1. **Problem statement:** Customers arrive at a bank at a rate of 10 per hour. Each customer is new or existing with probability 0.5. Given exactly 10 new customers arrived in one hour, find: (a) The probability exactly 10 existing customers also arrived. (b) The probability at least 20 customers arrived. 2. **Model and formulas:** The total arrivals follow a Poisson distribution with rate $\lambda=10$ per hour. Since each customer is new or existing with probability 0.5 independently, the number of new customers $N$ and existing customers $E$ are independent Poisson random variables with rates $\lambda_N=5$ and $\lambda_E=5$ respectively (by thinning property of Poisson processes). 3. **Given:** $N=10$ new customers arrived. 4. **(a) Probability exactly 10 existing customers arrived:** Since $E \sim \text{Poisson}(5)$, $$P(E=10) = \frac{5^{10} e^{-5}}{10!}$$ Calculate this value: - $10! = 3628800$ - $5^{10} = 9765625$ - $e^{-5} \approx 0.0067379$ So, $$P(E=10) = \frac{9765625 \times 0.0067379}{3628800} \approx \frac{65832.5}{3628800} \approx 0.0181$$ 5. **(b) Probability at least 20 customers arrived:** Total customers $T = N + E \sim \text{Poisson}(10)$. We want $P(T \geq 20) = 1 - P(T \leq 19)$. Using the Poisson CDF for $\lambda=10$: $$P(T \leq 19) = \sum_{k=0}^{19} \frac{10^k e^{-10}}{k!}$$ This sum can be computed using tables or software. Approximate value: $$P(T \leq 19) \approx 0.9999$$ Thus, $$P(T \geq 20) = 1 - 0.9999 = 0.0001$$ 6. **Summary:** - (a) $P(E=10) \approx 0.0181$ - (b) $P(T \geq 20) \approx 0.0001$ These probabilities reflect the Poisson nature of arrivals and the independence of new and existing customers.