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Two events A and B, with the conditions \(\Pr \left( B \right) > 0\) . It is required to prove the relation:

\(P\left( {{A^C}|B} \right) = 1 - P\left( {A|B} \right)\)

The Conditional probability of any event U was given any other event V (occurred in advance) is given by:

\(P\left( {U|V} \right) = \frac{{P\left( {U \cap V} \right)}}{{P\left( V \right)}};P\left( V \right) > 0\)

Considering the left-hand side of the relationship as:

\(\begin{array}{c}LHS = P\left( {{A^C}|B} \right)\\ = \frac{{P\left( {{A^C} \cap B} \right)}}{{P\left( B \right)}}\;\;\;...\left( 1 \right)\end{array}\)

Since,

\(P\left( {{A^C} \cap B} \right) = P\left( B \right) - P\left( {A \cap B} \right)\)

Therefore, equation (1) can be written as:

\(\begin{aligned}{}LHS &= \frac{{P\left( B \right) - P\left( {A \cap B} \right)}}{{P\left( B \right)}}\\ &= \frac{{P\left( B \right)}}{{P\left( B \right)}} - \frac{{P\left( {A \cap B} \right)}}{{P\left( B \right)}}\\ &= 1 - P\left( {A|B} \right)\\ &= RHS\end{aligned}\)

Hence, proved that,

\(P\left( {{A^C}|B} \right) = 1 - P\left( {A|B} \right)\)