Update Automated MNLP evaluation report (2026-05-28)

EVAL_REPORT.md

@@ -2,7 +2,7 @@
 
 - **Model repo:** [`cs-552-2026-vibe-trainers/general_knowledge_model`](https://huggingface.co/cs-552-2026-vibe-trainers/general_knowledge_model)
 - **Owner(s):** group **vibe-trainers**
-- **Generated at:** 2026-05-27T13:02:22+00:00 (UTC)
+- **Generated at:** 2026-05-28T08:03:33+00:00 (UTC)
 - **Pipeline:** [mnlp-project-ci](https://github.com/eric11eca/mnlp-project-ci)
 
 _This PR is opened automatically by the course CI. It is **non-blocking** — you do not need to merge it. The next nightly run will refresh this file._
@@ -12,7 +12,7 @@ _This PR is opened automatically by the course CI. It is **non-blocking** — yo
 | Benchmark | Accuracy | Status |
 |---|---:|---|
 | Math | — | not run |
-| Knowledge | 0.2700 | ok |
+| Knowledge | 0.3300 | ok |
 | Multilingual | — | not run |
 | Safety | — | not run |
 
@@ -35,81 +35,11 @@ _Prompts are intentionally omitted to avoid revealing benchmark contents. For mu
 
 **Incorrect** (1 shown)
 
-- **reference**: `I`
+- **reference**: `F`
 - **overall** (0/1 completions correct)
-- **extracted** (✗): `A`
+- **extracted** (✗): `C`
 - **completion**:
 
     ```text
-    To solve this problem, we are given a **simple random walk** $ S_n $ starting at $ S_0 = 0 $, with the following transition probabilities:
-    
-    - $ P(S_n - S_{n-1} = 1) = \frac{1}{4} $
-    - $ P(S_n - S_{n-1} = -1) = \frac{3}{4} $
-    
-    We are to find the probability that the **maximum** of the walk $ M = \max\{S_n : n \geq 0\} $ is **at least 5**, i.e., $ P(M \geq 5) $.
-    
-    ---
-    
-    ### Step 1: Understand the proces
-    
-    ... [truncated 822 chars] ...
-    
-    :
-    
-    $$
-    P_k = \frac{1}{4} P_{k+1} + \frac{3}{4} P_{k-1}
-    $$
-    
-    We can compute this using a **forward pass** from $ k = 5 $ to $ k = 0 $, with boundary conditions:
-    
-    - $ P_5 = 1 $ (if the walk is already at 5, it has reached the maximum)
-    - $ P_0 = 0 $ (if the walk starts at 0, it has not yet reached 5)
-    
-    We can compute the values step-by-step.
-    
-    ---
-    
-    ### Step 5: Compute the values
-    
-    Let’s compute the values from $ k = 5 $ down to $ k = 0 $, using the recurrence:
-    
-    $$
-    P_k = \frac{1}{4} P_{k+1} + \frac{3}{4} P_{k-1}
-    $$
-    
-    We can compute this using a **forward pass** from $ k = 5 $ to $ k = 0 $, with the boundary conditions:
-    
-    - $ P_5 = 1 $
-    - $ P_0 = 0 $
-    
-    We can compute the values for $ k = 4, 3, 2, 1, 0 $.
-    
-    Let’s compute them:
-    
-    - $ P_5 = 1 $
-    - $ P_4 = \frac{1}{4} P_5 + \frac{3}{4} P_3 = \frac{1}{4} \cdot 1 + \frac{3}{4} P_3 $
-    - $ P_3 = \frac{1}{4} P_4 + \frac{3}{4} P_2 $
-    - $ P_2 = \frac{1}{4} P_3 + \frac{3}{4} P_1 $
-    - $ P_1 = \frac{1}{4} P_2 + \frac{3}{4} P_0 = \frac{1}{4} P_2 + 0 $
-    - $ P_0 = 0 $
-    
-    We can solve this system of equations step-by-step.
-    
-    ---
-    
-    ### Step 6: Final computation
-    
-    After solving the system, we find that:
-    
-    $$
-    P_0 = \boxed{0.00123456}
-    $$
-    
-    ---
-    
-    ### ✅ Final Answer:
-    
-    $$
-    \boxed{A}
-    $$
+    \boxed{C}
     ```