Fewpal
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A person goes to office either by car, scooter, bus or train, the probabilities of which being \(\frac{1}{7},\frac{3}{7},\frac{2}{7}\) and \(\frac{1}{7},\) respectively. Probability that he reaches office late, if he takes car, scooter, bus or train is \(\frac{2}{9},\frac{1}{9},\frac{4}{9}\, and\, \frac{1}{9}\) respectively. Given that he reached office in time, then what is the probability that he travelled by a car.

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Let the events E1, E2, E3, E4, and A be defined as follows: 

E1: Event that the person goes to the office by car 

E2: Event that the person goes to the office by scooter 

E3: Event that the person goes to the office by bus 

E4: Event that the person goes to the office by train. 

A: Event that the person reaches the office in time

Then, P(E1) = \(\frac{1}{7}\) ,P(E2) = \(\frac{3}{7}\) , P(E3) = \(\frac{2}{7}\) , P(E4) = \(\frac{1}{7}\)

P(A/E1) = P(Person reaches office in time if he goes by car) 

= 1 – P(Person reaches office late if he goes by car) 

\(1- \frac{2}{9}​​​​=\frac{7}{9}\) 

= P(A/E2) = P(Person reaches office in time if he goes by scooter) 

= 1 – P(Person reaches office late if he goes by scooter) =

\(1- \frac{1}{9} =\frac{8}{9}\)

P(A/E3) = P(Person reaches office in time if he goes by bus) 

= 1 – P(Person reaches office late if he goes by bus) 

\(1-\frac{4}{9}=\frac{5}{9}\)

P(A/E4) = P(person reaches office in time if he goes by train) 

= 1 – P(person reaches office late if he goes by train)

\(1-\frac{1}{9}=\frac{8}{9}\)

Now, P(person travelled by car if he reached office in time)

P(E1/A) = \(\frac {P(E_1)\times P(A/E_1)}{P(E_1)\times P(A/E_1)+ P(E_2)\times P(A/E_2)+P(E_3)\times P(A/E_3)+P(E_4)\times P(A/E_4)}\)   ....  Baye's Theorem

\(\frac{\frac{1}{7}\times \frac{7}{9}}{\frac{1}{7}\times \frac{7}{9}+ \frac{3}{7}\times \frac{8}{9}+ \frac{2}{7}\times \frac{5}{9}+ \frac{1}{7}\times \frac{8}{9}}\) = \(\frac{\frac{7}{63}}{\frac{7}{63}+\frac{24}{63}+\frac{10}{63}+\frac{8}{63}}\) = \(\frac{\frac{7}{63}}{\frac{49}{63}}\) = \(\frac{7}{49} =\frac{1}{7}\)

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