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?

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    $\begingroup$ I think the problem is unclear. Under the poisson process, $8$ births is a real possibility ($.065$ or so). However, the problem appears to rule it out. Are we to assume that if the distribution yields a number greater than $7$, we simply assume $7$ births? If so, then your calculation looks ok to me. $\endgroup$ – lulu Sep 14 '16 at 19:22
  • $\begingroup$ @lulu: thanks for your comment. It's just due to the constraint on hospital's capacity, so we just asdume the case $8$ or more cannot occur. The calculation is the same but we would get the infinite series at the end $\endgroup$ – user177196 Sep 14 '16 at 20:45
  • $\begingroup$ Sure, as I say though, that's not entirely clear in the statement of the problem. For example, you get a different answer if, given the same underlying process (but no restriction on bed number), you say "Given that, on a certain night, you observe fewer than $8$ births, what is the probability that #girls = #boys". I agree with you that yours is the more natural reading, but I often find that probability problems are confusing precisely because the assumptions aren't made explicit. $\endgroup$ – lulu Sep 14 '16 at 21:03

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