SOLUTION: The University of America runs two bus lines on campus: red and green. The red line serves north
campus and the green line serves south campus with a transfer station linking the
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Question 1177981: The University of America runs two bus lines on campus: red and green. The red line serves north
campus and the green line serves south campus with a transfer station linking the two lines. Green
buses arrive randomly (according to a Poisson distribution) at the transfer station every 10 minutes.
Red buses also arrive randomly every 7 minutes.
(a) What is the probability that 2 buses will stop at the station during a 5 minute interval?
(b) A student whose dormitory is located next to the station has a class in 10 minutes. Either bus will
take the student to the classroom building. The ride takes 5 minutes, after which the student will walk
for around 3 minutes to reach the class room. What is the probability that the student will make it to
class on time?
Answer by CPhill(1959) (Show Source): You can put this solution on YOUR website!
Let's solve this problem step-by-step.
**Understanding the Problem**
* **Green Buses:** Poisson distribution, arrival every 10 minutes (λ_green = 10 minutes)
* **Red Buses:** Poisson distribution, arrival every 7 minutes (λ_red = 7 minutes)
**(a) Probability of 2 Buses in a 5-Minute Interval**
1. **Calculate Arrival Rates for 5 Minutes**
* Green buses: λ_green(5) = (5 minutes / 10 minutes) = 0.5 buses
* Red buses: λ_red(5) = (5 minutes / 7 minutes) ≈ 0.7143 buses
2. **Calculate Total Arrival Rate**
* Total arrival rate (λ_total) = λ_green(5) + λ_red(5) = 0.5 + 0.7143 = 1.2143 buses
3. **Use Poisson Distribution Formula**
* P(X = k) = (e^(-λ) * λ^k) / k!
We want to find P(X = 2), where λ = 1.2143.
* P(X = 2) = (e^(-1.2143) * 1.2143^2) / 2!
* P(X = 2) = (0.2969 * 1.4745) / 2
* P(X = 2) ≈ 0.4378 / 2
* P(X = 2) ≈ 0.2189
Therefore, the probability that 2 buses will stop at the station during a 5-minute interval is approximately 0.2189.
**(b) Probability of Making it to Class on Time**
1. **Determine Arrival Rates for 10 Minutes**
* Green buses: λ_green(10) = (10 minutes / 10 minutes) = 1 bus
* Red buses: λ_red(10) = (10 minutes / 7 minutes) ≈ 1.4286 buses
2. **Calculate Probabilities of No Buses Arriving in 10 Minutes**
* P(Green = 0) = (e^(-1) * 1^0) / 0! = e^(-1) ≈ 0.3679
* P(Red = 0) = (e^(-1.4286) * 1.4286^0) / 0! = e^(-1.4286) ≈ 0.2394
3. **Calculate Probability of No Buses Arriving (Both Lines)**
* P(Green = 0 and Red = 0) = P(Green = 0) * P(Red = 0) = 0.3679 * 0.2394 ≈ 0.0881
4. **Calculate Probability of at Least One Bus Arriving**
* P(At least one bus) = 1 - P(Green = 0 and Red = 0) = 1 - 0.0881 = 0.9119
5. **Conclusion**
If at least one bus arrives within the 10 minute period, the student will have 5 minutes of travel time and 3 minutes of walking time, which will get them to class on time.
Therefore, the probability that the student will make it to class on time is approximately 0.9119.
**Results**
(a) The probability that 2 buses will stop at the station during a 5-minute interval is approximately 0.2189.
(b) The probability that the student will make it to class on time is approximately 0.9119.
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