Poisson process worded problem (store with 2 products)

Question

A store has customers arrive according to a Poisson process with $\lambda=30$

where $p_1 = P($item $A$ ordered $)= {1\over 12}$

$p_2 = P($item $B$ ordered $)= {1\over 24}$

$p_3 = P($ $A$ and $B$ ordered$)= {e^{10-2t}\over 8} $

$p_4 = 1-p_1-p_2-p_3$

1. What is the probability that the store has 15 orders of item $A$ only over $t\in [10,15]$?

2. same as 1. but for any order with item A

(My only insight is that this is some kind of Poisson thinning process?)

Attempt at 1.

Let $(N_A(t), t\geq 0)$ be the Poisson process of the number of item $A$ ordered, with rate $\lambda_A = \lambda p_1={30\over 12} $

$\displaystyle P(N_A(10,15]=15)=\frac{(5\lambda_A)^{15}e^{-5\lambda_A}}{15!}$

Answer 1

Hint:

If you only count the customers that order item A then you are dealing with a Poisson process with rate $\frac{30}{12}$.

Answer 2

Let $T>0$ and $n$ be a nonnegative integer, and $\lambda$ the rate of the Poisson process, then for 1. we have $$ \mathbb P(N_1(T) = n) = e^{-p_1\lambda T}\frac{(p_1\lambda T)^n}{n!}. $$ In this case $p_1=\frac1{12}$, $T=5$, $n=15$ and $\lambda = 30$, so plugging in these values yields \begin{align} e^{-\frac1{12}\cdot30\cdot5}\frac{\left(\frac1{12}\cdot30\cdot5\right)^{15}}{15!} &= e^{-125}\frac{125^{15}}{15!}\\&=\frac{227373675443232059478759765625}{10461394944}e^{-125}\\ &\approx 1.122898\cdot 10^{-35}. \end{align}

For 2. I do not understand the question. Can you clarify?