conditional probability in sum of two Poisson RVs.

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Here we go ..... Serious accidents in a manufacturing factory are modeled by a Poisson distribution with a mean rate of 1.6 per week. What is the probability that in a four week period, there is exactly one week in which there are serious accidents?

I am finding this question not clear for:

a. It does not say how many accidents, can we just take 'at least one accident'? b. can we divide duration into one week and three weeks or three weeks and one week?

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Here we go ..... Serious accidents in a manufacturing factory are modeled by a Poisson distribution with a mean rate of 1.6 per week. What is the probability that in a four week period, there is exactly one week in which there are serious accidents?

I am finding this question not clear for:

a. It does not say how many accidents, can we just take 'at least one accident'?

The event is: exactly three weeks in which there are zero serious accidents, and exactly one which there are non-zero.

b. can we divide duration into one week and three weeks or three weeks and one week?

There are four weeks.   They are four independent periods.   You are counting the number of them which have non-zero serious accidents.


Hint: the count of "successes" in four independent trials (with identical success rate) is a what-is-the-name distributed random variable.