What is the probability that two or more persons' birthday is January 1st? A group of four people all have their birthday in January. What is the probability that at least one of them has their birthday on 1st January? My answer to the question:
Pr (No Birthday on First January for any of the 3 friends): $(\frac{30}{31})^4 $
Pr (At least one birthday on First January): $1-(\frac{30}{31})^4 \approx 0.12 = 12\%$
I am curious to know how can I calculate the probability of two or more persons birthday occur on First January. I think it must be $ 0.12 \times 0.12 \times 0.12 \times 0.12 = 0.00020736 \approx 0.02 \%$ But I am not sure. 
I Appreciate if you can confirm this is right or put me in the right direction.
 A: Note that the probability that two or more people have their birthday's on the first of January is $1$ minus the probability that at most one person does. So it suffices to find the probability that no people do (already found by you), then find the probability exactly one person does, add these, and subtract them from $1$.
The probability that no people have a birthday on the $1$st is $\left( \frac{30}{31}\right)^4$.
The probability that exactly one of the people has a birthday on the first is $$4\left( \frac{1}{31}\right)\left( \frac{30}{31}\right)^3.$$
To see this, note the probability that a particular person has their birthday on the $1$st is $\left( \frac{1}{31}\right)\left( \frac{30}{31}\right)^3$, then multiply by four because we have four people. (We can do this because the events are mutually exclusive.)
The final tally is then
$$1-\left( \frac{30}{31}\right)^4-4\left( \frac{1}{31}\right)\left( \frac{30}{31}\right)^3 .$$
A: $P($2 or more birthdays on Jan 1st$)$ = 1 - $P($no birthdays on Jan 1st$)$ - $P($exactly 1 birthday on Jan 1st$) = 1$ $-$ $(30/31)^4$ - $4\cdot P($friend 1 has a birthday on Jan 1 AND friends 2, 3, 4 do not$)$ $= 1 - (30/31)^4 - 4(1/31)(30/31)^3$.
The multiplication by 4 comes in because there are 4 ways that exactly 1 person can have a Jan 1st birthday.
