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A laboratory observes an average of five unaclaimed results in a day, what is the probability that it will observe 3 unaclaimed results?

User Bkillnest
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The average number of unclaimed results in a day is 5, which means that the Poisson parameter λ = 5. The Poisson distribution is used to model the probability of a given number of events occurring in a fixed interval of time or space, assuming that these events occur independently of each other and at a constant rate.

The probability of observing 3 unclaimed results in a day can be calculated using the Poisson probability mass function:

P(X = 3) = (e^(-λ) * λ^x) / x!

where X is the number of unclaimed results in a day, λ = 5, and x = 3.

Plugging in these values, we get:

P(X = 3) = (e^(-5) * 5^3) / 3! ≈ 0.140

Therefore, the probability that the laboratory will observe 3 unclaimed results in a day is approximately 0.140 or 14.0%.
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