Question

With an average of 2.5 per day following Poisson distribution,
what is the probability that the next day there is only 1?

What is the probability that there is at least 2 the next
day?

What is the probability that there are at most 2 the next day?

Answer #1

Let X ~ Poisson( = 2.5)

Then, P(X = x) = exp(-2.5) 2.5^{x} / x!

Probability that the next day there is only 1 = P(X = 1)

= exp(-2.5) * 2.5^{1} / 1!

= exp(-2.5) * 2.5

= 0.2052125

Probability that there is at least 2 the next day = P(X 2)

= 1 - P(X < 2)

= 1 - [P(X = 0) + P(X = 1)]

= 1 - [exp(-2.5) * 2.5^{0} / 0! + exp(-2.5) *
2.5^{1} / 1!]

= 1 - [exp(-2.5) + exp(-2.5) * 2.5]

= 1 - (0.082085 + 0.2052125)

= 0.7127025

Probability that there is at most 2 the next day = P(X 2)

= P(X = 0) + P(X = 1) + P(X = 2)

= exp(-2.5) * 2.5^{0} / 0! + exp(-2.5) * 2.5^{1}
/ 1! + exp(-2.5) * 2.5^{2} / 2!

= exp(-2.5) + exp(-2.5) * 2.5 + exp(-2.5) * 2.5^{2} /
2

= 0.082085 + 0.2052125 + 0.2565156

= 0.5438131

= 0.7127025

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