Question

Assume that the number of messages input to a communication channel in an Exponential distribution with 7 messages arriving in a 10 second period. Compute the probability that more than 3 messages will arrive during a 30-second interval.

Answer 1

The probability that more than 3 messages will arrive during a 30-second interval is $P(X=n)=\dfrac{e^{-0.3 \times 20} (0.3 \times 20)^n}{n!}$.

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:

$P(X=1)=\dfrac{e^{-0.3 \times 20} (0.3 \times 20)^1}{1!}\\P(X=2)=\dfrac{e^{-0.3 \times 20} (0.3 \times 20)^2}{2!}\\P(X=3)=\dfrac{e^{-0.3 \times 20} (0.3 \times 20)^3}{3!}$

And then substitute the values in it then

$Probability = \dfrac{e^{-0.3 \times 20} (0.3 \times 20)^1}{1!}\ +\dfrac{e^{-0.3 \times 20} (0.3 \times 20)^2}{2!}\ + \dfrac{e^{-0.3 \times 20} (0.3 \times 20)^3}{3!}$

This is the probability.

So, The probability that more than 3 messages will arrive during a 30-second interval is $P(X=n)=\dfrac{e^{-0.3 \times 20} (0.3 \times 20)^n}{n!}$.

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