91Ó°ÊÓ

Here given that A and B events are disjoint.

Where disjoint means that events A and B are mutually exclusive events.

It can be written as,

\(\begin{aligned}{c}A \cap B &= \phi \\P\left( {A \cap B} \right) &= 0\\P\left( \phi \right) &= 0\end{aligned}\)

The condition is as follows,

\(\begin{aligned}{c}P\left( {{A^c} \cap {B^c}} \right) &= P{\left( {A \cup B} \right)^c}\\ &= 1 - P\left( {A \cup B} \right)\\ &= 1 - \left[ {P\left( A \right) + P\left( B \right) - P\left( {A \cap B} \right)} \right]\\ &= 1 - P\left( A \right) - P\left( B \right) + 0\end{aligned}\)

Here it need to make\({A^c}\)and\({B^c}\)disjoint

i.e.,

\(\begin{aligned}{c}P\left( {{A^c} \cap {B^c}} \right) &= 0\\1 - P\left( A \right) - P\left( B \right) &= 0\\P\left( A \right) + P\left( B \right) &= 1\end{aligned}\)

Thus the condition of\({A^c}\)and\({B^c}\)disjoint is\(P\left( A \right) + P\left( B \right) = 1\)