Probability

If $A$ and $B$ are mutually exclusive events and $P\left(A\right) = 0.28$, $P \left(B\right)=0.44$, then find

1. $P \left(\overline{A}\right)$
2. $P \left(A \cup B\right)$
3. $P \left(A \cap \overline{B}\right)$
4. $P \left(\overline{A} \cap \overline{B}\right)$

$\because \;$ $A$ and $B$ are mutually exclusive events, $A \cap B = 0$

1. $P \left(\overline{A}\right) = 1 - P \left(A\right) = 1 - 0.28 = 0.72$



2. $P \left(A \cup B\right) = P \left(A\right) + P \left(B\right) - P \left(A \cap B\right) = P \left(A\right) + P \left(B\right)$
   
   i.e. $\;$ $P \left(A \cup B\right) = = 0.28 + 0.44 = 0.72 $



3. $P \left(A \cap \overline{B}\right) = P \left(A\right) - P \left(A \cap B\right) = P \left(A\right) = 0.28$



4. $P \left(\overline{A} \cap \overline{B}\right) = P \left(\overline{A \cup B}\right) = 1 - P \left(A \cup B\right) = 1 - 0.72 = 0.28$