MHB Probability of event modelled by poisson happening twice, consecutively

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To find the probability of receiving at least three telephone calls in two successive one-minute intervals modeled by a Poisson distribution with a mean of 3.5, first calculate the probability of receiving at least three calls in one minute. This can be done using the cumulative distribution function of the Poisson distribution. Once the probability for one minute is determined, it can be squared to find the probability for two successive intervals, as the events are independent. Therefore, the final probability is the square of the individual probability for at least three calls in one minute. Understanding the independence of the intervals is key to solving this problem correctly.
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I'm not great at statistics, so I don't know where to start with this problem. It is stated as follows:

The number of telephone calls, T, received each minute can be modeled by a Poisson distribution with a mean of 3.5.

Find the probability that at least three telephone calls are received in each of two successive one-minute intervals.

So, I understand we have T ~ Po(3.5), and using a calculator or formula, I could easily identify the probability of having at least three telephone calls, but I don't understand what to do about that two successive part. Been thinking on this one for a while, but I'm at a loss as to what to do. Can anyone give me any suggestions on how to think about this to arrive at an answer?
 
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You could multiply the probability of atleast 3 successive calls twice i.e P($$\ge$$ 3 phone calls) x P($$\ge$$ 3 phone calls), since our event gets completed only if we consider both of it's elements(in this case, the number of calls in two successive minutes).

This is just my suggestion as I, myself, am certainly not an ace in the area of statistics.
 
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