Assuming 30-day months, given 10 people in a room. What are the chances that 5 or more people are all born on the same day#? (i.e., 5 born on the 28th, or 5 born on the 6th, etc)
(EDIT: changed from chances of 5 to chances of 5 or more)
I have tried two answers so far.
In the first, you pick any person, and see what the chances are of the other 9, then 8, etc to match the first. This seems to be 10 * 9/30 * 8/30 * 7/30 * 6/30.
In the second, I suppose you could calculate the chances of 5 of the 10 having a birthday on day 1 + the chances of 5 having a birthday on day 2, etc.
These answers seem quite different. What do you all think?
Best Answer
The probability that at least five people were born on the first of the month (assuming uniform distribution) is
$$ p_1 = \sum_{k=5}^{10} \binom{10}{k} \left(\frac{1}{30}\right)^k \left(\frac{29}{30}\right)^{10-k} = \frac{886717229}{98415000000000} \doteq 0.0000090100 $$
The probability that at least five people were born on any of the thirty days of the month is almost
$$ 30p_1 = \frac{886717229}{3280500000000} \doteq 0.00027030 $$
However, this double-counts those cases where two sets of five people are born on two different days of the month. That happens with probability
$$ p_d = \frac{\binom{30}{2}\binom{10}{5}}{30^{10}} = \frac{203}{1093500000000} \doteq 0.00000000018564 $$
So the final probability is
$$ 30p_1-p_d = \frac{44335831}{164025000000} \doteq 0.00027030 $$
There is a difference made by $p_d$, but it's too small to show up in five significant digits.
ETA: If you want the probability that exactly five are born on the same day, then we'd set
$$ p_1 = \binom{10}{5} \frac{29^5}{30^{10}} = \frac{143578043}{16402500000000} \doteq 0.0000087534 $$
and then proceed as before.