An urn contains 7 red and 11white balls. Draw one ball at random from the urn. Let X=1 if a red ball drawn, and let X =-1 if a white ball is drawn. Give the pmf, mean, and variance of X
Please, help

Question

An urn contains 7 red and 11white balls. Draw one ball at random from the urn. Let X=1 if a red ball drawn, and let X =-1 if a white ball is drawn. Give the pmf, mean, and variance of X
Please, help

loser66 · Answer

Draw 1 ball means n =1
P( X =1) = 7/18 --> p = 7/18
P( X=-1) =11/18
How can I know which is the success drawing to give out the pmf?

loser66 · Answer

If I go to old way, I can define pmf by 
$$f(x) =\begin{cases}7/18~~~if~~x =1\11/18~~~if~~ x =-1\end{cases}$$

loser66 · Answer

But I am supposed to use Bernoulli trials to find pmf

loser66 · Answer

@kirbykirby

kirbykirby · Answer

this looks correct. It doesn't really matter what you define as the success here, like which one is 7/18 or which one is 11/18

loser66 · Answer

But if we define x =1 is correct one, then pmf, mean and variance totally different from X=-1, right?
How can I know which one is correct one to calculate others?

kirbykirby · Answer

No it should would the same, because you should associate the right value of x with its correct probability (rather than matching the x-value with the "success" probability, if you know what I mean o_O)
Say we define the success as drawing a red ball, so p = 7/18 
$$ E(X)=\sum_{x}x*f(x)=(-1)(11/18)+(1)(7/18)$$
if p was defined as 11/18, then you'd just re-associate the -1 with 11/18 and 1 with 7/18

kirbykirby · Answer

Beware that the expected value and Variance won't simply be $E(X)=p$ and $Var(X)=p(1-p)$ since that's when the bernoulli r.v. is defined with X=1 and X=0, not -1, you have to go through the definition

loser66 · Answer

I know, I have a problem with x =1 and x =0, it is quite easy, just apply B (n,p) because it is exactly the definition. I got stuck with this case. That 's why I make question. :)

loser66 · Answer

As above, I can find mean, variance by old way, just take E(x) = sum xf(x), and E(x^2)= x^2 f(x)and manipulate a little bit more.
But it is not the goal of the exercise which is applying Bernoulli :(

kirbykirby · Answer

it seems like this is a bernoulli distribution; 2 outcomes in one trial, with their success and failure probability,  although I am not 100% sure if it's only restricted to when the value the r.v. takes is 0 and 1 (but here you have -1 and 1)

loser66 · Answer

Ok, got you. Thanks for the help. I need you in other one. Please