Question 1177981
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.