QUESTION 2
On a randomly chosen day the probability that Bill travels to school by car, by bike or on foot is
3 $\frac{1}{2}, \frac{1}{6}$ and $\frac{1}{3}$ respectively. The probability of being late when using these methods of travel is $\frac{1}{5}$, $\frac{1}{5}$ and $\frac{1}{10}$ respectively.
Given that Bill is late, find the probability that he did not travel on foot.

Step 1 ：Given probabilities: \(P(\text{Car}) = \frac{7}{12}\), \(P(\text{Bike}) = \frac{1}{6}\), \(P(\text{Foot}) = \frac{1}{3}\), \(P(\text{Late} | \text{Car}) = \frac{1}{5}\), \(P(\text{Late} | \text{Bike}) = \frac{1}{5}\), and \(P(\text{Late} | \text{Foot}) = \frac{1}{10}\)

Step 2 ：Find the probability of not traveling on foot: \(P(\text{Not Foot}) = P(\text{Car}) + P(\text{Bike}) = \frac{7}{12} + \frac{1}{6} = \frac{8}{12} = \frac{2}{3}\)

Step 3 ：Find the probability of being late given not traveling on foot: \(P(\text{Late} | \text{Not Foot}) = \frac{P(\text{Late} | \text{Car}) \cdot P(\text{Car}) + P(\text{Late} | \text{Bike}) \cdot P(\text{Bike})}{P(\text{Not Foot})} = \frac{\frac{1}{5} \cdot \frac{7}{12} + \frac{1}{5} \cdot \frac{1}{6}}{\frac{2}{3}} = \frac{1}{5}\)

Step 4 ：Find the probability of being late: \(P(\text{Late}) = P(\text{Late} | \text{Car}) \cdot P(\text{Car}) + P(\text{Late} | \text{Bike}) \cdot P(\text{Bike}) + P(\text{Late} | \text{Foot}) \cdot P(\text{Foot}) = \frac{1}{5} \cdot \frac{7}{12} + \frac{1}{5} \cdot \frac{1}{6} + \frac{1}{10} \cdot \frac{1}{3} = \frac{11}{60}\)

Step 5 ：Use Bayes' theorem to find the probability of not traveling on foot given being late: \(P(\text{Not Foot} | \text{Late}) = \frac{P(\text{Late} | \text{Not Foot}) \cdot P(\text{Not Foot})}{P(\text{Late})} = \frac{\frac{1}{5} \cdot \frac{2}{3}}{\frac{11}{60}} = \frac{12}{11}\)

Step 6 ：\(\boxed{P(\text{Not Foot} | \text{Late}) \approx 0.8182}\)