#include <bits/stdc++.h>
#pragma GCC optimize("Ofast")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
using namespace std;
int n, T;
double dp[5005][5005];
double pw[5005];
double p[5005];
double sum[5005];
int t[5005];
inline double power(double x, int y) {
  double ret = 1;
  while (y) {
    if (y & 1) ret *= x;
    x *= x;
    y >>= 1;
  }
  return ret;
}
int main() {
  cin >> n >> T;
  for (int i = 1; i <= n; i++) scanf("%lf%d", &p[i], &t[i]), p[i] /= 100.0;
  dp[0][0] = 1;
  register double ans = 0;
  for (register int i = 0; i < n; ++i) {
    memset(sum, 0, sizeof(sum));
    double prob = power(1 - p[i + 1], t[i + 1]), prob2 = prob;
    for (register int j = 0; j <= T; ++j) {
      register int ed = min(T, j + t[i + 1]);
      if (T < j + t[i + 1]) prob2 /= (1 - p[i + 1]);
      if (dp[i][j] < 3e-10) continue;
      sum[j + 1] += dp[i][j];
      sum[ed + 1] -= dp[i][j] * prob2;
      if (j + t[i + 1] <= T) dp[i + 1][j + t[i + 1]] += dp[i][j] * prob;
    }
    for (register int j = 0; j <= T; ++j) {
      if (j) sum[j] += sum[j - 1] * (1 - p[i + 1]);
      dp[i + 1][j] += sum[j] * p[i + 1];
      ans += dp[i + 1][j];
    }
  }
  printf("%.12lf\n", ans);
  return 0;
}
