Question

A man takes either a Metrobus or Metrorail to work, with probabilities 0.4 and 0.6 respectively. When he 

takes the bus, he is late on 25% of the days. When he takes the train, he is late on 15% of the days. If the 

man is late for work on a particular day, what is the probability he took the bus?

Answer 1

To find that probability it is appropriate to use Bayesian formula:

$P(A|B)={\frac{P(B|A)*P(A)} {P(B)}}$

in our case A - man took the bus, B - man is late

Lets find the required terms

$P(B|A)=0.25$

$P(A)=0.4$

$P(B)=P(A)*P(B|A)+P(C)*P(B|C) = 0.4*0.25+0.6*0.15=0.19$

In the formula above C - man took train

Finally, the sought probability is

$P(A|B)={\frac{P(B|A)*P(A)} {P(B)}}={\frac{0.25*0.4} {0.19}}=0.526$