Brice is eating bowls of rice. He takes a random amount of time $t_1 \in (0,1)$ minutes to consume his first bowl, and every bowl thereafter takes $t_n = t_{n-1} + r_n$ minutes, where $t_{n-1}$ is the time it took him to eat his previous bowl and $r_n \in (0,1)$ is chosen uniformly and randomly. The probability that it takes Brice at least 12 minutes to eat 5 bowls of rice can be expressed as simplified fraction $\tfrac{m}{n}$. Compute $m+n$.
1. **Understanding the Problem:**
   Brice takes a random amount of time \( t_1 \in (0,1) \) minutes to consume his first bowl, and every subsequent bowl takes \( t_n = t_{n-1} + r_n \) minutes, where \( r_n \in (0,1) \) is chosen uniformly and randomly. We need to find the probability that it takes Brice at least 12 minutes to eat 5 bowls of rice.

2. **Reformulating the Problem:**
   Let \( X_i \) be the time taken to eat the \( i \)-th bowl of rice. Then, \( X_i \) are independent and uniformly distributed random variables in \( (0,1) \). We need to find the probability that the sum of these times is at least 12 minutes:
   \[
   P\left(\sum_{i=1}^{5} X_i \geq 12\right)
   \]

3. **Complementary Probability:**
   Instead of directly calculating the probability that the sum is at least 12, we can use the complementary probability:
   \[
   P\left(\sum_{i=1}^{5} X_i \geq 12\right) = 1 - P\left(\sum_{i=1}^{5} X_i < 12\right)
   \]

4. **Sum of Uniform Random Variables:**
   The sum of 5 independent uniform random variables \( X_i \) on \( (0,1) \) follows an Irwin-Hall distribution. The probability density function of the sum \( S = \sum_{i=1}^{5} X_i \) is given by:
   \[
   f_S(s) = \frac{1}{4!} \sum_{k=0}^{5} (-1)^k \binom{5}{k} (s-k)^4 \text{ for } k \leq s \leq k+1
   \]

5. **Calculating the Probability:**
   We need to find \( P(S < 12) \). However, since the maximum possible value of \( S \) is 5 (since each \( X_i \) is at most 1), \( P(S < 12) = 1 \). Therefore:
   \[
   P\left(\sum_{i=1}^{5} X_i \geq 12\right) = 1 - 1 = 0
   \]

6. **Conclusion:**
   Since the probability that the sum of 5 uniform random variables on \( (0,1) \) is at least 12 is 0, the simplified fraction is \( \frac{0}{1} \).

The final answer is \( \boxed{ 0 + 1 = 1 } \)