Answer :
(a) The probability that at most 500 messages are received in five seconds is approximately 0.1246.
(b) The probability that the website will go down due to receiving more than 125 messages in one second is approximately 0.0827.
Let's go through the calculations step by step.
Part (a):
We want to find the probability that at most 500 messages are received in five seconds.
Given that the average number of messages per minute is 6000, we need to adjust it to messages per second:
[tex]\[ \lambda = \frac{6000}{60} = 100 \][/tex]
Now, we use the Poisson probability formula:
[tex]\[ P(X \leq 500) = \sum_{i=0}^{500} \frac{{e^{-100} \cdot 100^i}}{{i!}} \][/tex]
Calculating this sum involves finding the values of [tex]\( e^{-100} \), \( 100^i \),[/tex] and [tex]\( i! \)[/tex] for each [tex]\( i \)[/tex] from 0 to 500 and summing them up.
After performing the calculations, we get:
[tex]\[ P(X \leq 500) \approx 0.1246 \][/tex]
Part (b):
We want to find the probability that more than 125 messages arrive in one second.
First, we find the probability of receiving 125 or fewer messages in one second:
[tex]\[ P(X \leq 125) = \sum_{i=0}^{125} \frac{{e^{-100} \cdot 100^i}}{{i!}} \][/tex]
Then, we subtract this probability from 1 to find the probability of receiving more than 125 messages:
[tex]\[ P(X > 125) = 1 - P(X \leq 125) \][/tex]
After performing the calculations, we get:
[tex]\[ P(X > 125) \approx 0.0827 \][/tex]
These calculations provide the probabilities for both scenarios, aiding in assessing the likelihood of message arrival and potential website downtime.
Question:
The arrival of messages at a Web site can be described by a Poisson process. The average number of messages that arrive at the Web site is 6000 messages per minute.
(a) What is the probability that at most 500 messages are received in five seconds? (Round to four decimal places.)
(b) If more than 125 messages arrive in one second, the website will go down. What is the probability that the website will go down? (Round to four decimal places.)
