The probability that more than 3 messages will arrive during a 30-second interval is
.
According to the statement
we have given that the an Exponential distribution with 7 messages arriving in a 10 second period and we have to find the probability that more than 3 messages will arrive during a 30-second interval.
So, For this purpose, we know that the
The probability is the measure of the likelihood of an event to happen. It measures the certainty of the event.
And the given information is that :
3 messages will arrive during a 30-second interval.
Then
Probability = P(X=1) + P(X=2) + P(X=3).Then
The probability become according to the exponential distribution:
![P(X=1)=\dfrac{e^{-0.3 \times 20} (0.3 \times 20)^1}{1!}\\P(X=2)=\dfrac{e^{-0.3 \times 20} (0.3 \times 20)^2}{2!}\\P(X=3)=\dfrac{e^{-0.3 \times 20} (0.3 \times 20)^3}{3!}\\](https://tex.z-dn.net/?f=P%28X%3D1%29%3D%5Cdfrac%7Be%5E%7B-0.3%20%5Ctimes%2020%7D%20%280.3%20%5Ctimes%2020%29%5E1%7D%7B1%21%7D%5C%5CP%28X%3D2%29%3D%5Cdfrac%7Be%5E%7B-0.3%20%5Ctimes%2020%7D%20%280.3%20%5Ctimes%2020%29%5E2%7D%7B2%21%7D%5C%5CP%28X%3D3%29%3D%5Cdfrac%7Be%5E%7B-0.3%20%5Ctimes%2020%7D%20%280.3%20%5Ctimes%2020%29%5E3%7D%7B3%21%7D%5C%5C)
And then substitute the values in it then
![Probability = \dfrac{e^{-0.3 \times 20} (0.3 \times 20)^1}{1!}\ +\dfrac{e^{-0.3 \times 20} (0.3 \times 20)^2}{2!}\ + \dfrac{e^{-0.3 \times 20} (0.3 \times 20)^3}{3!}\\](https://tex.z-dn.net/?f=Probability%20%3D%20%5Cdfrac%7Be%5E%7B-0.3%20%5Ctimes%2020%7D%20%280.3%20%5Ctimes%2020%29%5E1%7D%7B1%21%7D%5C%20%2B%5Cdfrac%7Be%5E%7B-0.3%20%5Ctimes%2020%7D%20%280.3%20%5Ctimes%2020%29%5E2%7D%7B2%21%7D%5C%20%2B%20%5Cdfrac%7Be%5E%7B-0.3%20%5Ctimes%2020%7D%20%280.3%20%5Ctimes%2020%29%5E3%7D%7B3%21%7D%5C%5C)
This is the probability.
So, The probability that more than 3 messages will arrive during a 30-second interval is
.
Learn more about probability here
brainly.com/question/24756209
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