Question

P(A) = P(B) + P(A ∩ ~B).
proof.?

Answer 1

$$
\text{ Let }B \subseteq A\\
\text{ then }A = \{x \in B\cup A\backslash B\}\\
\implies A= \{x\in B \cup x\in A\cup\overline{B}\}\\
\implies P(A) = P(B)+P(A\cup\overline{B})
$$

Answer 2

*A\B should be A AND Not(B)
$$
\text{ Let }B \subseteq A\\
\text{ then }A = \{x \in B\cup A\backslash B\}\\
\implies A= \{x\in B \cup x\in A\cap\overline{B}\}\\
\implies P(A) = P(B)+P(A\cap\overline{B})
$$

Answer 3

Note that this probability is possible because B and \(A\cap \overline{B}\) are mutually exclusive.