Solutions - 5030-06

(1) Solve for $P[A]$ using $P[B|A]$:

$$\begin{array}{c}P[B|A]=\frac{P[A\cap B]}{P[A]}\gg \gg P[A]=\frac{P[A\cap B]}{P[B|A]}\\ \\ \gg \gg \frac{0.036}{0.60}\gg \gg 0.06\end{array}$$

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(2) Solve for $P[B]$ using $P[A|B]$:

$$\begin{array}{c}P[A|B]=\frac{P[A\cap B]}{P[B]}\gg \gg P[B]=\frac{P[A\cap B]}{P[A|B]}\\ \\ \gg \gg \frac{0.036}{0.18}\gg \gg 0.2\end{array}$$

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(3) Apply inclusion-exclusion to find $P[A\cup B]$:

$$\begin{array}{c}P[A\cup B]=P[A]+P[B]-P[A\cap B]\\ \\ \gg \gg 0.06+0.2-0.036\gg \gg 0.224\end{array}$$