probability and urn 
An urn contains $6$ black balls numbered from $1$ through $6$ and $8$ white balls numbered from $1$ through $8$, and $3$ green balls numbered from $1$ through $3$. If a ball is drawn at random and it is numbered $3$, what is the probability that the ball is green.

My thoughts:
$$\mathbb{P}(" \text{ green } " | "3")=\mathbb{P}_{"3"}("green")=\dfrac{\mathbb{P}(" \text{ green } "\ \cap\ "3")}{\mathbb{P}("3")}=  $$


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*$\mathbb{P}("3")=\dfrac{3}{17}$

*$\mathbb{P}( " \text{ green } "\ \cap\ "3" )=\dfrac{1}{17}$


$$\mathbb{P}(" \text{ green } "/ "3")=\dfrac{1}{3}$$


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*Is my proof correct i'm interested in many way to solve it


Given that a $3$ has been drawn,
the restricted equiprobable sample space is black $3$, white $3$, and green $3$
Thus P[green $3$] $=\dfrac13$
