Question

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

Answer 1

As, the statem ent shows problem is to be related to Baye's law.
Let C,S,B,T be the events when person is going by car, scooter, bus or train, respectively.
$ \therefore P(C) = \frac{1}{7} , P(S) = \frac{3}{7} ,P(B) = \frac{2}{7} ,P(T) = \frac{1}{7}$
Again, $L$ be the event of the person reaching office late
$ \therefore \overline L $ be the event of the person reaching office in time.
Then ,$ P\bigg( \frac{\bar{L}}{C} \bigg) = \frac{7}{9} ,P\bigg( \frac{\bar{L}}{S} \bigg) = \frac{8}{9},P\bigg( \frac{\bar{L}}{B} \bigg) = \frac{5}{9} $ and $ P\bigg( \frac{\bar{L}}{T} \bigg) = \frac{8}{9} $
$ P\bigg( \frac{C}{L} \bigg) = \frac{ P\bigg( \frac{\bar{L}}{C} \bigg) . P(C) }{ P\bigg( \frac{\bar{L}}{C} \bigg) . P(C) +P\bigg( \frac{\bar{L}}{S} \bigg) .P(S) +P\bigg( \frac{\bar{L}}{B} \bigg).P(B) +P\bigg( \frac{\bar{L}}{T} \bigg) .P(T)} $
$ =\frac{ \frac{7}{9} \times \frac{1}{7} }{ \frac{7}{9} \times \frac{1}{7} +\frac{8}{9} \times \frac{3}{7} + \frac{5}{9} \times \frac{2}{7} \frac{8}{9} \frac{1}{7}} =\frac{1}{7}$