// Arup Guha
// 10/5/2019
// Solution to 2019 NAQ Problem G: Research Productivity Index
// Note: Written in contest, commented afterwards.

import java.util.*;

public class g {
	
	public static void main(String[] args) {
	
		// Read in the list and sort in reverse order.
		// We can greedily consider in this order.
		Scanner stdin = new Scanner(System.in);
		ArrayList<Double> list = new ArrayList<Double>();
		int n = stdin.nextInt();
		for (int i=0; i<n; i++)
			list.add(stdin.nextInt()/100.0);
		Collections.sort(list);
		Collections.reverse(list);
	
		// Do it!
		System.out.println(solve(list));
	}
	
	public static double solve(ArrayList<Double> list) {
		
		// DP array prob[i] is probability of i papers accepted.
		double max = 0;
		double[] prob = new double[1];
		prob[0] = 1;
		
		// Loop through the list, building off the old prob array.
		for (int i=0; i<list.size(); i++) {
			
			// Add event i in - consider both cases 
			// (accept, not accept) as you build off prob.
			double[] newprob = new double[i+2];
			for (int j=0; j<i+1; j++) {
				newprob[j] += prob[j]*(1-list.get(i));
				newprob[j+1] += prob[j]*list.get(i);
			}
			
			// Update stuff.
			double tmp = getExp(newprob);
			max = Math.max(max, tmp);
			
			// Reassign this for next iteration.
			prob = newprob;
		}
		
		return max;
	}
	
	// Calculates the desired formula
	public static double getExp(double[] p) {
		int size = p.length-1;
		double res = 0;
		for (int i=1; i<=size; i++) {
			res += (p[i]*Math.pow(i, 1.0*i/size));
		}
		return res;
	}
}