Odds at least 1 person is born in January? There are 100 people in a room.
What are the odds at least 1 person is born in January?
What is the best way to calculate this?
I used Binomial Distribution.
$E(x) = np = 100 \frac{1}{12} = 8.3$
$std.dev. = \sqrt{npq} = 2.76$
$Z(\text{1 person}) = (1 - 8.3) / 2.76 = -2.64$
$p = P(Z<-2.64) = .004$ or $.4%$ odds that less than 1 person is born in Jan
Can a confidence interval calc apply here?  If so, how?
(With 95% certainty, the number of people born in Jan is b/w x & y)
 A: Just to add few comments to Thylacoleo excellent answer. From the CDC’s National Vital Statistics System recent report, we can estimate the probability of being born on January 1st with a bit more precision. 
Month
From page 50 of the report, we the probability of being born in a certain month:
#R code
#Month totals - page 50
m_totals = c(329803, 307248, 336920, 330106, 346754, 337425, 
             364226, 360103, 359644, 354048, 320094, 343579)

m_probs = m_totals/sum(m_totals)

This gives the probability of being born in January as 0.0806 which is a bit lower that 31/365=0.085. For info, Feb and Nov have probabilities around 0.076(ish). 
Day of the week
For fun, (and since it was on the next page), from page 51 of the report we get the probability of being born on a certain day:
#Starting on a Sunday - page 51
> d_totals = c(7563, 11733, 13001, 12598, 12514, 12396, 8605)
> d_probs = d_totals/sum(d_totals)
> round(d_probs,4)
[1] 0.0965 0.1496 0.1658 0.1607 0.1596 0.1581 0.1097

Looks like Sunday is a popular day!
Notes


*

*This page pointed me in the direction of the CDC report.

A: I will shamelessly propose using simulation (in R) to answer your questions.
rbinom(n = 100000, size = 100, prob = 1/12) draws 100 trials from the binomial distribution 100,000 times and returns a vector of the number of 'successes' (i.e., January births, using the naive 1/12 probability).
You should be able to approximate whatever you want from the results.
A: Do you mean odds or probability?
$$p(\text{at least 1 person born in January}) = 1 - p(\text{no one born in January})$$
Assuming Independence:
\begin{align}
p(\text{no one born in January}) &= P(\text{person 1 not born in Jan}) \\
                                &\cdot P(\text{person 2 not born in Jan})\\
                                &\cdot \dotsm P(\text{person 100 not born in Jan})
\end{align}
Assume $P(\text{not born in Jan}) = \frac{365-31}{365}$
$$p(\text{no one born in January}) = \frac{334}{365}^{100} = 0.00013975
$$
So, $p(\text{at least 1 person born in January})  = 0.99986025$
This is a binomial calculation.
A few points.

*

*Your question is about probability it is not about statistics, so it is not meaningful to ask for a confidence interval on a probability calculation. A confidence interval is a measure of uncertainty about a parameter estimate. By framing the question as you have you enable a direct calculation of the parameter value (p = 334/365).

*Not withstanding comment (1) you have attempted to use a normal approximation to the binomial distribution for your calculation. This will produce a serious error when p is close to 0 or 1 (as here) and is not necessary as exact binomial solutions are trivial in this case.

*Your calculation of the mean or expected value (and Std. dev.) of the binomial distribution is correct. We can use this and the binomial distribution to estimate the probability of seeing a range of results.  I will use p(born Jan) = 31/365 ~ 0.085.

We expect to see 8.5 people born in January in a class of 100 people. Of course this is impossible. But the probability of seeing 9 people born in January is:
$$\begin{align} 
p(K = k|n = 100)&= \binom{n}{k} * p^k*(1-p)^{(n-k)} \\\\
&= \binom{100}{9} * (0.085)^9(1 - 0.085)^{(100-9)}  \\\\
&= 1.902*10^{12}*(2.316*10^{-10})*(0.00030855)  \\\\
&= 0.13591726  \\\\
\end{align}$$
Where $\binom{n}{k}$ is the binomial coefficient.
Similarly the probability of seeing 8 people born in January is: 0.14315621. These two outcomes cover nearly 28% of the probability.
We can construct an approximately symmetrical bound on the mean of 0.085 by using the cumulative binomial distribution functions. I'll use one in Stata, but I think Excel has them. It turns out the probability of seeing 3 or fewer January birthdays is 2.53% and the probability of seeing 15 or more January birthdays is 2.19%. So the probability of seeing 4, 5, 6, 7, 8, 9, 10, 11, 12, 13 or 14 birthdays is 95.3% which is a reasonable estimate of a 95% "coverage probability", although coverage probability has a slightly different meaning in statistics.
