Given the births of babies follow Poisson process with $\lambda = 5$ per day. Assume a hospital handling these births only have 7 beds for moms. Find the probability that the number of boys equals number of girls on a given day, given that $P(\text{baby = girl})\ = P(\text{baby = boy} = 0.5$.
My attempt: Let $Y = \text{number of babies who are boys}$. Then number of babies who are boys = number of babies who are girls if the total number of births are even. Let $X = \text{total number of births}$ on a given day. So $X\leq 7$, and $X$ is even, which means $X = \left\{0,2,4,6\right\}$. Thus, $\text{number of boys} = \text{number of girls} = \left\{0,1,2,3\right\}$.
Now using conditional probabilities, we get: $P(\text{number of boys} = \text{number of girls} = 0) = P(X=0) = e^{-5},
$P(\text{number of boys} = \text{number of girls} = 1) = P(X=2|Y=1)P(Y=1) = e^{-5}5^2/2!\times 0.5^2\times {2\choose 1}$
$P(\text{number of boys} = \text{number of girls} = 2) = P(X=4|Y=2)P(Y=2) = e^{-5}5^4/4!\times 0.5^4\times {4\choose 2}$
$P(\text{number of boys} = \text{number of girls} = 3) = P(X=6|Y=3)P(Y=3) = e^{-5}5^6/6!\times 0.5^6\times {6\choose 3}$
Thus, the $P(\text{number of boys} = \text{number of girls}) = e^{-5}(1+\frac{5^2\times 2}{2!}+\frac{5^4\times {4\choose 2}}{4!} + \frac{5^6\times {6\choose 3}}{6!})\approx 0.1603$
My question: Could someone help review my solution above to see if it's correct?