3. Customers arrive at a bank according to a Poisson process with rate of one customer every two minutes. Find the probability that the fifth customer arrives at the bank within 10 minutes after the bank opens. (A) .560 (B) .440 (C) .500 (D) .616 (E) .384 4. Consider the situation described in problem #3. If no customers have arrived within the first two minutes after the bank has opened, what is the probability that it will be at least two more minutes before the first arrival? (A) e^-2 (B) e^-4 (C) 1 - e^-4 (D) e^-1 (E) 1 - e^-1 5. The length of time that I wait for my bus in the morning is exponentially distributed with mean ? = 10 minutes. What is the probability that the minimum of my waiting times over the next 5 days is more than 6 minutes? Assume that the waiting times on different days are independent. (A) e^-0.6 (B) e^-3.0 (C) 1 - e^-0.6 (D) 1 - e^-3.0 (E) (1 - e^-0.6)^5
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The rate of customers arriving is 1 every 2 minutes, so the rate is 0.5 customers per minute. The time until the kth event in a Poisson process follows a gamma distribution, which in the case of k=5 simplifies to an Erlang distribution. The probability density Show more…
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