Question

Say cars crossing a certain multilane bridge take either 3, 4 or 5 minutes for the trip. 50% take 3 minutes, with a 25% figure each for the 4- and 5-minute trips. We will consider the traversal by three cars, named A, B and C, that simultaneously start crossing the bridge. They are in different lanes, and operate independently.

(a) Find the probability that the first arrival to the destination is at the 4-minute mark.

(b) Find the probability that the total trip time for the three cars is 10 minutes.

(c) An observer reports that the three cars arrived at the same time. Find the probability that the cars each took 3 minutes to make the trip.

Answer 1

a)P( probability that the first arrival to the destination is at the 4-minute mark)=P(one car reach at 4 minute and other 2 on five minute)+P(two car reach at 4 minute and other one five minute)+P(all three reach on 4 minute)

=(3!/(1!*2!))*(0.25)*(0.25)²+(3!/(2!*1!))*(0.25)²*(0.25)⁰+(3!/(3!*0!))*(0.25)³ =0.109375

b)

probability that the total trip time for the three cars is 10 minutes =P(2 cars take 3 minutes and 1 car takes 4 minutes)

=(3!/(2!*1!))*(0.5)²*(0.25) =0.1875

c)

P(each took 3 minutes given all arrive at same time) =P(each took 3 min)/(P(each took 3 min+ 4 min +5 min))

=0.5³/(0.5³+0.25³+0.25³) =0.8