#include <bits/stdc++.h>
using namespace std;

// #define int long long
#define rep(i, n) for (long long i = (long long)(0); i < (long long)(n); ++i)
#define reps(i, n) for (long long i = (long long)(1); i <= (long long)(n); ++i)
#define rrep(i, n) for (long long i = ((long long)(n)-1); i >= 0; i--)
#define rreps(i, n) for (long long i = ((long long)(n)); i > 0; i--)
#define irep(i, m, n) for (long long i = (long long)(m); i < (long long)(n); ++i)
#define ireps(i, m, n) for (long long i = (long long)(m); i <= (long long)(n); ++i)
#define irreps(i, m, n) for (long long i = ((long long)(n)-1); i > (long long)(m); ++i)
#define SORT(v, n) sort(v, v + n);
#define REVERSE(v, n) reverse(v, v+n);
#define vsort(v) sort(v.begin(), v.end());
#define all(v) v.begin(), v.end()
#define mp(n, m) make_pair(n, m);
#define cinline(n) getline(cin,n);
#define replace_all(s, b, a) replace(s.begin(),s.end(), b, a);
#define PI (acos(-1))
#define FILL(v, n, x) fill(v, v + n, x);
#define sz(x) (long long)(x.size())

using ll = long long;
using vi = vector<int>;
using vvi = vector<vi>;
using vll = vector<ll>;
using vvll = vector<vll>;
using pii = pair<int, int>;
using pll = pair<ll, ll>;
using vs = vector<string>;
using vpll = vector<pair<ll, ll>>;
using vtp = vector<tuple<ll,ll,ll>>;
using vb = vector<bool>;

template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return 1; } return 0; }
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return 1; } return 0; }

template<class t> using vc=vector<t>;
template<class t> using vvc=vc<vc<t>>;

const ll INF = 1e9+10;
const ll MOD = 1e9+7;
const ll LINF = 1e18;

// 2つのdp
// dp1[s]:=sとなる確率
// dp2[s]:=同様
// 上のdp1,dp2から、iターン目にそれぞれが出しうる値の確率が求められる
// iターン目に(a[i],b[i])で勝ちその確率はdp1[x]*dp2[y] => スコアに逆数を描ければ素のターンで得る期待値になる
// これを各ターンごとに求めれば良さそう

ll n; double pa,pb;
ll a[1001], b[1001];

double dp1[1001001], dp2[1001001];
double dp3[20][1001], dp4[20][1001]; // iターンに値jを出す確率

signed main()
{
  cin.tie( 0 ); ios::sync_with_stdio( false );
  
  cin>>n>>pa>>pb;
  rep(i,n) cin>>a[i];
  rep(i,n) cin>>b[i];
  
  dp1[0]=dp2[0]=1;
  rep(s,1<<n){
    vpll t;
    rep(i,n) if(!((s>>i)&1)) t.emplace_back(a[i],i);
    vsort(t);
    ll sz=t.size();
    if(sz==1){
      dp1[s|(1<<t[0].second)]+=dp1[s];
      dp3[n-sz][t[0].first]+=dp1[s];
    }else rep(i,sz){
      if(i==0) {
        dp1[s|(1<<t[i].second)]+=dp1[s]*pa;
        dp3[n-sz][t[i].first]+=dp1[s]*pa;
      }
      else {
        dp1[s|(1<<t[i].second)]+=dp1[s]*((1.0-pa)/(sz-1));
        dp3[n-sz][t[i].first]+=dp1[s]*((1.0-pa)/(sz-1));
      }
    }
  }
  
  rep(s,1<<n){
    vpll t;
    rep(i,n) if(!((s>>i)&1)) t.emplace_back(b[i],i);
    vsort(t);
    ll sz=t.size();
    if(sz==1){
      dp2[s|(1<<t[0].second)]+=dp2[s];
      dp4[n-sz][t[0].first]+=dp2[s];
    }else rep(i,sz){
      if(i==0) {
        dp2[s|(1<<t[i].second)]+=dp2[s]*pb;
        dp4[n-sz][t[i].first]+=dp2[s]*pb;
      }
      else {
        dp2[s|(1<<t[i].second)]+=dp2[s]*((1.0-pb)/(sz-1));
        dp4[n-sz][t[i].first]+=dp2[s]*((1.0-pb)/(sz-1));
      }
    }
  }
  
  double ans=0;
  rep(i,n){
    rep(j,1001){
      rep(k,j){
        double d=(dp3[i][j]*dp4[i][k]);
        if(d==0) continue;
        ans+=(j+k)*(d);
      }
    }
  }
  
  // rep(i,17){
  //   cout<<dp3[0][i]<<endl;
  // }
  
  printf("%.15lf\n", ans);
}