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:
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)=‌

1

5

,‌‌P(E2)=‌

2

5

,‌‌P(E3)=‌

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(A∣E1)=‌

2

7

‌ leading to ‌P(A∣E1)=1−‌

2

7

=‌

5

7

‌P(A∣E2)=‌

4

7

‌ leading to ‌P(A∣E2)=1−‌

4

7

=‌

3

7

‌P(A∣E3)=‌

1

7

‌ leading to ‌P(A∣E3)=1−‌

1

7

=‌

6

7

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:
=‌

( 5 7 )×( 1 5 )

(‌ 5 7 ×‌ 1 5 )+(‌ 3 7 ×‌ 2 5 )+(‌ 6 7 ×‌ 3 5 )

Calculate the numerator and the denominator:
=‌

‌ 1 5 × 5 7

1 5 ×‌ 5 7 +‌ 2 5 ×‌ 3 7 +‌ 3 5 ×‌ 6 7

Simplify:
=‌

‌ 5 35

5 35 +‌ 6 35 +‌ 18 35

=‌

‌ 5 35

29 35

=‌

5

29

Thus, the probability that the person traveled by car, given they arrived on time, is ‌

5

29

.