A new bus runs on Sunday Tuesday Thursday and Saturday while

A new bus runs on Sunday, Tuesday, Thursday, and Saturday while an older bus runs on the other days. The new bus has a probability of being on time of 0.9 while the older bus has a probability of only 0.4. If a passenger chooses an arbitrary day of the week to ride the bus:

A) Whait is the probability that the bus will be on time?

B) What is the probability that the bus will be on time, given that the passenger chose a business day (Monday to Friday)?

Solution

Probability of choosing any day out of the 7 days in a week = 1/7

Let Mo, Tu, We, Th, Fr, Sa, Su denote the events of the 7 days in a week be chosen, in order from Monday to Sunday.

P(older bus) = P(Mo) + P(We) + P(Fr) = 3/7

P(new bus) = P(Tu) + P(Th) + P(Sa) + P(Su) = 4/7

Using total probability theorem,

P(on time) = P(on time | old bus) * P(old bus) + P(on time | new bus) * P(new bus)

= 0.4*4/7 + 0.9*3/7

= 4.3/7 = 0.6143

b. Now, Probability of choosing a business day = 1/5

P(older bus) = P(Mo) + P(We) + P(Fr) = 3/5

P(new bus) = P(Tu) + P(Th) = 2/5

Using total probability theorem,

P(on time) = P(on time | old bus) * P(old bus) + P(on time | new bus) * P(new bus)

= 0.4*2/5 + 0.9*3/5

= 0.70