If $A,B,C$ are events,$P(A)=0.85,P(B)=0.9.P(C)=0.95$.Then find the ranges of $P(A\cap B\cap C),P(A\mid(B\cap C)),P(B\mid(C\cap A)),P(B\mid(A\cap C))$ If $A,B,C$ are events,$P(A)=0.85,P(B)=0.9.P(C)=0.95$.Then find the ranges of $P(A\cap B\cap C),P(A\mid(B\cap C)),P(B\mid(C\cap A)),P(B\mid(A\cap C))$.

I know the formula $P(A\cup B\cup C)=P(A)+P(B)+P(C)-P(A\cap B)-P(B\cap C)-P(C\cap A)+P(A\cap B\cap C)$
EDIT 1:
In my book,$P(A\cap B\cap C)\in (0.7,0.85)$.I understood how they found the maximum value of $P(A\cap B\cap C)$ but the book finds the minimum value of $P(A\cap B\cap C)\geq 0.85+0.9+0.95-2$.Why book puts $''-2''$ is not clear to me.Please elaborate this concept.Thanks
 A: Hint: Consider $P(A\cap B)$.  We know that $P(A\cup B)=P(A)+P(B)-P(A\cap B)$.  
Since probabilities are always between $0$ and $1$, the smallest value for $P(A\cap B)$ occurs when $P(A\cup B)=1$.  In this case, $P(A\cap B)=0.85+0.9-1=0.75$.
Since $P(A\cap B)$ can be, at most, the probability of the least likely event, $P(A\cap B)$ is at most $0.85$.
There are many other cases that you must consider, in general, but this gives you a place to start.
A: Hint
Note that
$$0,85=P(A)\ge P(A\cap B)=P(A)+P(B)-P(A\cup B)\ge P(A)+P(B)-1=0,75.$$ Proceeding in a similar way you can bound $P(A\cap C)$ and $P(B\cap C).$ 
In particular, you have the upper bound $P(A\cap B\cap C)\le P(A)=0,85.$
The lower bound can be obtained as follows:
$$\begin{align}P(A\cap B\cap C)&\\ &=P(A\cup B\cup C)+P(A\cap B)+P(A\cap C)+P(B\cap C)-P(A)-P(B)-P(C)&\\ &= P(A\cup B\cup C)+P(A)+P(B)-P(A\cup B)+P(A)+P(C)-P(A\cup C)\\& \qquad+P(B)+P(C)-P(B\cup C)-P(A)-P(B)-P(C)&\\ &=P(A\cup B\cup C)+P(A)+P(B)+P(C)-P(A\cup B)-P(A\cup C)-P(B\cup C)\\&  \underbrace{\ge}_{P(A\cup B\cup C)\ge P(B\cup C)} P(A)+P(B)+P(C)-P(A\cup B)-P(A\cup C)\\&  \underbrace{\ge}_{P(A\cup B),P(A\cup C)\le 1} P(A)+P(B)+P(C)-1-1\\&  = P(A)+P(B)+P(C)-2.\end{align}$$
