The probabilities that a government servant goes to work by car, bus or train on a certain day are 1/10, 2/5 and 1/2 respectively. The probabilities of his being late for work by these modes of transport are 1/5,1/2 and 3/10 respectively. If he was late on this particular day, using Bayes’ Theorem, calculate the probability that he travelled be train.

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Let define a notation to answer this question.

P(C)=probability traveled by car

P(B) = probability travelled by Bus

P(T)= probabilty travelled by train

P(L/C)=prabability being late travelled by car

and similarly we can define P(L/B) and P(L/T).

We know

`P(X//Y)=(P(XnnY))/(P(Y))`

`P(X//Y)P(Y)=P(XnnY)`

Thus

`P(LnnC)=P(L//C)P(C)`

`=1/10xx1/5=1/50`

`P(LnnB)=P(L//B)P(B)`

`=1/2xx2/5=1/5`

`P(LnnT)=P(L//T)P(T)`

`=3/10xx1/2=3/20`

Thus by Bayes Theorem

`P(T//L)=(P(TnnL))/{P(LnnC)+P(LnnB)+P(LnnT)}`

`=(3/20)/{1/50+1/5+3/20}`

`=(3/20)/{(2+20+15)/100}`

`=15/37`

`` Hence he was late on this particular day, the probability that he travelled be train is 15/37.

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