In my town, it’s rainy one third of the days. Given that it is rainy, there will

Question

In my town, it’s rainy one third of the days. Given that it is rainy, there will be heavy traffic with probability 12, and given that it is not rainy, there will be heavy traffic with probability 14. If it’s rainy and there is heavy traffic, I arrive late for work with probability 12. On the other hand, the probability of being late is reduced to 18 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?

b. What is the probability that I am late?

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

Explanation / Answer

Here we are given that

P( rainy ) = 1/3 and therefore P( non rainy ) = 2/3

Also, we are given that:

P( heavy traffic | rainy ) = 1/2 = 0.5
P( heavy trafffic | non rainy ) = 1/4 = 0.25
P( late | rainy and heavy traffic ) = 1/2 = 0.5,
P( late | non rainy and no heavy traffic ) = 1/8 = 0.125,
P( late | rainy and no heavy traffic ) = P( late | non rainy and heavy traffic ) = 0.25

a) Now the probability that it’s not raining and there is heavy traffic and I am not late is computed as:

= 1 - P( late | not rainy and heavy traffic ) P( not rainy and heavy traffic )

= 1 - P( late | not rainy and heavy traffic ) P( heavy trafffic | not rainy ) P( not rainy )

= 1 - 0.25*0.25*(2/3)

= 0.9583

Therefore 0.9583 is the required probability here.

b) the probability of being late is computed as:

P( late ) = P( late | rainy and heavy traffic ) P( heavy traffic | rainy )P( rainy ) + P( late | non rainy and no heavy traffic)P( no heavy traffic | not rainy )P( not rainy ) + P( late | rainy and no heavy traffic ) P( no heavy traffic | rainy ) P( rainy ) + P( late | non rainy and heavy traffic ) P( heavy traffic | not rainy ) P( not rainy )

P( late ) = 0.5*0.5*(1/3) + 0.25*0.75*(2/3) + 0.25*0.5*(1/3) + 0.25*0.25*(2/3)

P( late ) = 0.0833 + 0.125 + 0.04167 + 0.4167 = 0.2917

Therefore 0.2917 is the required probability here.

c) now given that I arrived late at work, probability that it rained that day is computed as:

= [ P( late | rainy and heavy traffic ) P( heavy traffic | rainy )P( rainy ) + P( late | rainy and no heavy traffic ) P( no heavy traffic | rainy ) P( rainy ) ] / P( late )

= [ 0.5*0.5*(1/3) + 0.25*0.5*(1/3) ] / 0.2917 = 0.4285

Therefore 0.4285 is the required probability here.

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