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You have to pass through an obstacle course. The probabilities that you make a mistake on each of the 4 obstacles is (respectively) $0.2,0.3,0.25$ and $0.5$. You pass the course if you make no more than 2 mistakes. What is the probability that you pass the course?
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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 $)=1-(.0075+.0775)=.915$
by Diamond (40,719 points)

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