On a given day, machine A has a 10% chance of malfunctioning and machine B has a 7% chance of the same. Given that at least one of the machines malfunctioned today, what is the chance that machine B malfunctioned?

Question

anonymous · Answer

They are asking for $\Pr(B|A\cup B)$.
They are given you $\Pr(A)$ and $\Pr(B)$.

anonymous · Answer

You're going to want to use: $$
\Pr(B|A\cup B) = \frac{\Pr(B\cap (A\cup B))}{\Pr(A\cup B)}
$$

anonymous · Answer

Now, since $B\subseteq A\cup B$, we can say $B\cap (A\cup B)=B$.
Thus we can say: $$
\Pr(B|A\cup B) = \frac{\Pr(B)}{\Pr(A\cup B)}
$$

anonymous · Answer

Next we use our formula for disjoint probability: $$
\Pr(A\cup B) = \Pr(A)+\Pr(B)-\Pr(A\cap B)
$$Since we have no reason not to think that the machines fail independently, we can use the formula for independent random events: $$
\Pr(A\cap B) = \Pr(A)\Pr(B)
$$Therefore: $$
\Pr(A\cup B) = \Pr(A)+\Pr(B)-\Pr(A)\Pr(B)
$$

anonymous · Answer

Finally, $$
\Pr(B|A\cup B) = \frac{\Pr(B)}{\Pr(A)+\Pr(B)-\Pr(A)\Pr(B)}
$$Since they gave us $\Pr(A)$ and $Pr(B)$, we have our answer.