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|L) [∵ reaches in time ≡ not late] Using Baye's theorem
P (C|L) = P (L|C) P (C)/ P(L|C) P(C) + P (L| S) P(S) . . . . . . . . . . . . . . . . . . . . . . . . (i)
+ P (L|B) P(B) + A (L|T)P (T)
Now, P (L|C) = 1 – 2/9 = 7/9; P(L| S) = 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.