Answer
The percentage of programs that did comply with the group’s standards is 41%.
Work Step by Step
Let A denote the programs that were offensive language and let event B denote the programs that were violence. Then,
\[\begin{align}
& P(A)=0.42 \\
& P(B)=0.27 \\
& P(A\cap B)=0.10 \\
\end{align}\]
We have to find the percentage of programs which did comply with the group’s standard; that means the percentage of programs that were not offensive and not violent (we need to find\[P({{A}^{C}}\cap {{B}^{C}})\]
By using DeMorgan’s law \[P({{A}^{C}}\cap {{B}^{C}})\]
\[P({{A}^{C}}\cap {{B}^{C}})=P({{A}^{C}})+P({{B}^{C}})-P({{A}^{C}}\cup {{B}^{C}})\]
Since,
\[\begin{align}
& P({{A}^{C}})=1-P(A) \\
& P({{B}^{C}})=1-P(B) \\
& P({{A}^{C}}\cup {{B}^{C}})=1-P(A\cap B) \\
\end{align}\]
Therefore,
\[\begin{align}
& P({{A}^{C}}\cap {{B}^{C}})=P({{A}^{C}})+P({{B}^{C}})-P{{(A\cap B)}^{C}} \\
& P({{A}^{C}}\cap {{B}^{C}})=(1-P(A))+(1-P(B))-(1-P(A\cap B)) \\
& P({{A}^{C}}\cap {{B}^{C}})=(1-0.42)+(1-0.27)-(1-0.10) \\
& P({{A}^{C}}\cap {{B}^{C}})=1-0.42-0.27+0.10 \\
& P({{A}^{C}}\cap {{B}^{C}})=0.41 \\
\end{align}\]
Hence, the percentage of programs which did comply with the group’s standards is 41%.
