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로그인 회원가입 You are solving a Korean CSAT mathematics problem. Reason carefully, but the final response must contain exactly one final answer inside \boxed{}.

For multiple-choice problems, put only one of ①, ②, ③, ④, ⑤ inside \boxed{}. For short-answer problems, put only the final integer inside \boxed{}. Do not use external tools.

Problem 28 There are sixteen balls and six empty boxes labeled with the integers 1,2,3,4,5,6. Using a single die, we perform the following experiment. Roll the die once. If the outcome isk, then – ifkis odd, put one ball into each of the boxes labeled 1, 3, and 5; – ifkis even, put one ball into each box whose label is a positive divisor ofk. This experiment is repeated 4 times. After the 4 repetitions, suppose that the sum of the numbers of balls in all six boxes is odd. Under this condition, find the probability that the number of balls in the box labeled 3 is exactly one greater than the number of balls in the box labeled 2. [4 points] ① 1 8 ② 3 16 ③ 1 4 ④ 5 16 ⑤ 3 8 Numerical answer 더보기

We are given 16 balls and 6 empty boxes labeled 1 through 6. A fair die is rolled 4 times. For each roll with outcome 𝑘 k:

If 𝑘 k is odd, put one ball into each of boxes 1, 3, 5. If 𝑘 k is even, put one ball into each box whose label is a positive divisor of 𝑘 k.

After the 4 rolls, the total number of balls in all six boxes is odd. Under this condition, we need the probability that the number of balls in box 3 is exactly one greater than the number in box 2.

Step 1: Effect of a single roll on each box. For each 𝑘 ∈ { 1 , 2 , 3 , 4 , 5 , 6 } k∈{1,2,3,4,5,6}:

𝑘

1 , 3 , 5 k=1,3,5 (odd): boxes 1,3,5 get +1. Boxes 2,4,6 get 0. 𝑘

2 k=2: divisors 1,2 → boxes 1,2 get +1. 𝑘

4 k=4: divisors 1,2,4 → boxes 1,2,4 get +1. 𝑘

6 k=6: divisors 1,2,3,6 → boxes 1,2,3,6 get +1.

Thus the increments for box 2 and box 3 are:

Box 2: +1 if 𝑘 ∈ { 2 , 4 , 6 } k∈{2,4,6} (even), else 0. Box 3: +1 if 𝑘 ∈ { 1 , 3 , 5 , 6 } k∈{1,3,5,6} (odd or 6), else 0.

Define the difference 𝐷

𝐵 3 − 𝐵 2 D=B 3 ​

−B 2 ​

after 4 rolls. For a single roll, the contribution to 𝐷 D is:

𝑘 ∈ { 1 , 3 , 5 } k∈{1,3,5}: Δ 𝐵 3

1 , Δ 𝐵 2

0 ⇒ Δ 𝐷

1 ΔB 3 ​

=1,ΔB 2 ​

=0⇒ΔD=+1 𝑘 ∈ { 2 , 4 } k∈{2,4}: Δ 𝐵 3

0 , Δ 𝐵 2

1 ⇒ Δ 𝐷

− 1 ΔB 3 ​

=0,ΔB 2 ​

=1⇒ΔD=−1 𝑘

6 k=6: Δ 𝐵 3

1 , Δ 𝐵 2

1 ⇒ Δ 𝐷

0 ΔB 3 ​

=1,ΔB 2 ​

=1⇒ΔD=0

Step 2: Total balls parity condition. The total number of balls added in a roll is:

𝑘

1 , 3 , 5 k=1,3,5: 3 balls (odd) 𝑘

2 k=2: 2 balls (even) 𝑘

4 k=4: 3 balls (odd) 𝑘

6 k=6: 4 balls (even)

So the total after 4 rolls is odd exactly when the number of rolls that add an odd number of balls is odd. The odd-add rolls are 𝑘 ∈ { 1 , 3 , 4 , 5 } k∈{1,3,4,5}.

Step 3: Categorize the 6 outcomes. Group the outcomes by ( Δ 𝐷 , parity of total balls ) (ΔD,parity of total balls):

P: 𝑘 ∈ { 1 , 3 , 5 } k∈{1,3,5} → Δ 𝐷

1 ΔD=+1, odd total. (3 outcomes) Q: 𝑘

4 k=4 → Δ 𝐷

− 1 ΔD=−1, odd total. (1 outcome) R: 𝑘

2 k=2 → Δ 𝐷

− 1 ΔD=−1, even total. (1 outcome) S: 𝑘

6 k=6 → Δ 𝐷

0 ΔD=0, even total. (1 outcome)

Let 𝑎 , 𝑏 , 𝑐 , 𝑑 a,b,c,d be the number of rolls (out of 4) that fall into P, Q, R, S respectively. Then:

𝐷

𝑎 − 𝑏 − 𝑐 D=a−b−c Number of odd-total rolls = 𝑎 + 𝑏 a+b. The condition “total odd” means 𝑎 + 𝑏 a+b is odd.

We need 𝐷

1 D=1 (box 3 has exactly one more ball than box 2) and 𝑎 + 𝑏 a+b odd, with 𝑎 + 𝑏 + 𝑐 + 𝑑

4 a+b+c+d=4.

Step 4: Find all valid ( 𝑎 , 𝑏 , 𝑐 , 𝑑 ) (a,b,c,d). From 𝐷

1 D=1: 𝑐

𝑎 − 𝑏 − 1 c=a−b−1. Since 𝑐 ≥ 0 c≥0, we need 𝑎 ≥ 𝑏 + 1 a≥b+1. Also 𝑐 ≤ 4 − 𝑎 − 𝑏 c≤4−a−b gives 𝑎 ≤ 2 a≤2. So 𝑎 ∈ { 1 , 2 } a∈{1,2}.

𝑎

1 a=1: then 𝑏 ≤ 0 ⇒ 𝑏

0 b≤0⇒b=0, 𝑐

0 c=0, 𝑑

3 d=3. 𝑎 + 𝑏

1 a+b=1 (odd) ✓. 𝑎

2 a=2: 𝑏 ∈ { 0 , 1 } b∈{0,1}. 𝑏

0 b=0: 𝑐

1 c=1, 𝑑

1 d=1. 𝑎 + 𝑏

2 a+b=2 (even) ✗. 𝑏

1 b=1: 𝑐

0 c=0, 𝑑

1 d=1. 𝑎 + 𝑏

3 a+b=3 (odd) ✓.

Thus only two count patterns:

( 𝑎 , 𝑏 , 𝑐 , 𝑑 )

( 1 , 0 , 0 , 3 ) (a,b,c,d)=(1,0,0,3) ( 𝑎 , 𝑏 , 𝑐 , 𝑑 )

( 2 , 1 , 0 , 1 ) (a,b,c,d)=(2,1,0,1)

Step 5: Count sequences and probabilities. Total possible sequences of 4 rolls: 6 4

1296 6 4 =1296. For a given ( 𝑎 , 𝑏 , 𝑐 , 𝑑 ) (a,b,c,d), the number of sequences is 4 ! 𝑎 !   𝑏 !   𝑐 !   𝑑 ! ⋅ 3 𝑎 a!b!c!d! 4! ​

⋅3 a (since P has 3 choices per roll, others have 1).

Case 1: ( 1 , 0 , 0 , 3 ) (1,0,0,3): 4 ! 1 ! 0 ! 0 ! 3 ! ⋅ 3 1

4 ⋅ 3

12 1!0!0!3! 4! ​

⋅3 1 =4⋅3=12 sequences. Case 2: ( 2 , 1 , 0 , 1 ) (2,1,0,1): 4 ! 2 ! 1 ! 0 ! 1 ! ⋅ 3 2

12 ⋅ 9

108 2!1!0!1! 4! ​

⋅3 2 =12⋅9=108 sequences.

Total favorable sequences (with 𝐷

1 D=1 and total odd): 12 + 108

120 12+108=120.

Step 6: Sequences with total odd (condition). Total odd means 𝑎 + 𝑏 a+b is odd. Enumerate all ( 𝑎 , 𝑏 , 𝑐 , 𝑑 ) (a,b,c,d) with 𝑎 + 𝑏 + 𝑐 + 𝑑

4 a+b+c+d=4 and 𝑎 + 𝑏 a+b odd. Possible odd 𝑎 + 𝑏 a+b: 1 or 3.

𝑎 + 𝑏

1 a+b=1: ( 𝑎 , 𝑏 )

( 1 , 0 ) (a,b)=(1,0) or ( 0 , 1 ) (0,1).

( 1 , 0 ) (1,0): 𝑐 + 𝑑

3 c+d=3. 𝑐 c can be 0,1,2,3. For each, sequences = 4 ! 1 ! 0 ! 𝑐 ! 𝑑 ! ⋅ 3 1 1!0!c!d! 4! ​

⋅3 1 . 𝑐

0 , 𝑑

3 c=0,d=3: 4 ! 1 ! 0 ! 0 ! 3 ! ⋅ 3

4 ⋅ 3

12 1!0!0!3! 4! ​

⋅3=4⋅3=12 𝑐

1 , 𝑑

2 c=1,d=2: 4 ! 1 ! 0 ! 1 ! 2 ! ⋅ 3

12 ⋅ 3

36 1!0!1!2! 4! ​

⋅3=12⋅3=36 𝑐

2 , 𝑑

1 c=2,d=1: 4 ! 1 ! 0 ! 2 ! 1 ! ⋅ 3

12 ⋅ 3

36 1!0!2!1! 4! ​

⋅3=12⋅3=36 𝑐

3 , 𝑑

0 c=3,d=0: 4 ! 1 ! 0 ! 3 ! 0 ! ⋅ 3

4 ⋅ 3

12 1!0!3!0! 4! ​

⋅3=4⋅3=12 Sum = 12 + 36 + 36 + 12

96 12+36+36+12=96. ( 0 , 1 ) (0,1): 𝑐 + 𝑑

3 c+d=3. Sequences = 4 ! 0 ! 1 ! 𝑐 ! 𝑑 ! ⋅ 3 0

4 ! 1 ! 𝑐 ! 𝑑 ! 0!1!c!d! 4! ​

⋅3 0

1!c!d! 4! ​

. 𝑐

0 , 𝑑

3 c=0,d=3: 4 4 𝑐

1 , 𝑑

2 c=1,d=2: 12 12 𝑐

2 , 𝑑

1 c=2,d=1: 12 12 𝑐

3 , 𝑑

0 c=3,d=0: 4 4 Sum = 4 + 12 + 12 + 4

32 4+12+12+4=32. Total for 𝑎 + 𝑏

1 a+b=1: 96 + 32

128 96+32=128.

𝑎 + 𝑏

3 a+b=3: ( 𝑎 , 𝑏 )

( 3 , 0 ) , ( 2 , 1 ) , ( 1 , 2 ) , ( 0 , 3 ) (a,b)=(3,0),(2,1),(1,2),(0,3).

( 3 , 0 ) (3,0): 𝑐 + 𝑑

1 c+d=1. Sequences = 4 ! 3 ! 0 ! 𝑐 ! 𝑑 ! ⋅ 3 3

4 ⋅ 27

108 3!0!c!d! 4! ​

⋅3 3 =4⋅27=108 for 𝑐

0 , 𝑑

1 c=0,d=1; and 𝑐

1 , 𝑑

0 c=1,d=0: 4 ⋅ 27

108 4⋅27=108. Sum = 216. ( 2 , 1 ) (2,1): 𝑐 + 𝑑

1 c+d=1. Sequences = 4 ! 2 ! 1 ! 𝑐 ! 𝑑 ! ⋅ 3 2

12 ⋅ 9

108 2!1!c!d! 4! ​

⋅3 2 =12⋅9=108 for each 𝑐 , 𝑑 c,d. Sum = 216. ( 1 , 2 ) (1,2): 𝑐 + 𝑑

1 c+d=1. Sequences = 4 ! 1 ! 2 ! 𝑐 ! 𝑑 ! ⋅ 3 1

12 ⋅ 3

36 1!2!c!d! 4! ​

⋅3 1 =12⋅3=36 for each. Sum = 72. ( 0 , 3 ) (0,3): 𝑐 + 𝑑

1 c+d=1. Sequences = 4 ! 0 ! 3 ! 𝑐 ! 𝑑 ! ⋅ 3 0

4 0!3!c!d! 4! ​

⋅3 0 =4 for each. Sum = 8. Total for 𝑎 + 𝑏

3 a+b=3: 216 + 216 + 72 + 8

512 216+216+72+8=512.

Total sequences with total odd: 128 + 512

640 128+512=640.

Step 7: Conditional probability.

𝑃 ( 𝐷

1 ∣ total odd )

120 640

3 16 . P(D=1∣total odd)= 640 120 ​

= 16 3 ​

.

The answer is 3 16 16 3 ​

​

.

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