About the value of $\mathbb{P}(N(t)=10,N(t/2)=5|N(t/4)=3)$ where $(N(t))$ is a Poisson process

Question

Let $\left \{N(t) \ | t \geq 0 \right \}$ be a poisson process with rate $\lambda > 0$ and let $t > 0$. I need to calculate the following probability \begin{equation} \mathbb{P}(N(t)=10,N(t/2)=5|N(t/4)=3) \end{equation} I came to the conclusion that this probability equals \begin{equation} \mathbb{P}(N(t)-N(t/2)=5)\mathbb{P}(N(t/2)-N(t/4)=2) = e^{-\lambda\frac{t}{2}}\frac{(t\lambda/2)^5}{5!}e^{-\lambda\frac{t}{4}}\frac{(t\lambda/4)^2}{2!} \end{equation} However in the final answer they state that the probability equals \begin{equation} \mathbb{P}(N(t)-N(t/2)=5)\mathbb{P}(N(t/2)-N(t/4)=2) = e^{-\lambda t}\frac{(t\lambda)^5}{5!}e^{-\lambda t}\frac{(t\lambda)^2}{2!} \end{equation} I assumed that because the intervals $(t/4,t/2)$ and $(t/2,t)$ are disjoint the probability of $x$ events in each interval had a $poi(\lambda(t/4))$, $poi(\lambda(t/2))$ distribution. Can someone explain to me where my reasoning is wrong? Thanks in advance!