A claims process has one claim every ten hours. This is modeled as a Poisson process with $\displaystyle \lambda = 0.1$

What is the probability that the 3rd claim will be processed in the 5th hour?

How can I get the correct solution using the Erlang PDF function? I tried:

$\displaystyle \int_4^5{\frac{0.1^3 \cdot y^{2} \cdot e^{-0.1y}}{2!} \, dy} \approx 0.0065$

My simpler solution using the Poisson probability mass function is this: The probability of the third claim being processed during the fifth hour is the probability of exactly two claims in four hours combined with the independent probability of at least one in the next hour:

$\displaystyle p_X(2; 0.1 \cdot 4) \cdot (1 - p_X(0; 0.1 \cdot 1))$

$\displaystyle = \frac{0.4^2 \cdot e^{-0.4}}{2!} \cdot \left(1 - \frac{0.1^0 \cdot e^{-0.1}}{0!}\right) \approx 0.0051$

Are both of these approaches correct? Why do these two approaches not match up?