Answer to Question #174794 in Statistics and Probability for Shyam jain jain

Question

Statistics and ProbabilityA secretary goes to work following one of three routes A, B, C. Her choice of route is independent of the weather. If it rains, the probabilities of arriving late, following A, B, C are 0.06, 0.15, 0.12 respectively. The corresponding probabilities, if it does not rain, are 0.05, 0.10, 0.15 respectively. Given that on a sunny day she arrives late, what is the probability that she took route C? Assume that on an average one in every four days is rainy.

Answer 1

Let event A be that the secretary was late on a sunny day.

Let the events A, B, C consist in the fact that she chose the route A, B, C, respectively.

If on average every fourth day is rainy, then there are 91 rainy days in a year.

Then the number of sunny days is 365-91=274.

whence we have

$p(A) = p(B) = p(C) = \frac{1}{3}$

$p(H|A) = \frac{{274}}{{365}} \cdot 0.05,\,p(H|B) = \frac{{274}}{{365}} \cdot 0.10,\,\,\,p(H|C) = \frac{{274}}{{365}} \cdot 0.15$

Then, by the formula of total probability

$p(H) = p(A)p(H|A) + p(B)p(H|B) + p(C)p(H|C) = \frac{1}{3} \cdot \frac{{274}}{{365}}\left( {0.05 + 0.1 + 0.15} \right) = \frac{{137}}{{1825}}$

Whence, according to Bayes' formula, the wanted probability is

$p(C|A) = \frac{{p(C)p(H|C)}}{{p(H)}} = \frac{{\frac{1}{3} \cdot \frac{{274}}{{365}} \cdot 0.15}}{{\frac{{137}}{{1825}}}} = 0.5$

Answer: 0.5