Variation on the birthday problem? I'm not certain if this would classify as a variation of the birthday problem or not but I have an interesting problem.
Say you have a group of 600 people, what is the approximate probability that at least 75 of them were born in September?
The problem further provides these assumptions for part a and b.
a) Months are equally likely to contain birthdays
b) Days are equally likely to be birthdays
The emphasis on approximate is mine. This leads be to believe I could set this up as a binomial variable with $\frac{1}{12}$ probability of birthday being in September for part a and solve using the normal approximation to the binomial.
But for part b, would I just set the probability equal to $\frac{1}{365}$ and multiply the final answer by 30? That somehow doesn't sound right to me.
Any ideas? 
 A: We could use a binomial distribution. Think of this like pulling $600$ people out of thin air, and one by one, you assign them a September birthday, or a non-September birthday, because if months and days  are equally likely to contain birthdays, then the probability of "success" is constant. 
We have that $n$ is the "trials" (600 people) and $p$ is the probability $\displaystyle \frac{30}{365}=0.0829$ (for September).
However, this method is seriously inconvenient, because this is a discrete (whole number) variable and individual values have to be calculated for $X=75$, $X=76$, etc. 
We should use a normal approximation. A normal approximation is good when we have $np>10$ and $n(1-p)>10$.
The standard deviation would be $\sqrt{np(1-p)}=6.72$, and the mean would be $np=49.3$.
The probability that $75$ or more people are born in September corresponds to a z-score of $z=\displaystyle \frac{75-49.3}{6.72}=3.82$. 
Z scores above $3$ are pretty rare!
Using a probability table, we have $P(z>3.82)=\boxed{0.0000667}$, or $0.0066\%$
