#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
#define X first
#define Y second
#define ALL(v) v.begin(), v.end()
#define pb push_back
#define SZ(a) ((int)a.size())

double ans[105][17], dp[105][1 << 17];
int r[105], c[105];
double p[105];

int main() {
    ios::sync_with_stdio(0), cin.tie(0);
    int n, t;
    cin >> n >> t;
    for (int i = 0; i < n; ++i)
        cin >> r[i] >> c[i] >> p[i];
    dp[0][0] = 1;
    for (int i = 0; i < t; ++i) {
        for (int j = 0; j < (1 << n); ++j) {
            if (dp[i][j] == 0) continue;
            double total = 0;
            int cnt = 0;
            for (int k = 0; k < n; ++k)
                if (!(j >> k & 1)) {
                    total += ans[i][k];
                    ++cnt;
                }
            for (int k = 0; k < n; ++k)
                if (!(j >> k & 1)) {
                    double prob = 0;
                    if (total == 0) prob = (double)1 / cnt;
                    else if (ans[i][k] > 0) prob = ans[i][k] / total;
                    // solved
                    if (i + r[k] + c[k] <= t) {
                        dp[i + r[k] + c[k]][j | (1 << k)] += dp[i][j] * prob * p[k];
                        ans[i + r[k] + c[k]][k] += dp[i][j] * prob * p[k];
                    }
                    // unsolved
                    if (i + r[k] <= t)
                        dp[i + r[k]][j | (1 << k)] += dp[i][j] * prob * (1 - p[k]);
                }
        }
        for (int j = 0; j < n; ++j)
            ans[i + 1][j] += ans[i][j];
    }
    for (int i = 0; i < n; ++i)
        cout << fixed << setprecision(15) << ans[t][i] << "\n";
}