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:
E1 : The event that a person travels by car.
E2 : The event that a person travels by bus.
E3 : The event that a person travels by train.
The probabilities for choosing each mode of transport are:
P(E1)=‌,‌‌P(E2)=‌,‌‌P(E3)=‌Let
A be the event of reaching the college on time. The given probabilities of being late for each mode of transport are:
‌P(A∣E1)=‌‌ leading to ‌P(A∣E1)=1−‌=‌‌P(A∣E2)=‌‌ leading to ‌P(A∣E2)=1−‌=‌‌P(A∣E3)=‌‌ leading to ‌P(A∣E3)=1−‌=‌Using Bayes' Theorem, we calculate
P(E1∣A), the probability that the person traveled by car, given they arrived on time:
P(E1∣A)=‌| P(A∣E1)⋅P(E1) |
| P(A∣E1)⋅P(E1)+P(A∣E2)⋅P(E2)+P(A∣E3)⋅P(E3) |
Substituting the known values:
=‌| ()×() |
| (‌×‌)+(‌×‌)+(‌×‌) |
Calculate the numerator and the denominator:
=‌| ‌× |
| ×‌+‌×‌+‌×‌ |
Simplify:
=‌=‌=‌Thus, the probability that the person traveled by car, given they arrived on time, is
‌.