Question

In my friend’s town (not Vancouver), it's rainy one third of the days. Given that it is rainy, there will be heavy traffic with probability 1/2, and given that it is not rainy, there will be heavy traffic with probability 1/4. If it's rainy and there is heavy traffic, I arrive late for work with probability 1/2. On the other hand, the probability of being late is reduced to 1/8 if it is not rainy and there is no heavy traffic. In other situations (rainy and no traffic, not rainy and traffic) the probability of being late is 0.25. You pick a random day.

a) What is the probability that it's not raining and there is heavy traffic and I am not late? Use both tree diagram and application of chain rule to show your result.

b) What is the probability that I am late? Use both tree diagram and the law of total probability to show your result.

c) Given that I arrived late at work, what is the probability that it rained that day?

Answer 1

given that probability of rainy with heavy traffic = 1/2

probability of not rainy with traffic = 1/4

probability of arriving late due to traffic = 1/2

probability of arriving late without traffic = 1/8

a)

probability that its not raining and there is heavy traffic and im not late = (2/3)*(1/4)*(1-0.25)

= 0.125

b)

probability of being late = (1/3)(1/2)(1/2) + (1/3)(1/2)(0.25) + (2/3)(1/4)(0.25) + (2/3)(3/4)(1/8)

= 0.229

c)

probability (rainy day + arrived late) = ((1/3)*(1/2)*(1/2) + (1/3)*(1/2)*(0.25)) / (0.229) = 0.5454