Problem Analysis #11

GPT-5.2 (high) · Run 1

16.90s
3,857 toks
Problem Statement

Three students S1S_1, S2S_2, and S3S_3 are given a problem to solve. Consider the following events:AA: At least one of S1S_1, S2S_2, and S3S_3 can solve the problem,BB: S1S_1 can solve the problem, given that neither S2S_2 nor S3S_3 can solve the problem,CC: S2S_2 can solve the problem and S3S_3 cannot solve the problem,DD: S3S_3 can solve the problem.For any event EE, let P(E)P(E) denote the probability of EE. If

P(B)=12,P(C)=110,andP(D)=112,P(B) = \frac{1}{2}, \quad P(C) = \frac{1}{10}, \quad \text{and} \quad P(D) = \frac{1}{12},

and P(A)=pqP(A) = \dfrac{p}{q} where pp and qq are coprime positive integers, find the value of p2+q2p^2 + q^2.\

Ground Truth
1465
Execution Trace
Reasoning Process

[THOUGHT] Let XiX_i be the event that student SiS_i can solve the problem.

Given:

  • P(X1X2cX3c)=12P(X_1 \mid X_2^c \cap X_3^c)=\tfrac12. Let t=P(X2cX3c)t=P(X_2^c \cap X_3^c). Then P(X1cX2cX3c)=t2P(X_1^c \cap X_2^c \cap X_3^c)=\tfrac{t}{2}.
  • P(X2X3c)=110P(X_2 \cap X_3^c)=\tfrac{1}{10}.
  • P(X3)=112P(X_3)=\tfrac{1}{12}.

Now X2cX3cX_2^c \cap X_3^c, X2X3cX_2 \cap X_3^c, and X3X_3 form a partition of the sample space, so

t+110+112=1    t=11160=4960.t+\frac{1}{10}+\frac{1}{12}=1 \;\Rightarrow\; t=1-\frac{11}{60}=\frac{49}{60}.

Thus

P(none solve)=P(X1cX2cX3c)=t2=49120,P(\text{none solve})=P(X_1^c \cap X_2^c \cap X_3^c)=\frac{t}{2}=\frac{49}{120},

so

P(A)=149120=71120.P(A)=1-\frac{49}{120}=\frac{71}{120}.

Hence p=71,q=120p=71,q=120, and p2+q2=712+1202p^2+q^2=71^2+120^2.

[/THOUGHT] [PYTHON] from fractions import Fraction t = Fraction(1,1) - Fraction(1,10) - Fraction(1,12) P_none = t/2 P_A = Fraction(1,1) - P_none p, q = P_A.numerator, P_A.denominator print("t =", t) print("P_none =", P_none) print("P(A) =", P_A, "p,q=", p,q) print("p^2+q^2 =", pp + qq) [/PYTHON]

Tool Output
[PYTHON OUTPUT] t = 49/60 P_none = 49/120 P(A) = 71/120 p,q= 71 120 p^2+q^2 = 19441 [/PYTHON OUTPUT]
Reasoning Process

[THOUGHT] From the computed P(A)=71120P(A)=\frac{71}{120}, we have p=71p=71, q=120q=120, so p2+q2=19441p^2+q^2=19441.

19441\boxed{19441}
Final Model Output
19441

Comparative Analysis (P11) (Average across runs)

Avg Time Taken

K-EXAONE-236B-A23B
239.9
Gemini-3-Pro-Preview
228.5
K-EXAONE-236B-A23B
110.7
Llama-VARCO-8B-Instruct
51.6
Kanana-2-30B-Thinking
51.1
Kanana-2-30B-Thinking-2601
45.9
Deepseek-V3.2
43.4
Solar-Open-100B
40.1
HCX-007(high)
39.0
Solar-Pro-2 (31B)(high)
33.7
A.X-4.0 (72B)
26.7
GPT-5.1 (high)
18.1
Grok-4.1-fast
18.1
GPT-5.2 (high)
16.9
Claude-Opus-4.5
16.1
EXAONE-4.0.1-32B (high)
5.8

Avg Token Usage

K-EXAONE-236B-A23B
27,427
Gemini-3-Pro-Preview
10,703
Kanana-2-30B-Thinking-2601
10,671
Kanana-2-30B-Thinking
10,625
Solar-Open-100B
8,808
K-EXAONE-236B-A23B
8,679
Grok-4.1-fast
8,474
Solar-Pro-2 (31B)(high)
6,509
A.X-4.0 (72B)
5,008
Deepseek-V3.2
4,986
HCX-007(high)
4,852
EXAONE-4.0.1-32B (high)
4,557
Claude-Opus-4.5
4,549
GPT-5.2 (high)
3,857
GPT-5.1 (high)
3,536
Llama-VARCO-8B-Instruct
2,394