Probability

To find the probability that a person traveled to college by car, given that they arrived on time, let's denote the events as follows:$E_1$ : The event that a person travels by car.$E_2$ : The event that a person travels by bus.$E_3$ : The event that a person travels by train.The probabilities for choosing each mode of transport are:$P (E_1) =\;\frac{1}{5}, \;\; P (E_2) =\;\frac{2}{5}, \;\; P (E_3) =\;\frac{3}{5}$Let $A$ be the event of reaching the college on time. The given probabilities of being late for each mode of transport are:$\;P (\overline{A} \mid E_1) =\;\frac{2}{7} \;\text{ leading to }\; P (A \mid E_1) =1-\;\frac{2}{7}=\;\frac{5}{7}$$\;P (\overline{A} \mid E_2) =\;\frac{4}{7} \;\text{ leading to }\; P (A \mid E_2) =1-\;\frac{4}{7}=\;\frac{3}{7}$$\;P (\overline{A} \mid E_3) =\;\frac{1}{7} \;\text{ leading to }\; P (A \mid E_3) =1-\;\frac{1}{7}=\;\frac{6}{7}$Using Bayes' Theorem, we calculate $P (E_1 \mid A)$, the probability that the person traveled by car, given they arrived on time:$P (E_1 \mid A) =\;\frac{P (A \mid E_1) \cdot P (E_1)}{P (A \mid E_1) \cdot P (E_1)+P (A \mid E_2) \cdot P (E_2)+P (A \mid E_3) \cdot P (E_3)}$Substituting the known values:$=\;\frac{ \left(\frac{5}{7}\right) \times \left(\frac{1}{5}\right) }{ \left(\;\frac{5}{7} \times \;\frac{1}{5}\right) + \left(\;\frac{3}{7} \times \;\frac{2}{5}\right) + \left(\;\frac{6}{7} \times \;\frac{3}{5}\right) }$Calculate the numerator and the denominator:$=\;\frac{\;\frac{1}{5} \times \frac{5}{7}}{\frac{1}{5} \times \;\frac{5}{7}+\;\frac{2}{5} \times \;\frac{3}{7}+\;\frac{3}{5} \times \;\frac{6}{7}}$Simplify:$=\;\frac{\;\frac{5}{35}}{\frac{5}{35}+\;\frac{6}{35}+\;\frac{18}{35}}=\;\frac{\;\frac{5}{35}}{\frac{29}{35}}=\;\frac{5}{29}$Thus, the probability that the person traveled by car, given they arrived on time, is $\;\frac{5}{29}$.