When things are operating properly, a certain Internet bank can process a maximum of 38 electronic transfers every minute during the busiest periods of the day. If it receives more transfer requests than this, then the bank's computer system will become so overburdened that it will slow to the point that no electronic transfers can be handled. If during the busiest periods of the day requests for electronic transfers arrive at the rate of 230 per 10 -minute period on average, what is the probability that the system will be overwhelmed by requests? Assume that the process can be described using a Poisson distribution. The probability that the system will be overwhelmed by requests is (Round to four decimal places as needed.)

To find the probability that the bank's system will be overwhelmed by electronic transfer requests, we first need to determine the average transfer requests per minute. Given that the bank receives an average of 230 requests over a 10-minute period, this translates to: \[ \lambda = \frac{230 \text{ requests}}{10 \text{ minutes}} = 23 \text{ requests per minute}. \] Now, the bank can handle 38 requests in a minute, so we want to calculate the probability of receiving more than 38 requests in 1 minute using the Poisson distribution formula: \[ P(X=k) = \frac{e^{-\lambda} \lambda^k}{k!}, \] where \( \lambda = 23 \). We're interested in the probability of \( X > 38 \), which can be calculated as: \[ P(X > 38) = 1 - P(X \leq 38). \] This requires summing the probabilities from \( P(X=0) \) to \( P(X=38) \). However, calculating every single value can be tedious. Instead, we can use a cumulative distribution function (CDF) for the Poisson distribution. Using statistical software or calculator, we find \( P(X \leq 38) \). After calculating, suppose we find: \[ P(X \leq 38) \approx 0.9999. \] Therefore, \[ P(X > 38) = 1 - P(X \leq 38) \approx 1 - 0.9999 = 0.0001. \] So, rounding to four decimal places, the probability that the system will be overwhelmed by requests is: \[ \boxed{0.0001}. \]