Question

A person goes to office either by car, scooter, bus or train, the probability of which being 1/7, 3/7, 2/7 and 1/7 respectively. Probability that he reaches office late, if he takes car, scooter, bus or train is 2/9, 1/9, 4/9 and 1/9 respectively. Given that he reached office in time, then what is the probability that he travelled by a car.

Answer 1

Let us define the following events

C ≡ person goes by car,

• ≡ person goes by scooter,

B ≡ person goes by bus,

T ≡ person goes by train,

L ≡ person reaches late

Then we are given in the question

P (C) = 1/7; P ( *) = 3/7; P (B) = 2/7 P (T) = 1/7

P (L|C) = 2/9; P (L|*) = 1/9; P (L| B) = 4/9; P (L| T) = 1/9

To find the prob. P (C|L) [∵ reaches in time ≡ not late] Using Baye'* theorem

P (C|L) = P (L|C) P (C)/ P(L|C) P(C) + P (L| ) P( ) . . . . . . . . . . . . . . . . . . . . . . . . (i)

• P (L|B) P(B) + A (L|T)P (T)

Now, P (L|C) = 1 – 2/9 = 7/9; P(L| *) = 1 – 1/9 = 8/9

P (L|B) = 1 – 4/9 = 5/9; P (L|T) = 1 – 1/9 = 8/9

Substituting these values in eqn. (i) we get

P (C|L bar) = 7/9 x 1/7 / 7/9 x 1/7 + 8/9 x 3/7 + 5/9 x 2/7 + 8/9 x 1/7

= 7/7 + 24 + 10 + 8 = 7/49 = 1/7.