# What's the probability that 2 of 4 people were born in the same month

I have this problem to solve:

"Calculate what's the probability that in the four-person family at least 2 people were born in the same month".

I know that I can calculate opposite event in very simple way but I'd like to know what's wrong with my reasoning.

We need to at least 2 people so we can make 3 cases:

1. Two people were born at the same month.
2. Three people were born at the same month.
3. Four people were born at the same month.

So in the first case firstly we choose 2 people of 4, which were born at the same month, then multiple by 12 possible months and then by 11 and 10 i.e. by possible months for respectively third and fourth person. In other cases I do it analogously.

So the result is:
$\binom{4}{2}*12*11*10 + \binom{4}{3}*12*11+\binom{4}{4}*12 = 8640$ But the result should be 8856.

Edit: I've got it. As you said, there are four cases. I've wrote them symbolically.

1. X X Y Z
2. X X Y Y
3. X X X Y
4. X X X X

First, third and fourth case I've described before. In the second case we need to first chose two people then a month for them and then a month for two other people. But in this case we must notice that the results double each other.For example in first step we chose person '1' and '3' and a month '6' for them. For person '2' and '4' we chose month '5'. So we have the result: 6 5 6 5. But we can alson first chose person '2' and '4' and a month '5' for them and for person '1' and '3' chose month '6'. So we have 6 5 6 5 again.

So final result is: $\binom{4}{2}*12*11*10 + \binom{4}{2}*12*11*\frac 12 + \binom{4}{3}*12*11+\binom{4}{4}*12 = 8856$

• Did you consider the case that, for example, two persons are born in January and two in February ? Feb 25, 2015 at 22:42