To get from Berkeley to the San Francisco airport (SFO) one can either take a sh

Question

To get from Berkeley to the San Francisco airport (SFO) one can either take a shuttle van or drive. Either way, one must cross the San Francisco Bay Bridge. The shuttle van drops its passengers at the airport departure terminal. If one drives, he or she must park in a lot near the airport, then take a parking shuttle from the parking lot to the airport departure terminal. There is a 50% chance that the Bay Bridge will be congested with traffic. If it is, it takes 1 hour to drive to the parking lot. If not, it takes 40 minutes to drive to the parking lot. The parking shuttle runs every 10 minutes and takes 10 minutes to get to the airport departure terminal from the parking lot. Suppose it is equally likely that one must wait 0, 1, 2, .., or 9 minutes for the parking shuttle once getting to the parking lot, and that the amount of time one must wait is independent of the amount of time it takes to drive to the parking lot from Berkeley. If one takes the shuttle van from Berkeley directly to the airport, it will take an hour, plus 10 minutes per stop for other passengers. Suppose that it is equally likely that the shuttle van will make 0, 1, 2, or 3 stops to pick up other passengers. The shuttle van can take the carpool lane; assume that the time it takes the shuttle van to go from Berkeley to the airport is not affected by traffic on the bridge. Answer the following questions The chance it takes less than one hour to get to the airport by driving and taking the parking shuttle is (Q7) The chance it takes more than 79 minutes to get to the airport by driving and taking the parking shuttle is (Q8) The chance it takes more than 79 minutes to get to the airport taking the shuttle van directly from Berkeley to the airport is (Q9) Suppose John takes the shuttle van directly from Berkeley to the airport, and Jane drives from Berkeley and takes the parking shuttle. They leave at the same time, and their travel times are independent. The chance that John gets to the airport before Jane is (Q10)

Explanation / Answer

Considering only driving

There is 0.5 probability of each 60 minutes due to traffic or only 40 minutes.

There is 0.1 probability of each 0, 1, 2...9 minute waiting time.

So there are 20 possible combinations:

40+1, 40+2, ....40+9 without traffic at bridge and 60+1, 60+2....60+9 with traffic at bridge.

Adding 10 more minutes due to parking shuttle travel, we have 50, 51,52,53...59 minutes ad 70, 71, 72, 73.... 79 minutes.

Assuming the traffic and the shuttle waiting time being independent, the probability of each time is = 0.5*0.1 = 0.05.

Now consider the shuttle route:

There are four possibilities of 60, 70, 80 and 90 due to the number of stops possible and each have a probability of 0.25.

Ans 9)Final Solution = 0.5

The possibilitties of taking more than 79 minutes can happen only when it takes 80 minutes or 90 minutes.

So, P(X>79) = P(X=80) + P(X=90) = 025 + 025 = 0.5

Ans to 10) Final solution = 0.2375

For john to reach before jane, we have following possibilities:

Let X1 be john's time and X2 be jane's time.

P(X1 < X2) = P(X1= 60)*P(X2>60) + P(X1= 70)*P(X2>70)

Jane will take more than 60 minutes whenever there is traffic on the bridge

P(X2 > 60) = 0.05*10 = 0.5

Similarly, P(X2 > 70) = P(71)+P(72)+P(73)+P(74)+P(75)+P(76)+P(77)+P(78)+P(79) = 0.05*9 = 0.45

Therefore P(X1 < X2) = 0.5*0.25 + 0.45*0.25 = 0.2375

Ans 7) Chance of taking less than 60 minutes by driving = 0.5.

Ans 8) Chance of taking more than 79 minutes by driving = 0.0. (maximum time taken is = 79 minutes and more than 79 is not possible)

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