Originally Posted by

**Plato** First, note that you mean $\displaystyle B\backslash A$ which is set difference.

What you typed looks like conditional probability.

The first one is just an exercise in sets.

Looking at the diagram, it is easy to see three disjoint sets. Their union is the whole.

$\displaystyle A \cup \left[ {B\backslash A} \right] \cup \left[ {C\backslash \left( {A \cup B} \right)} \right] = A \cup B \cup C$.

So by the additive property $\displaystyle P(A) + P(B) + P(C) = P(A) + P\left[ {B\backslash A} \right] + P\left[ {C\backslash \left( {A \cup B} \right)} \right]$

For #2, use the monotone property of probability.

$\displaystyle T \subseteq S \Rightarrow \quad P(T) \leqslant P(S)$ (here again you typed < when it should be $\displaystyle \le$).

$\displaystyle \left[ {B\backslash A} \right] \subseteq B \Rightarrow \quad P\left[ {B\backslash A} \right] \leqslant P(B)$.

A similar statement for the third term. Make the substitutions it follows.