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let y1, y2, . . . , yk be k independent observed random variables, each having a poisson distribution given by

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Final answer:

The question involves the Poisson distribution which models the probability of a number of events in a fixed interval. The Exponential distribution is related, where the time between events defines the event rate for the Poisson distribution. Probabilities are calculated using the Poisson probability mass function.

Step-by-step explanation:

The student's question pertains to the Poisson distribution, which is a probability distribution used to model the number of events that occur in a fixed interval of time or space, under the condition that these events occur with a known average rate and independently from each other. If y1, y2, ..., yk represent k independent observed random variables each following a Poisson distribution, then for each variable, the probability of observing exactly k events is given by P(X = k) = λ^k * e^(-λ) / k!, where λ is the average rate (mean) of occurrence of events per unit interval, and k! is the factorial of k.

The Exponential distribution is closely related to the Poisson distribution, with the connection being that if the time between events follows an Exponential distribution with mean μ, then the number of events per unit time follows a Poisson distribution with mean λ = 1/μ. When applying this in practice, one can use statistical software or calculators with built-in functions like 'poissonpdf' or 'poissoncdf' to compute the exact probabilities.

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