Message arrives at an electronic message center at random times, with an average of 9 messages per hour. What is the probability of receiving at least four messages during the next hour? What is the probability of receiving at most three messages during the next hour?

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Message arrives at an electronic message center at random times, with an average of 9 messages per hour.  What is the probability of receiving at least four messages during the next hour? What is the probability of receiving at most three messages during the next hour?

Suresh Chand · Accepted Answer

let, x = no of messages arriving per hour Given, Average no. of messages per hour (λ) = 9 we know, p(x) = (frac{e^{-λ} . λ^x}{x!}) Solution a): Probability of receiving at least four messages during the next how, p(x ≥ 4) = 1 - p(x ≤ 4) = 1 - {P(x = 0) + p(x = 1) + p(x = 2) + p(x = 3)} = (1 - left { frac{e^{-9}.9^0}{0!} + frac{e^{-9}.9^1}{1!} + frac{e^{-9}.9^2}{2!} + frac{e^{-9}.9^3}{3!} right }) = 1 - {0.00123 + 0.0011 + 0.0149 + 0.00499} = 0.9778 Solution b): Probability of receiving at least messages during the next hour p(x ≤ 3) = p(x = 0) + p(x = 1) + p(x = 2) + p(x = 3) = 0.00123 + 0.0011 + 0.0149 + 0.00499 = 0.0222

Message arrives at an electronic message center at random times, with an average of 9 messages per hour.

What is the probability of receiving at least four messages during the next hour?

What is the probability of receiving at most three messages during the next hour?

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Message arrives at an electronic message center at random times, with an average of 9 messages per hour. What is the probability of receiving at least four messages during the next hour? What is the probability of receiving at most three messages during the next hour?

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Message arrives at an electronic message center at random times, with an average of 9 messages per hour.

What is the probability of receiving at least four messages during the next hour?

What is the probability of receiving at most three messages during the next hour?