TITLE: Are most cubic plane curves over the rationals elliptic?
QUESTION [29 upvotes]: %This is a new version of the original question modified in the light of the answers and comments.
The word 'most' in the title is ambiguous. The following is one way of making it precise.
Question1: (This seems to be open. See Poonen's answer below)
A cubic projective curve over $\mathbb{Q}$ is given by ten relatively prime  integers (the coefficients of its equation after clearing the denominators). Suppose we take a ten dimensional box $[-N,N]^{10}$ and choose points with integer coordinates with respect to the uniform measure and form the equation of the associated cubic curve. Suppose the number of points which give rise to a curve with a rational point is $E(N)$. Then what can we say about $E(N)/(2N+1)^{10}$ as $N\rightarrow \infty$?
Should the limit exist and if it does, should it be one, zero, or some other number?
Another question of interest is:
Question 2: (There is a satisfactory answer to this. See Voloch's response below.)
Are either of the sets {cubics with no rational point} and {cubics with at least one rational point} Zariski dense?

REPLY [12 votes]: Manjul Bhargava has answered this question yesterday : 
A positive proportion of plane cubics fail the Hasse principle
Manjul Bhargava
(Submitted on 5 Feb 2014)
When all ternary cubic forms over $\mathbf{Z}$ are ordered by the heights of their coefficients, we show that a positive proportion of them fail the Hasse principle, i.e., they have a zero over every completion of $\mathbf{Q}$ but no zero over $\mathbf{Q}$. We also show that a positive proportion of all ternary cubic forms over $\mathbf{Z}$ nontrivially satisfy the Hasse principle, i.e., they possess a zero over every completion of $\mathbf{Q}$ and also possess a zero over $\mathbf{Q}$. Analogous results are proven for other genus one models, namely, for equations of the form $z^2=f(x,y)$ where $f$ is a binary quartic form over $\mathbf{Z}$, and for intersections of pairs of quadrics in $\mathbf{P}_3$.﻿