Answer
The smallest possible value for \[P{{(A\cup B\cup C)}^{C}}\] would be 0.4.
Work Step by Step
Let events A, B and C be defined on a sample space, S. Then,
\[\begin{align}
& P(A)=0.2 \\
& P(B)=0.1 \\
& P(C)=0.3 \\
\end{align}\]
We know that,
\[\begin{align}
& P(A\cup B\cup C)=P(A)+P(B)+P(C)-P(A\cap B)-P(B\cap C)-P(A\cap C)+P(A\cap B\cap C) \\
& P(A\cup B\cup C)=0.2+0.1+0.3-[P(A\cap B)+P(B\cap C)+P(A\cap C)-P(A\cap B\cap C)] \\
\end{align}\]
Suppose, we put \[P(A\cap B)+P(B\cap C)+P(A\cap C)-P(A\cap B\cap C)=X\]
Then,
\[P(A\cup B\cup C)=0.6-X\]
Therefore,
\[\begin{align}
& P{{(A\cup B\cup C)}^{C}}=1-P(A\cup B\cup C) \\
& P{{(A\cup B\cup C)}^{C}}=1-(0.6-X) \\
& P{{(A\cup B\cup C)}^{C}}=0.4+X \\
\end{align}\]
The above value of \[P{{(A\cup B\cup C)}^{C}}\] must be greater than or equal to 0.4
Therefore the smallest possible value for \[P{{(A\cup B\cup C)}^{C}}\]will be 0.4.
