# Probability question

• Aug 19th 2013, 05:26 PM
kepler
Probability question
Hi,

I made a survey where n people answered several questions; one of the questions was "What is your sign?"
A person can be of one of the 12 signs of the zodiac.
My question is: what is the probability of x people of n be of one specific sign?

Kind regards,

Kepler
• Aug 19th 2013, 05:49 PM
Plato
Re: Probability question
Quote:

Originally Posted by kepler
I made a survey where n people answered several questions; one of the questions was "What is your sign?"
A person can be of one of the 12 signs of the zodiac.
My question is: what is the probability of x people of n be of one specific sign?

You must assume that this is binomial probability.
$\displaystyle \mathcal{P}(X=k)=\binom{N}{k}(p)^k(1-p)^{N-k}$ which means there are $\displaystyle k$ happenings in $\displaystyle N$ tries and $\displaystyle p$ is the probability is occurs.

So in your query, $\displaystyle N$ is the number of people you survey and $\displaystyle p=\tfrac{1}{12}~.$
• Aug 19th 2013, 06:16 PM
kepler
Re: Probability question
Quote:

Originally Posted by Plato
You must assume that this is binomial probability.
$\displaystyle \mathcal{P}(X=k)=\binom{N}{k}(p)^k(1-p)^{N-k}$ which means there are $\displaystyle k$ happenings in $\displaystyle N$ tries and $\displaystyle p$ is the probability is occurs.

So in your query, $\displaystyle N$ is the number of people you survey and $\displaystyle p=\tfrac{1}{12}~.$

Hi,

So, it's:

k!/(N!*(N-k)!)*(p^k)*((1-p)^(N-k))

?

Kind regards,

Kepler
• Aug 19th 2013, 06:33 PM
kepler
Re: Probability question
Sorry...I mean:

N!/(k!*(N-k)!)*(p^k)*((1-p)^(N-k))

Kepler