Question

A chartered financial analyst can choose any one of Routes A, B, or C to get to work. The probabilities that he/she will arrive at work on time using Routes A, B, and C are 0.45, 0.51, and 0.62, respectively. Assuming that the analyst is equally likely to choose any one of the routes, and supposing that he/she has just arrived at work on time, what is the probability that he/she chose Route A?

Answer 1

We are given here that:

P( arrive on time | A) = 0.45,
P( arrive on time | B) = 0.51,
P( arrive on time | C) = 0.62

Also, we are given here that:
P(A) = P(B) = P(C) = 1/3

Therefore using law of total probability, we get here:
P( arrive on time) = P( arrive on time | A) P(A) + P( arrive on time | B) P(B) + P( arrive on time | C) P(C)

P( arrive on time) = (1/3)*(0.45 + 0.51 + 0.62) = 0.2832

Given that the analyst arrives on time, probability that he / she took route A is computed using bayes theorem here as:

P(A | arrive on time) =  P( arrive on time | A) P(A) / P( arrive on time) = (0.45/3) / 0.2832 = 0.5297

Therefore 0.5297 is the required probability here.