Question
In a city, a person owns independently a sedan car with probability $\frac{3}{10}$ and a SUV with probability $\frac{4}{10}$. If he has sedan only, then he keeps a driver with probability $\frac{6}{10}$, whereas if he owns SUV only, then he keeps a driver with probability $\frac{7}{10}$, whereas if he keeps both type of cars then his probability of keeping a driver is $\frac{9}{10}$. Then
Options
Solution
Sol. [Key Concept: Independent ownership; condition on the events for the driver-keeping probability.]
$D \rightarrow$ event that person keeps driver
$E_{1} \rightarrow$ event that person own sedan
$E_{1} \rightarrow$ event that person own SUV
$$\begin{matrix} P(D) = & P\left( E_{1} \cap E_{2} \right)P\left( D/E_{1} \cap E_{2} \right) + P\left( E_{1} \cap E_{2} \right)P\left( D/E_{1} \cap E_{2} \right) + P\left( E_{1} \cap E_{2} \right)P\left( D/E_{1} \cap E_{2} \right) \\ & \ \\ = & \frac{3}{10} \times \frac{6}{10} \times \frac{6}{10} + \frac{7}{10} \times \frac{4}{10} \times \frac{7}{10} + \frac{3}{10} \times \frac{4}{10} \times \frac{9}{10} \\ = & \frac{103}{250} \\ P\left( E_{1}/D \right) = & \frac{P\left( E_{1} \cap {\bar{E}}_{2} \right)P\left( D/E_{1} \cap {\bar{E}}_{2} \right) + P\left( E_{1} \cap E_{2} \right)P\left( D/E_{1} \cap E_{2} \right)}{P(D)} \\ = & \frac{\left( \frac{54}{250} \right)}{\left( \frac{103}{250} \right)} = \frac{54}{103} \end{matrix}$$
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