Calls are received at a company call center according to a Poisson process at the rate of five calls per minute.
(a) Find the probability that no call occurs over a 30-second period.
(b) Find the probability that exactly four calls occur in the 1st minute, and six calls occur in the second minute.
(c) Find the probability that 25 calls are received in the 1st 5 minutes and six of those calls occur in the 1st minute.

My solution:

(b) $P(N_1=4, N_2=6) \\= P(N_1=4)\cdot P(N_2=6) \\= ...$

As far as I understand, the problem talks about two different time spans. So, $(N_1=4)$ and $(N_2=6)$ should be independent.

Another notion could be:

$P(N_1=4, N_2=6) \\= P(N_1=4, N_2 - N_1 = 6) \\= P(N_1=4, N_1=6) \\=P(N_1=4) \cdot P(N_1=6)$

Which one is correct and why?


closed as off-topic by mkt, mdewey, Michael Chernick, Peter Flom Apr 16 at 19:10

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We first need to clear out your notation. $N_t$ is the number of calls by time $t$. If the time unit is taken as minutes, $N_1$ refers to number of calls in the first minute. So, four calls in the first minute can be formulated as $N_1=4$, and six calls in the second minute can be formulated as $N_2-N_1=6$. The question is (b) is to find $P(N_1=4\cap N_2-N_1=6)=P(N_1=4\cap N_2=10)$, not $P(N_1=4\cap N_2=6)$. So, the second line in your second method is the correct question. Number of events occurring in non-overlapping time intervals are independent by the definition of Poisson Process. So, $N_1$ and $N_2-N_1$ are independent (not $N_1$ and $N_2$), which yields:

$$P(N_1=4\cap N_2-N_1=6)=P(N_1=4)P(N_2-N_1=6)$$

Here, we can write $P(N_2-N_1=6)=P(N_1=6)$ because we're interested in $6$ calls occurring in some one-minute interval, which is equivalent to asking for probability of $6$ calls in the first minute. Your final result in the second method is correct, although the way you reach it is wrong due to the reasons I've provided above.


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