1. ## Question Help!

There are 12 students. Compute the probability that at least 3 students are born in the same month.

Am I right with:

${12 \choose 3} \times (\frac{1}{12})^3$

OR

$(\frac{1}{3} \times \frac{1}{12}) + (\frac{1}{3} \times \frac{1}{12}) + (\frac{1}{3} \times \frac{1}{12})$

2. Originally Posted by zepher
There are 12 students. Compute the probability that at least 3 students are born in the same month.

Am I right with:

${12 \choose 3} \times (\frac{1}{12})^3$
You have counted the probability of exactly 3 people born in the same month.
A better way is to solve the problem the other way.

Pr(at least 3 students are born in the same month) = 1 - Pr(exactly 2 people are born in the same month)

3. Thanks!
But actually, why does Pr(at least 3 students are born in the same month) = 1 - Pr(exactly 2 people are born in the same month) ?

1 - Pr(exactly 2 people are born in the same month) can still have other cases besides having at least 3 students born in the same month right?

4. Please look at the webpage: Birthday Problem -- from Wolfram MathWorld
You will see that this is far from an easy problem.
P.S. For this problem n=12 & d=12.

5. I guess I should reduce the numbers a little.
Say the question now is:
There are 5 students. Compute the probability that AT LEAST 2 students are born in the same month.

I will take:
Pr(AT LEAST 2 students born in same month) = 1 - Pr(NONE of them born in same month)

which is

1 - (12_P_5 / 12^5)

Please tell me I am correct =P