Probability that a random graph will remain planar after adding an edge According to this answer, a random graph on $n$ vertices is a graph which has each of the $n\choose2$ edges independently with probability $1/2$ each. The probability of at most $3n-6$ edges (which is a necessary condition for planarity) is: $$2^{-n(n-1)/2}\sum_{k=0}^{3n-6}{n(n-1)/2\choose k}$$ 

Therefore, the plot for this equation is:




  If I choose a random graph with 10 vertices, what is the probability
  that it will remain planar after adding an edge between two randomly
  chosen vertices?


To be more clear:

Consider event $P_{1}$, the event of choosing a planar graph $G_{1}$ in the universe $U_{10}$ of graphs with 10 vertices.

Consider a subset $S$ of the universe of graphs with 11 vertices $U_{11}$ conditionally defined by the graph $G_{1}$ we got in event $P_{1}$ in this way: $S$ is formed by all possible graphs that can result from adding one edge to $G_{1}$.

And then finally consider event $P_{2}$, the event of choosing a planar graph in the universe $S$ defined in the previous step.

Therefore I want to know what is the probability of event $P_{2}$.  

Possible solution:

I followed D.W.'s advice below and wrote a small simulation. According to the code, the probability of choosing a planar graph of 10 vertices is 0.0915 ( 9,15% ) and the probability of a given planar graph of 10 vertices remaining planar after adding one random edge is 0.6140 ( 61,40% ).

#!/usr/bin/python3
import planarity
from itertools import combinations
import random
from multiprocessing import Process, Manager, Pool, cpu_count
import json

nodes = [str(x) for x in range(10)]
edges = list(combinations(nodes,2))
tests = 10**6
filename = 'results.json'
runs = 100

def make_test():
    while 1:
        chosen_edges = list(filter(lambda x: x != None, [x if random.choice([0,1]) else None for x in edges]))
        if planarity.is_planar(chosen_edges):
            break
    extra_edge = random.choice( list(set(edges) - set(chosen_edges)) )
    chosen_edges += [extra_edge]
    return planarity.is_planar(chosen_edges)

if __name__ == '__main__':

    for i in range(runs):
        with Pool(processes=cpu_count()) as pool:
            results = [ pool.apply_async(make_test, ()) for i in range(tests) ]
            lst = [ res.get(timeout=1) for res in results ]
        pool.close()
        pool.join()

        n = len(lst)
        p = len(list(filter(lambda x: x != False, lst)))
        prob = ( p / n )
        print(prob)

        try:
            with open(filename,'r') as f:
                file_lst = json.load(f)
        except:
            open(filename,'w+').close()
            file_lst = []
        file_lst += [prob]
        with open(filename,'w') as f:
            json.dump(file_lst,f)
Let $S$ be the set of planar graphs on $n$ vertices, and let $E(S,\overline{S})$ be the set of edges between $S$ and its complement, that is, the set of pairs $(G,e)$ where $G \in S$ and $G \cup \{e\} \notin S$. The edge-isoperimetric inequality states that
$$
\frac{|E(S,\overline{S})|}{|S|} \geq \binom{n}{2} - \log_2 |S|.
$$
It follows that the probability that you're after is at most
$$
\frac{\log_2 |S|}{\binom{n}{2}-(3n-6)}.
$$
It is known that $\log_2 |S| = O(n \log n)$ (see for example Gimenez and Noy, The number of planar graphs and properties of random
planar graphs, and so we conclude that your probability is at most $O\left(\frac{\log n}{n}\right)$.

On the other hand, the same paper shows that with constant probability, a random planar graph has an isolated vertex. When this happens, there is $\Omega\left(\frac{1}{n}\right)$ probability to add an edge adjacent to it, leaving the graph planar. Hence your probability is at least $\Omega\left(\frac{1}{n}\right)$.