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.

Question

Aditi Chauhan · Answer

Hello Student,

Please find the answer to your question

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
Aditi Chauhan
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