(A\cap B\cap C)=0.2$ $P(B\cap C)=\beta, \quad P(A\cup B\cup C)=\alpha$ with $0.85\le \alpha \le 0.95$

We need the interval in which $\beta$ lies.

2. Find $P(A\cap B)$

Using $P(A\cup B)=P(A)+P(B)-P(A\cap B)$ we get $0.8=0.6+0.4-P(A\cap B)$ $0.8=1.0-P(A\cap B)$ $P(A\cap B)=0.2$

So, $P(A\cap B)=0.2$

3. Use inclusion-exclusion for three events

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

Substitute the known values: $\alpha=0.6+0.4+0.5-0.2-\beta-0.3+0.2$

Now simplify: $\alpha=1.5-0.5-\beta+0.2$ $\alpha=1.2-\beta$

Hence, $\beta=1.2-\alpha$

4. Use the given range of $\alpha$

Since $0.85\le \alpha \le 0.95$ we have $\beta=1.2-\alpha$

So the maximum value of $\beta$ occurs when $\alpha$ is minimum: $\beta_{\max}=1.2-0.85=0.35$

And the minimum value of $\beta$ occurs when $\alpha$ is maximum: $\beta_{\min}=1.2-0.95=0.25$

Therefore, $0.25\le \beta \le 0.35$

So $\beta$ lies in the interval $[0.25,0.35]$

5. Match with options

The correct option is:

C: $[0.25, 0.35]$