Let \(A, B, C\) and \(D\) represent the events "make a mistake on obstacle \(A\) " and so forth.
\(P(\) pass \()=P(0,1\), or 2 mistakes \()=1-P(3\) or 4 mistakes \()\).
\[
P(4 \text { mistakes })=P(A \cap B \cap C \cap D)=(.2)(.3)(.25)(.5)=.0075
\]
\[
\begin{aligned}
P(3 \text { mistakes })=& P\left(A^{\prime} \cap B \cap C \cap D\right)+P\left(A \cap B^{\prime} \cap C \cap D\right) \\
&+P\left(A \cap B \cap C^{\prime} \cap D\right)+P\left(A \cap B \cap C \cap D^{\prime}\right) \\
=&(.8)(.3)(.25)(.5)+(.2)(.7)(.25)(.5)+(.2)(.3)(.75)(.5)+(.2)(.3)(.25)(.5) \\
=& .0775
\end{aligned}
\]
So \(P(\) pass \()=1-(.0075+.0775)=.915\)