ents 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}, \quad P(E_2) = \frac{2}{5}, \quad 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(\bar{A} \mid E_1) = \frac{2}{7} $ leading to $ P(A \mid E_1) = 1 - \frac{2}{7} = \frac{5}{7} $

$ P(\bar{A} \mid E_2) = \frac{4}{7} $ leading to $ P(A \mid E_2) = 1 - \frac{4}{7} = \frac{3}{7} $

$ P(\bar{A} \mid E_3) = \frac{1}{7} $ 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 $\frac{5}{29}$.

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