Poisson Process with Random Variables

Question

I am dealing with the following problem. Problem Definitions

Problem itself

I converted P( N2 + 2N5 = 5 ) to P( N2 + 2(N5 - N2 + N2 ) = 5 ), which is equal to P( 3N2 + 2(N5 - N2) = 5 ).

Since N2 is Poisson (2λ), 3N2 is Poisson(6λ) ( I think the problem lies here, since N2 's are not independent, their summation is not Poisson ~(6λ) but I am not sure.)

Since N5 - N2 is Poisson (3λ), 2(N5 - N2) is Poisson(6λ)

So in total 3N2 + 2(N5 - N2) = T ( call it T) is Poisson (12λ)

Therefore I think P ( T = 5) , where T is Poisson ~(12λ) should gave the correct answer, however it is not. Could you please tell where am I making a mistake ?

Answer 1

$3N_2$ does not have Poisson distribution since it attains only values which are multiples of $3$.

Note that $3k+2j=5$ ($j,k\in \{0,1,2...\}$) is possible only with $k=j=1$. Hence, $P(3N_2+2(N_5-N_2)=5)=P(N_2=1)P(N_5-N_2=1)=(2\lambda e^{-2\lambda}) ( 3\lambda e^{-3\lambda}) =6\lambda^{2}e^{-5\lambda}$.