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A person goes to office either by car, scooter, bus or train 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?

Text Solution

Verified by Experts

Let us define the following events
`{:("C: person goes by car",),("S:person goes by scooter",),("B: person goes by bus",),("T: person gies 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,(L//S)=1/9,P(L//B)=4/9,P(L//T)=1/9`
We have to find `P(C//barL)` [Since reaches in time=not late]. using Bayes's theorem,
`P(C//barL)`
`(P(barL//C)P(C))/(P(barL//C)P(C)+P(barL//S)P(S)(barL//B)+A(barL//T)P(T))`
`Now, " "(1)`
`P(barL//C)=1-2/9=7/9,P(barL//S)=1-1/9=8/9`
`P(barL//B)=1-4/9=5/9,P(barL//T)=1-1/9=8/9`
Substituating these values in Eq. (1), we get
`P(C//barL)=(7/9xx1/7)/(7/9xx1/7+8/9xx3/7+5/9xx2/7+8/9xx1/7)`
`=(7)/(7+24+10+8)=(7)/(49)=1/7`
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