#include <bits/stdc++.h>

using namespace std;

#define FOR(i, a, b) for(int i = (a); i < (b); i++)
#define RFOR(i, a, b) for(int i = (a) - 1; i >= (b); i--)
#define SZ(a) int(a.size())
#define ALL(a) a.begin(), a.end()
#define PB push_back
#define MP make_pair
#define F first
#define S second

typedef long long LL;
typedef vector<int> VI;
typedef pair<int, int> PII;
typedef double db;

const int N = 17;
const int T = 107;
const db EPS = 1e-9;

db dp[T][1 << N];
db f[T][N];
int r[N], c[N];
db p[N];

bool hasBit(int mask, int i)
{
	return (mask >> i) & 1;
}

int main()
{
	ios::sync_with_stdio(0);
	cin.tie(0);
	cout << fixed << setprecision(15);
	int n, t;
	cin >> n >> t;
	FOR(i, 0, n)
		cin >> r[i] >> c[i] >> p[i];
	dp[0][0] = 1;
	FOR(i, 0, t)
	{
		FOR(mask, 0, 1 << n)
		{
			db s = 0;
			int cntUnused = 0;
			FOR(k, 0, n)
				if (!hasBit(mask, k))
				{
					s += f[i][k];
					cntUnused++;
				}
			// choose problem k
			FOR(k, 0, n)
			{
				if (hasBit(mask, k))
					continue;
				db probK = s < EPS ? (1.0 / cntUnused) : (f[i][k] / s);
				if (i + r[k] <= t)
				{
					dp[i + r[k]][mask | (1 << k)] += dp[i][mask] * probK * (1 - p[k]);
				}
				if (i + r[k] + c[k] <= t)
				{
					dp[i + r[k] + c[k]][mask | (1 << k)] += dp[i][mask] * probK * p[k];
					f[i + r[k] + c[k]][k] += dp[i][mask] * probK * p[k];
				}
			}
		}
		FOR(k, 0, n)
			f[i + 1][k] += f[i][k];
	}
	FOR(k, 0, n)
		cout << f[t][k] << "\n";
	return 0;
}