Call centre incoming calls are randomly distributed following poisson distribution with an average rate of 12 per hour. Calculate the probability that for seven one hour periods there will be one hour with no calls, three hours with one call and three hours with 2 or more calls.
I used f(0)={mu^0 (e)^-mu]/0! for no calls and then similarly for one call f(1) to the power of 3 as there are 3 hours and also for the last part but the answer is not the same
Once you know the probability for $0$ calls, $1$ call, and more than one call in an hour, say these are $p_0$, $p_1$, and $p_2$ respectively, you are halfway done.
Then you have to choose one of the seven hours to receive no calls ($7$ ways);
And choose $3$ of the remaining $6$ hours to receive $1$ call (${6\choose 3}$ ways);
That leaves the remaining three hours for more than one call each.
Then your desired probability is $7\cdot{6\choose 3}\cdot p_0\cdot p_1^3\cdot p_2^3$.