Problem Analysis #probability_29

Motif 3 · Run 1

Python ● execution succeeded
Correct
67.09s
20,906 toks

Infron · Motif 3 · 46문항 × 3회 · Python TAR. 131,072 output tokens · 기본 1,800초 · 6+1 turns. ACC는 전체 138회, Pass@1은 각 문항의 첫 유효 시도, Pass@3는 3회 내 정답으로 계산합니다.

Problem Statement

6 이하의 자연수 a에 대하여 한 개의 주사위와 한 개의 동전을 사용하여 다음 시행을 한다. 주사위를 한 번 던져 나온 눈의 수가 a보다 작거나 같으면 동전을 5번 던져 앞면이 나온 횟수를 기록하고, 나온 눈의 수가 a보다 크면 동전을 3번 던져 앞면이 나온 횟수를 기록한다. 이 시행을 19200번 반복하여 기록한 수가 3인 횟수를 확률변수 X라 하자. E(X)=4800E(X)=4800일 때, P(X≤4800+30a)P(X\le4800+30a)의 값을 오른쪽 표준정규분포표를 이용하여 구한 값이 k이다. 1000×k1000\times k의 값을 구하시오. 표준정규분포표: z | P(0≤Z≤z) 0.5 | 0.191 1.0 | 0.341 1.5 | 0.433 2.0 | 0.477 2.5 | 0.494 3.0 | 0.499

최종 정수 답을 \boxed{} 안에 쓰세요.

Ground Truth
977
Recorded model response / solution
Reasoning Process

Recorded Python call:

import math

# per-trial probability of recording 3
def p(a):
    return (a/6)*(math.comb(5,3)/2**5) + ((6-a)/6)*(math.comb(3,3)/2**3)

# solve p(a)=1/4 for a in 1..6
candidates=[a for a in range(1,7) if abs(p(a)-1/4)<1e-12]
candidates

Tool Output
Execution succeeded
Reasoning Process

Recorded Python call:

import math

def p(a):
    return (a/6)*(math.comb(5,3)/2**5) + ((6-a)/6)*(math.comb(3,3)/2**3)

print([a for a in range(1,7) if abs(p(a)-1/4)<1e-12])
print([p(a) for a in range(1,7)])

Tool Output
Execution succeeded [4] [0.15625, 0.1875, 0.21875, 0.24999999999999997, 0.28125, 0.3125]
Reasoning Process

Recorded Python call:

import math

a=4
n=19200
p=1/4
mu=n*p
sigma=math.sqrt(n*p*(1-p))
threshold=4800+30*a
z=(threshold-mu)/sigma
print(mu, sigma, threshold, z)

# table value at z=2.0: P(0<=Z<=z)
table={0.5:0.191,1.0:0.341,1.5:0.433,2.0:0.477,2.5:0.494,3.0:0.499}
z_table=2.0
k=0.5+table[z_table]
print(k, 1000*k)

# normal approx with continuity correction
z_cc=(threshold+0.5-mu)/sigma
print("z_cc", z_cc, "Phi", 0.5+ (0.5*(1+math.erf(z_cc/math.sqrt(2)))) )

Tool Output
Execution succeeded 4800.0 60.0 4920 2.0 0.977 977.0 z_cc 2.0083333333333333 Phi 1.4776960590006878
Reasoning Process

Recorded Python call:

z_cc=2.0083333333333333
phi_cc=0.5*(1+math.erf(z_cc/math.sqrt(2)))
print("Phi cc", phi_cc, 1000*phi_cc)
phi_no=0.5*(1+math.erf(2/math.sqrt(2)))
print("Phi no cc", phi_no, 1000*phi_no)

Tool Output
Execution failed name 'math' is not defined
Reasoning Process

Recorded Python call:

import math
z_cc=2.0083333333333333
phi_cc=0.5*(1+math.erf(z_cc/math.sqrt(2)))
print("Phi cc", phi_cc, 1000*phi_cc)
phi_no=0.5*(1+math.erf(2/math.sqrt(2)))
print("Phi no cc", phi_no, 1000*phi_no)

Tool Output
Execution succeeded Phi cc 0.9776960590006879 977.6960590006879 Phi no cc 0.9772498680518208 977.2498680518208
Reasoning Process
pa=a6⋅(53)25+6−a6⋅(33)23=a6⋅1032+6−a6⋅18=a+432.p_a=\frac a6\cdot\frac{\binom53}{2^5}+\frac{6-a}{6}\cdot\frac{\binom33}{2^3} =\frac a6\cdot\frac{10}{32}+\frac{6-a}{6}\cdot\frac18 =\frac{a+4}{32}.

Since E(X)=19200pa=4800E(X)=19200p_a=4800, we need pa=14p_a=\frac14, so

a+432=14⇒a=4.\frac{a+4}{32}=\frac14 \Rightarrow a=4.

Then X∼Binomial(19200,14)X\sim\mathrm{Binomial}(19200,\frac14), with

μ=4800,σ2=19200⋅14⋅34=3600,σ=60.\mu=4800,\qquad \sigma^2=19200\cdot\frac14\cdot\frac34=3600,\qquad \sigma=60.

The threshold is

4800+30a=4800+120=4920=4800+2⋅60.4800+30a=4800+120=4920=4800+2\cdot60.

Using the normal approximation,

z=4920−480060=2.0.z=\frac{4920-4800}{60}=2.0.

From the table, P(0≤Z≤2.0)=0.477P(0\le Z\le 2.0)=0.477, so

k≈P(Z≤2.0)=0.5+0.477=0.977.k\approx P(Z\le 2.0)=0.5+0.477=0.977.

Thus,

1000k=1000⋅0.977=977.1000k=1000\cdot0.977=977. 977\boxed{977}
Final Model Output
977

Comparative Analysis (Pprobability_29) (Average across runs)

Averages use reported measurements only. Missing usage is not treated as zero.

Avg Time Taken

Union Alpha
206.5
Solar Pro 4
162.5
MiMo V2.6 Pro
71.9
Grok 4.7
56.0
Motif 3
55.1
Muse Spark 1.3
16.2
DeepSeek V4.1 Flash
14.3
Solar Mini 4 Preview
4.0

Avg Token Usage

Solar Pro 4
20,003
Grok 4.7
19,390
Motif 3
12,271
MiMo V2.6 Pro
11,157
DeepSeek V4.1 Flash
7,376
Union Alpha
6,413
Muse Spark 1.3
5,639
Solar Mini 4 Preview
3,109
    CSAT 2026 · Problem probability_29 · Motif 3 | EntropyMath