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.

8 years ago
Hello Student,
Let us define the following events
C ≡ person goes by car,
S ≡ 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 (S) = 3/7; P (B) = 2/7 P (T) = 1/7
P (L|C) = 2/9; P (L| S) = 1/9; P (L| B) = 4/9; P (L| T) = 1/9
To find the prob. P (C|$\underset{L}{\rightarrow}$) [∵ reaches in time ≡ not late] Using Baye’s theorem
P (C |$\underset{L}{\rightarrow}$) = P ($\underset{L}{\rightarrow}$| C) P (C)/ P ($\underset{L}{\rightarrow}$| C) P (C) + P ($\underset{L}{\rightarrow}$| S) P (S) . . . . . . . . . . . . . . . . . . . . . . . . (i)
+ P ($\underset{L}{\rightarrow}$| B) P (B) + A ($\underset{L}{\rightarrow}$| T) P (T)
Now, P ($\underset{L}{\rightarrow}$| C) = 1 – 2/9 = 7/9; P ($\underset{L}{\rightarrow}$| S) = 1 – 1/9 = 8/9
P ($\underset{L}{\rightarrow}$| B) = 1 – 4/9 = 5/9; P ($\underset{L}{\rightarrow}$| T) = 1 – 1/9 = 8/9
Substituting these values in eqn. (i) we get
P (C| $\underset{L}{\rightarrow}$)= 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.

Thanks