com; All Rights Reserved. \) The equality holds only for mutually exclussive events. \( \Rightarrow S = A \cup B \cup C\)Therefore, using the axioms of probability, we get\(P\left( A \right) \geqslant 0,P\left( B \right) \geqslant 0,P\left( C \right) \geqslant 0\,{\text{and}}\,P\left( {A \cup B \cup C} \right) = P\left( A \right) + P\left( B \right)9 + P\left( C \right)\)\( = P\left( S \right) = 1\)\( \Rightarrow \frac{4}{7} + \frac{1}{7} + \frac{2}{7} = \frac{{4 + 1 + 2}}{7} = \frac{7}{7} = 1\)So, the given probabilities are permissible.
A
c
=
P
(
A
)
=
1
P
(
A
)
{\displaystyle P\left(A^{c}\right)=P(\Omega -A)=1-P(A)}
Given
A
{\displaystyle A}
and
A
c
{\displaystyle A^{c}}
are mutually exclusive and that
A
c
=
other
{\displaystyle A\cup A^{c}=\Omega }
:
P
(
A
A
c
)
=
P
(
A
)
+
P
(
A
c
)
{\displaystyle P(A\cup A^{c})=P(A)+P(A^{c})}
. .