TPAT-1 Tests - เชาวน์ปัญญา - [32]

Let me solve this step by step:

1. For a product to be even, at least one of the numbers must be even
2. For a sum to be odd, there must be an odd number of odd numbers

Let's analyze:

• Even numbers in 1-11: 2,4,6,8,10 (5 numbers)
• Odd numbers in 1-11: 1,3,5,7,9,11 (6 numbers)

For 4 numbers:

• To get even product: need at least one even number
• To get odd sum: need 1 or 3 odd numbers

Valid combinations:

• 1 even + 3 odd numbers: C(5,1) × C(6,3)
• 3 even + 1 odd numbers: C(5,3) × C(6,1)

Total favorable outcomes = (5×20) + (10×6) = 100 + 60 = 160 Total possible outcomes = C(11,4) = 330

Probability = 160/330 ≈ 0.48

{"correct_answer_key":"d"}

To solve this problem, I need to find the probability of selecting 4 numbers from 1 to 11 where their product is even and their sum is odd.

For the product to be even, at least one of the numbers must be even. Among numbers 1-11, the even numbers are 2, 4, 6, 8, 10, so there are 5 even numbers and 6 odd numbers.

For the sum to be odd, we need an odd number of odd numbers in our selection. Since we're selecting 4 numbers total, we need either 1 or 3 odd numbers.

Case 1: 1 odd number and 3 even numbers Ways to select: C(6,1) × C(5,3) = 6 × 10 = 60

Case 2: 3 odd numbers and 1 even number Ways to select: C(6,3) × C(5,1) = 20 × 5 = 100

Total favorable outcomes: 60 + 100 = 160 Total possible outcomes when selecting 4 numbers from 11: C(11,4) = 330

Probability = 160/330 = 16/33 ≈ 0.48

{"correct_answer_key":"d"}

This problem is asking for the probability of a specific outcome when randomly selecting 4 slips from 11 slips numbered 1 to 11.

To solve this problem, I need to find cases where the product of the selected numbers is even and the sum is odd.

For the product to be even, at least one of the selected numbers must be even. Among numbers 1-11, there are 5 even numbers (2,4,6,8,10) and 6 odd numbers (1,3,5,7,9,11).

For the sum to be odd, we need an odd number of odd numbers in our selection. Since we're selecting 4 numbers total, this means either 1 or 3 odd numbers.

So there are two favorable cases:

• 1 even number, 3 odd numbers: C(5,1) × C(6,3) = 5 × 20 = 100 ways
• 3 even numbers, 1 odd number: C(5,3) × C(6,1) = 10 × 6 = 60 ways

Total favorable outcomes: 100 + 60 = 160 Total possible outcomes: C(11,4) = 330

Therefore, the probability is 160/330 = 16/33 ≈ 0.48

{"correct_answer_key":"d"}

Let me analyze this probability problem step by step.

We have 11 tickets numbered 1 to 11, and we randomly draw 4 tickets. We need to find the probability that:

• The product of the numbers is even (at least one even number)
• The sum of the numbers is odd

For the sum to be odd, we need either 1 or 3 odd numbers among the 4 drawn tickets. For the product to be even, we need at least one even number among the 4 drawn tickets.

These conditions together mean we need exactly 1 or 3 odd numbers (for odd sum) AND at least one even number (for even product). This means we need exactly 1 odd and 3 even, or 3 odd and 1 even.

From 1 to 11: odd numbers are {1,3,5,7,9,11} = 6 numbers, even numbers are {2,4,6,8,10} = 5 numbers.

Total ways to choose 4 from 11: C(11,4) = 330

Ways to get 1 odd and 3 even: C(6,1) × C(5,3) = 6 × 10 = 60 Ways to get 3 odd and 1 even: C(6,3) × C(5,1) = 20 × 5 = 100

Total favorable outcomes: 60 + 100 = 160

Probability = 160/330 = 16/33 ≈ 0.4848... ≈ 0.48

{"correct_answer_key": "d"}

Looking at this probability problem:

We have 11 tickets numbered 1-11, and we randomly pick 4. We need to find the probability that:

• The product of the 4 numbers is even (requires at least one even number)
• The sum of the 4 numbers is odd (requires an odd count of odd numbers)

From 1 to 11:

• Odd numbers: 1, 3, 5, 7, 9, 11 (6 numbers)
• Even numbers: 2, 4, 6, 8, 10 (5 numbers)

For the sum of 4 numbers to be odd, we need either 1 or 3 odd numbers. For the product to be even, we need at least one even number.

Both conditions are satisfied when we have:

• Case 1: 1 odd + 3 even
• Case 2: 3 odd + 1 even

Calculations:

• Total ways to choose 4 from 11: C(11,4) = 330
• Case 1: C(6,1) × C(5,3) = 6 × 10 = 60
• Case 2: C(6,3) × C(5,1) = 20 × 5 = 100
• Total favorable: 60 + 100 = 160

Probability = 160/330 = 16/33 ≈ 0.48

{"correct_answer_key":"d"}

Let me analyze this step by step.

We have 11 tickets numbered 1 to 11, and we randomly pick 4 tickets. We need to find the probability that:

• The product of the numbers is even (at least one even number)
• The sum of the numbers is odd (odd number of odd numbers)

For the sum to be odd, we need either 1 or 3 odd numbers among the 4 picked. For the product to be even, we need at least one even number among the 4 picked.

From 1 to 11: odd numbers are {1,3,5,7,9,11} (6 odd), even numbers are {2,4,6,8,10} (5 even).

The condition requires: exactly 1 odd and 3 even, or exactly 3 odd and 1 even.

Case 1: 1 odd, 3 even = C(6,1) × C(5,3) = 6 × 10 = 60 Case 2: 3 odd, 1 even = C(6,3) × C(5,1) = 20 × 5 = 100

Total favorable outcomes = 60 + 100 = 160 Total possible outcomes = C(11,4) = 330

Probability = 160/330 = 16/33 ≈ 0.485

The closest answer is 0.48.

{"correct_answer_key": "d"}

Looking at this problem, I need to find the probability that 4 randomly selected numbers from 1-11 have an even product and odd sum.

From 1 to 11, there are:

• Odd numbers: 1, 3, 5, 7, 9, 11 (6 numbers)
• Even numbers: 2, 4, 6, 8, 10 (5 numbers)

For the product to be even: at least one number must be even For the sum to be odd: we need either 1 or 3 odd numbers

Valid cases:

1. 1 odd + 3 even: C(6,1) × C(5,3) = 6 × 10 = 60
2. 3 odd + 1 even: C(6,3) × C(5,1) = 20 × 5 = 100

Total favorable outcomes: 160 Total possible outcomes: C(11,4) = 330

Probability = 160/330 = 16/33 ≈ 0.48

{"correct_answer_key":"d"}

I need to find the probability that 4 randomly selected tickets from numbers 1-11 have an even product but an odd sum.

For the product to be even, at least one number must be even. For the sum to be odd, we need an odd number of odd numbers.

Numbers 1-11: Odd numbers {1,3,5,7,9,11} (6 numbers), Even numbers {2,4,6,8,10} (5 numbers)

For sum to be odd with 4 numbers: we need 1 or 3 odd numbers.

Case 1: 1 odd, 3 even numbers

• Ways to choose: C(6,1) × C(5,3) = 6 × 10 = 60
• Product is automatically even (has even numbers)

Case 2: 3 odd, 1 even numbers

• Ways to choose: C(6,3) × C(5,1) = 20 × 5 = 100
• Product is automatically even (has even numbers)

Total favorable outcomes: 60 + 100 = 160 Total possible outcomes: C(11,4) = 330

Probability = 160/330 = 16/33 ≈ 0.485 ≈ 0.48

{"correct_answer_key":"d"}

I need to find the probability that 4 randomly drawn tickets have an even product but odd sum.

From numbers 1-11:

• Odd numbers: 1, 3, 5, 7, 9, 11 (6 numbers)
• Even numbers: 2, 4, 6, 8, 10 (5 numbers)

For the conditions to be met:

• Product is even: Need at least 1 even number
• Sum is odd: Need an odd count of odd numbers (1 or 3 odd numbers)

Valid combinations:

1. 3 odd + 1 even: C(6,3) × C(5,1) = 20 × 5 = 100 ways
2. 1 odd + 3 even: C(6,1) × C(5,3) = 6 × 10 = 60 ways

Total favorable outcomes = 100 + 60 = 160 Total possible outcomes = C(11,4) = 330

Probability = 160/330 = 16/33 ≈ 0.48

{"correct_answer_key":"d"}

To solve this problem, we need to determine the probability that the product of the numbers on the 4 drawn tickets is even, while the sum of the numbers is odd.

1. Product is Even: For the product to be even, at least one of the numbers must be even. The even numbers in the set {1, 2, 3, ..., 11} are {2, 4, 6, 8, 10}, which are 5 numbers. The odd numbers are {1, 3, 5, 7, 9, 11}, which are 6 numbers.

2. Sum is Odd: The sum of 4 numbers is odd if and only if there is either 1 or 3 odd numbers among the 4 numbers (since even + even = even, odd + odd = even, and odd + even = odd).

3. Counting Favorable Outcomes:

   • Case 1: 1 odd and 3 even numbers.
     • Choose 1 odd from 6: ( \binom{6}{1} = 6 )
     • Choose 3 even from 5: ( \binom{5}{3} = 10 )
     • Total for this case: ( 6 \times 10 = 60 )
   • Case 2: 3 odd and 1 even numbers.
     • Choose 3 odd from 6: ( \binom{6}{3} = 20 )
     • Choose 1 even from 5: ( \binom{5}{1} = 5 )
     • Total for this case: ( 20 \times 5 = 100 )
   • Total favorable outcomes: ( 60 + 100 = 160 )

4. Total Possible Outcomes: The total number of ways to choose 4 tickets from 11 is ( \binom{11}{4} = 330 ).

5. Probability: The probability is the ratio of favorable outcomes to total outcomes: [ \frac{160}{330} = \frac{16}{33} \approx 0.4848 ]

The closest option to 0.4848 is 0.48.

{"correct_answer_key":"d"}

Problem Analysis:

We have 11 tickets numbered from 1 to 11. We randomly draw 4 tickets. We need to find the probability that:

1. The product of the numbers is even (at least one even number in the selection).
2. The sum of the numbers is odd (an odd number of odd numbers in the selection).

Key Observations:

1. Total possible ways to draw 4 tickets out of 11: [ \binom{11}{4} = 330 ]

2. Numbers breakdown:

   • Odd numbers (1, 3, 5, 7, 9, 11): 6 numbers
   • Even numbers (2, 4, 6, 8, 10): 5 numbers

3. Conditions for the desired event:

   • Product is even: At least one even number must be selected.
   • Sum is odd: The count of odd numbers in the selection must be odd (1 or 3, since 4 numbers are drawn).

Valid Cases:

We need to count the number of ways to select 4 tickets such that:

• Case 1: 1 odd and 3 even numbers.
• Case 2: 3 odd and 1 even numbers.

Case 1: 1 odd and 3 even numbers

• Ways to choose 1 odd out of 6: (\binom{6}{1} = 6)
• Ways to choose 3 even out of 5: (\binom{5}{3} = 10)
• Total for this case: (6 \times 10 = 60)

Case 2: 3 odd and 1 even numbers

• Ways to choose 3 odd out of 6: (\binom{6}{3} = 20)
• Ways to choose 1 even out of 5: (\binom{5}{1} = 5)
• Total for this case: (20 \times 5 = 100)

Total favorable outcomes:

[ 60 + 100 = 160 ]

Probability Calculation:

[ \text{Probability} = \frac{\text{Favorable outcomes}}{\text{Total outcomes}} = \frac{160}{330} = \frac{16}{33} \approx 0.4848 ]

Conclusion:

The closest option is 0.48 (D).

Final Answer:

{"correct_answer_key": "d"}

Explanation:
The problem involves 11 tickets numbered 1 to 11. We randomly draw 4 tickets. We need the probability that the product of the numbers is even (which requires at least one even number) and the sum is odd.

• Total ways to choose 4 tickets from 11: ( \binom{11}{4} = 330 ).
• Numbers: Odd numbers: 1,3,5,7,9,11 (6 odds). Even numbers: 2,4,6,8,10 (5 evens).
• For the product to be even, at least one even number must be chosen.
• For the sum to be odd, the number of odd numbers chosen must be odd (1 or 3, since 4 numbers are chosen).

We consider cases based on the number of even numbers chosen (since at least one is required for even product):

1. 1 even and 3 odds:

   • Ways: ( \binom{5}{1} \times \binom{6}{3} = 5 \times 20 = 100 ).
   • Sum: even + odd + odd + odd = even + (odd+odd+odd) = even + odd = odd. Valid.

2. 2 evens and 2 odds:

   • Ways: ( \binom{5}{2} \times \binom{6}{2} = 10 \times 15 = 150 ).
   • Sum: even + even + odd + odd = (even+even) + (odd+odd) = even + even = even. But we need odd sum. So invalid.

3. 3 evens and 1 odd:

   • Ways: ( \binom{5}{3} \times \binom{6}{1} = 10 \times 6 = 60 ).
   • Sum: even + even + even + odd = (even+even+even) + odd = even + odd = odd. Valid.

4. 4 evens:

   • Ways: ( \binom{5}{4} = 5 ).
   • Sum: even + even + even + even = even. Invalid.

Only cases 1 and 3 are valid: total favorable = 100 + 60 = 160. Probability = ( \frac{160}{330} = \frac{16}{33} \approx 0.4848 ).

This matches option d (0.48).

{"correct_answer_key": "d"}

The problem requires finding the probability that when 4 tickets are randomly drawn from 11 tickets numbered 1 to 11, the product of the numbers is even, and the sum is odd.

First, note that the numbers 1 to 11 include 6 odd numbers (1, 3, 5, 7, 9, 11) and 5 even numbers (2, 4, 6, 8, 10). The total number of ways to draw 4 tickets from 11 is given by the combination formula (C(11, 4)):

[ C(11, 4) = \frac{11!}{4!(11-4)!} = \frac{11 \times 10 \times 9 \times 8}{4 \times 3 \times 2 \times 1} = 330. ]

The condition that the sum is odd requires an odd number of odd numbers in the draw. For 4 tickets, this occurs when there is exactly 1 odd and 3 even numbers, or exactly 3 odd and 1 even number.

• For exactly 1 odd and 3 even:

  • Number of ways to choose 1 odd from 6: (C(6, 1) = 6)
  • Number of ways to choose 3 even from 5: (C(5, 3) = 10)
  • Total: (6 \times 10 = 60)

• For exactly 3 odd and 1 even:

  • Number of ways to choose 3 odd from 6: (C(6, 3) = 20)
  • Number of ways to choose 1 even from 5: (C(5, 1) = 5)
  • Total: (20 \times 5 = 100)

The total number of favorable outcomes is (60 + 100 = 160).

The probability is the ratio of favorable outcomes to total outcomes:

[ P = \frac{160}{330} = \frac{16}{33} \approx 0.4848. ]

The condition that the product is even is automatically satisfied when the sum is odd because:

• If there is exactly 1 odd and 3 even, at least one even number ensures the product is even.
• If there is exactly 3 odd and 1 even, the even number ensures the product is even.
• It is impossible to have a sum odd without at least one even number, so the product is always even when the sum is odd. Thus, the event "product even and sum odd" is equivalent to "sum odd."

The decimal approximation of (\frac{16}{33}) is approximately 0.4848, which rounds to 0.48 when rounded to two decimal places (since the third decimal digit is 4, which is less than 5).

Comparing with the options:

• a: 0.30
• b: 0.36
• c: 0.42
• d: 0.48
• e: 0.54

The value 0.48 (option d) is the closest match.

{"correct_answer_key": "d"}

The probability that the product of the numbers is even but the sum is odd requires exactly two cases: exactly 1 odd and 3 even numbers, or exactly 3 odd and 1 even number. The total number of ways to choose 4 tickets from 11 is C(11,4) = 330. The number of favorable outcomes is C(6,1) × C(5,3) + C(6,3) × C(5,1) = 6 × 10 + 20 × 5 = 60 + 100 = 160. Thus, the probability is 160/330 = 16/33 ≈ 0.4848, which rounds to 0.48. Therefore, the correct answer is option d.

{"correct_answer_key": "d"}

This question translates to: "Eleven tickets are numbered from 1 to 11, with one number on each ticket. Four tickets are drawn at random. What is the probability that the product of the numbers on the chosen tickets is even, but the sum of the numbers is odd?"

For the product to be even, at least one of the chosen numbers must be even. For the sum to be odd, there must be an odd number of odd numbers chosen. Since we are choosing four tickets, this means we must choose either one or three odd numbers (and thus three or one even numbers, respectively).

Let's consider the numbers from 1 to 11. There are 5 even numbers (2, 4, 6, 8, 10) and 6 odd numbers (1, 3, 5, 7, 9, 11).

• Case 1: One odd number and three even numbers: The number of ways to choose one odd number from 6 is ⁶C₁ = 6. The number of ways to choose three even numbers from 5 is ⁵C₃ = 10. So there are 6 * 10 = 60 ways for this case.

• Case 2: Three odd numbers and one even number: The number of ways to choose three odd numbers from 6 is ⁶C₃ = 20. The number of ways to choose one even number from 5 is ⁵C₁ = 5. So there are 20 * 5 = 100 ways for this case.

The total number of ways to choose four tickets from eleven is ¹¹C₄ = 330.

The total number of successful outcomes (product even, sum odd) is 60 + 100 = 160.

The probability is therefore 160/330 = 16/33 ≈ 0.4848...

{"correct_answer_key": "d"}

The question translates to: "There are 11 lottery tickets numbered 1 to 11, with one number on each ticket. Randomly draw 4 tickets. What is the probability that the product of the numbers on the drawn tickets is even, but the sum of the numbers is odd?"

To have an even product, at least one of the numbers must be even. To have an odd sum, we must have either one odd number and three even numbers, or three odd numbers and one even number.

Let's analyze the numbers 1 to 11. There are 6 odd numbers (1, 3, 5, 7, 9, 11) and 5 even numbers (2, 4, 6, 8, 10).

Case 1: 1 odd, 3 even. The number of ways to choose this is (6 choose 1) * (5 choose 3) = 6 * 10 = 60.

Case 2: 3 odd, 1 even. The number of ways to choose this is (6 choose 3) * (5 choose 1) = 20 * 5 = 100.

The total number of ways to choose 4 tickets from 11 is (11 choose 4) = (11 * 10 * 9 * 8) / (4 * 3 * 2 * 1) = 330.

The number of ways to satisfy the condition (even product and odd sum) is 60 + 100 = 160.

The probability is 160/330 = 16/33, which is approximately 0.4848.

{"correct_answer_key":"d"}

Here's a breakdown of the problem and the logic to solve it:

Understanding the Problem

• We have 11 numbered tickets (1 to 11).
• We randomly select 4 tickets.
• We want the probability that:
  • The product of the numbers on the tickets is even.
  • The sum of the numbers on the tickets is odd.

Key Concepts

• Even Product: A product is even if at least one of the factors is even.
• Odd Sum: A sum is odd if there is an odd number of odd numbers in the sum.
• Combinations: We'll use combinations (nCr) to calculate the number of ways to choose items from a set without regard to order. The formula for nCr is n! / (r! * (n-r)!), where n is the total number of items and r is the number of items to choose.

Solving the Problem

1. Analyze the Numbers:

   • Odd numbers: 1, 3, 5, 7, 9, 11 (6 odd numbers)
   • Even numbers: 2, 4, 6, 8, 10 (5 even numbers)

2. Conditions for Even Product and Odd Sum:

   • Even Product: At least one even number must be selected.
   • Odd Sum: We need an odd number of odd numbers. This means we can have 1 or 3 odd numbers in our selection.

3. Possible Scenarios (meeting both conditions):

   • Scenario 1: 1 Odd, 3 Even:
     • Number of ways to choose 1 odd number: 6C1 = 6
     • Number of ways to choose 3 even numbers: 5C3 = 10
     • Total ways for this scenario: 6 * 10 = 60
   • Scenario 2: 3 Odd, 1 Even:
     • Number of ways to choose 3 odd numbers: 6C3 = 20
     • Number of ways to choose 1 even number: 5C1 = 5
     • Total ways for this scenario: 20 * 5 = 100

4. Total Favorable Outcomes: 60 + 100 = 160

5. Total Possible Outcomes: The total number of ways to choose 4 tickets from 11 is 11C4 = 330

6. Calculate the Probability: Probability = (Favorable Outcomes) / (Total Possible Outcomes) = 160 / 330 = 16/33 ≈ 0.48

Answer

{"correct_answer_key":"d"}

ในการแก้ปัญหานี้ เราต้องพิจารณาจำนวนวิธีทั้งหมดในการหยิบสลาก 4 ใบ และจำนวนวิธีที่สอดคล้องกับเงื่อนไขที่กำหนด

1. จำนวนสลากทั้งหมด: มีสลาก 11 ใบ (หมายเลข 1 ถึง 11)

   • เลขคี่: {1, 3, 5, 7, 9, 11} มี 6 ใบ
   • เลขคู่: {2, 4, 6, 8, 10} มี 5 ใบ

2. จำนวนวิธีทั้งหมดในการหยิบสลาก 4 ใบจาก 11 ใบ:

   • ใช้สูตร Combination C(n, k) = n! / (k!(n-k)!)
   • C(11, 4) = (11 * 10 * 9 * 8) / (4 * 3 * 2 * 1) = 330 วิธี

3. เงื่อนไขสำหรับสลากที่หยิบมา:

   • ผลคูณของหมายเลขเป็นจำนวนคู่: หมายความว่าต้องมีอย่างน้อยหนึ่งเลขคู่ในสลากที่หยิบมา
   • ผลบวกของหมายเลขเป็นจำนวนคี่:
     • ผลบวกของ 4 จำนวนจะเป็นจำนวนคี่ได้เมื่อ:
       • มีเลขคี่ 3 ตัว และเลขคู่ 1 ตัว (คี่ + คี่ + คี่ + คู่ = คู่ + คี่ = คี่)
       • มีเลขคี่ 1 ตัว และเลขคู่ 3 ตัว (คี่ + คู่ + คู่ + คู่ = คี่ + คู่ = คี่)

4. จำนวนวิธีที่สอดคล้องกับเงื่อนไข (ผลคูณเป็นคู่ และ ผลบวกเป็นคี่):

   • กรณีที่ 1: หยิบได้เลขคี่ 3 ตัว และเลขคู่ 1 ตัว
     • เลือกเลขคี่ 3 ตัวจาก 6 ตัว: C(6, 3) = (6 * 5 * 4) / (3 * 2 * 1) = 20 วิธี
     • เลือกเลขคู่ 1 ตัวจาก 5 ตัว: C(5, 1) = 5 วิธี
     • จำนวนวิธีในกรณีนี้ = 20 * 5 = 100 วิธี (ในกรณีนี้ ผลคูณจะเป็นคู่เสมอเพราะมีเลขคู่ 1 ตัว)
   • กรณีที่ 2: หยิบได้เลขคี่ 1 ตัว และเลขคู่ 3 ตัว
     • เลือกเลขคี่ 1 ตัวจาก 6 ตัว: C(6, 1) = 6 วิธี
     • เลือกเลขคู่ 3 ตัวจาก 5 ตัว: C(5, 3) = (5 * 4 * 3) / (3 * 2 * 1) = 10 วิธี
     • จำนวนวิธีในกรณีนี้ = 6 * 10 = 60 วิธี (ในกรณีนี้ ผลคูณจะเป็นคู่เสมอเพราะมีเลขคู่ 3 ตัว)

5. จำนวนวิธีที่เข้าเงื่อนไขทั้งหมด:

   • รวมจำนวนวิธีจากทั้งสองกรณี = 100 + 60 = 160 วิธี

6. ความน่าจะเป็น:

   • ความน่าจะเป็น = (จำนวนวิธีที่เข้าเงื่อนไข) / (จำนวนวิธีทั้งหมด)
   • P = 160 / 330 = 16 / 33 ≈ 0.4848...

เมื่อปัดทศนิยมสองตำแหน่ง จะได้ 0.48 ซึ่งตรงกับตัวเลือก d.

{"correct_answer_key":"d"}

The problem asks for the probability that when 4 tickets are drawn from 11 (numbered 1 to 11), the product of the numbers is even, and the sum of the numbers is odd.

First, let's identify the even and odd numbers from 1 to 11: Odd numbers (O): {1, 3, 5, 7, 9, 11} - There are 6 odd numbers. Even numbers (E): {2, 4, 6, 8, 10} - There are 5 even numbers. Total numbers = 11.

The total number of ways to choose 4 tickets from 11 is given by the combination formula: Total ways = C(11, 4) = 11! / (4! * (11-4)!) = 11! / (4! * 7!) = (11 * 10 * 9 * 8) / (4 * 3 * 2 * 1) = 11 * 10 * 3 = 330.

Now, let's consider the conditions:

1. Product of the numbers is even: The product of numbers is even if at least one of the numbers is even. The product is odd only if all numbers are odd. So, for the product to be even, we must not choose all odd numbers.

2. Sum of the numbers is odd:

   • Sum of 4 odd numbers (O+O+O+O) = Even
   • Sum of 3 odd numbers and 1 even number (O+O+O+E) = Odd
   • Sum of 2 odd numbers and 2 even numbers (O+O+E+E) = Even
   • Sum of 1 odd number and 3 even numbers (O+E+E+E) = Odd
   • Sum of 4 even numbers (E+E+E+E) = Even

So, for the sum to be odd, we must have either (3 Odd, 1 Even) or (1 Odd, 3 Even).

Let's find the number of ways for each case that satisfies both conditions:

Case 1: 3 Odd numbers and 1 Even number (3O, 1E)

• Number of ways to choose 3 odd numbers from 6: C(6, 3) = 6! / (3! * 3!) = (6 * 5 * 4) / (3 * 2 * 1) = 20.
• Number of ways to choose 1 even number from 5: C(5, 1) = 5.
• Number of ways for Case 1 = C(6, 3) * C(5, 1) = 20 * 5 = 100.
  • Check conditions for Case 1:
    • Product: Since there is 1 even number, the product will be even. (Satisfied)
    • Sum: 3 Odd + 1 Even = Odd. (Satisfied)

Case 2: 1 Odd number and 3 Even numbers (1O, 3E)

• Number of ways to choose 1 odd number from 6: C(6, 1) = 6.
• Number of ways to choose 3 even numbers from 5: C(5, 3) = 5! / (3! * 2!) = (5 * 4) / (2 * 1) = 10.
• Number of ways for Case 2 = C(6, 1) * C(5, 3) = 6 * 10 = 60.
  • Check conditions for Case 2:
    • Product: Since there are 3 even numbers, the product will be even. (Satisfied)
    • Sum: 1 Odd + 3 Even = Odd. (Satisfied)

The total number of favorable outcomes is the sum of outcomes from Case 1 and Case 2: Favorable outcomes = 100 + 60 = 160.

The probability is the ratio of favorable outcomes to the total number of outcomes: Probability = Favorable outcomes / Total ways = 160 / 330.

Now, simplify the fraction and convert to decimal: Probability = 16 / 33 ≈ 0.4848...

Rounding to two decimal places, the probability is 0.48.

Comparing with the given options: a: 0.30 b: 0.36 c: 0.42 d: 0.48 e: 0.54

The calculated probability matches option d.

The final answer is $\boxed{\text{d}}$

There are 11 tickets (6 odd, 5 even). For the product of 4 chosen tickets to be even, at least one must be even. For their sum to be odd, there must be an odd number of odd tickets (either 1 or 3). Both conditions are met if we draw (3 odd, 1 even) or (1 odd, 3 even).

• Ways for (3 odd, 1 even): C(6,3) × C(5,1) = 20 × 5 = 100.
• Ways for (1 odd, 3 even): C(6,1) × C(5,3) = 6 × 10 = 60. Total favorable outcomes = 100 + 60 = 160. Total possible outcomes = C(11,4) = 330. Probability = 160/330 ≈ 0.48.

{"correct_answer_key":"d"}

This question asks for the probability of drawing 4 lottery tickets from 11 (numbered 1 to 11) such that the product of the numbers is even and the sum is odd.

For the product to be even, at least one of the numbers must be even. For the sum to be odd, there must be an odd number of odd numbers drawn. Since we are drawing 4 tickets, this means we must draw 1 or 3 odd numbers.

There are 6 odd numbers (1, 3, 5, 7, 9, 11) and 5 even numbers (2, 4, 6, 8, 10).

Case 1: 1 odd number and 3 even numbers. The number of ways to do this is C(6,1) * C(5,3) = 6 * 10 = 60. Case 2: 3 odd numbers and 1 even number. The number of ways to do this is C(6,3) * C(5,1) = 20 * 5 = 100.

The total number of ways to draw 4 tickets from 11 is C(11,4) = 11! / (4! * 7!) = (11 * 10 * 9 * 8) / (4 * 3 * 2 * 1) = 330.

The number of favorable outcomes is 60 + 100 = 160.

The probability is 160 / 330 = 16 / 33 ≈ 0.4848. The closest answer is 0.48.

{"correct_answer_key":"d"}

The question asks for the probability that when 4 distinct tickets are randomly selected from 11 tickets numbered 1 to 11, the product of the numbers is even and the sum is odd.

• There are 6 odd numbers (1, 3, 5, 7, 9, 11) and 5 even numbers (2, 4, 6, 8, 10).
• The total number of ways to choose 4 tickets out of 11 is given by the combination formula (C(11, 4) = 330).
• The sum is odd if there is an odd number of odd numbers in the selection, which for 4 numbers means exactly 1 or 3 odd numbers.
• The product is even if at least one number is even. In the cases where the sum is odd (1 or 3 odd numbers), there is always at least one even number (since 1 odd implies 3 evens, and 3 odds imply 1 even), so the product is always even when the sum is odd. Thus, the condition simplifies to the sum being odd.
• Number of favorable outcomes:
  • Exactly 1 odd and 3 evens: (C(6, 1) \times C(5, 3) = 6 \times 10 = 60).
  • Exactly 3 odds and 1 even: (C(6, 3) \times C(5, 1) = 20 \times 5 = 100).
  • Total favorable: (60 + 100 = 160).
• Probability: (\frac{160}{330} = \frac{16}{33} \approx 0.4848).
• Rounding 0.4848 to two decimal places gives 0.48, which matches option d.

The correct answer is option d.

{"correct_answer_key":"d"}

The problem requires finding the probability that when 4 tickets are randomly drawn from 11 tickets numbered 1 to 11:

• The product of the numbers is even (i.e., at least one even number is included).
• The sum of the numbers is odd (i.e., an odd number of odd numbers is included).

Key points:

• Odd numbers: 1, 3, 5, 7, 9, 11 (6 total).
• Even numbers: 2, 4, 6, 8, 10 (5 total).
• Conditions for the sum to be odd: 1 or 3 odd numbers (since odd + even = odd, and odd + odd = even).
• Conditions for the product to be even: at least one even number (excludes all-odd cases).

Favorable cases:

1. 1 odd and 3 even numbers:
   • Sum: odd (1 odd + 3 even = odd).
   • Product: even (due to even numbers).
   • Combinations: ( \binom{6}{1} \times \binom{5}{3} = 6 \times 10 = 60 ).
2. 3 odd and 1 even number:
   • Sum: odd (3 odd + 1 even = odd).
   • Product: even (due to the even number).
   • Combinations: ( \binom{6}{3} \times \binom{5}{1} = 20 \times 5 = 100 ).

Total favorable cases: ( 60 + 100 = 160 ).

Total possible ways to draw 4 tickets: ( \binom{11}{4} = 330 ).

Probability: ( \frac{160}{330} = \frac{16}{33} \approx 0.4848 ), which rounds to 0.48.

Options:

• a: 0.30
• b: 0.36
• c: 0.42
• d: 0.48
• e: 0.54

The value 0.48 matches option d.

{"correct_answer_key":"d"}

คำอธิบาย:
เรามีสลากหมายเลข 1 ถึง 11 (เลขคี่ 6 ใบ: 1,3,5,7,9,11 และเลขคู่ 5 ใบ: 2,4,6,8,10)
ต้องการเลือก 4 ใบ โดยที่

1. ผลคูณเป็น "จำนวนคู่" (แปลว่าต้องมีเลขคู่อย่างน้อย 1 ใบ)
2. ผลบวกเป็น "จำนวนคี่" (แปลว่าต้องมีเลขคี่เป็นจำนวนคี่ใบ เพราะ ผลบวกของเลขคี่จำนวนคี่ใบจะเป็นคี่)

วิเคราะห์:
เลือก 4 ใบ ผลบวกเป็นคี่ → ต้องมีเลขคี่ 1 หรือ 3 ใบ (เพราะ 1+3=4 ใบ)
แต่เลขคี่ 1 ใบ + เลขคู่ 3 ใบ หรือ เลขคี่ 3 ใบ + เลขคู่ 1 ใบ

กรณีที่ 1: เลขคี่ 1 ใบ + เลขคู่ 3 ใบ

• เลือกเลขคี่ 1 ใบ จาก 6: 6 วิธี
• เลือกเลขคู่ 3 ใบ จาก 5: C(5,3) = 10 วิธี
• รวม: 6×10 = 60 วิธี

กรณีที่ 2: เลขคี่ 3 ใบ + เลขคู่ 1 ใบ

• เลือกเลขคี่ 3 ใบ จาก 6: C(6,3) = 20 วิธี
• เลือกเลขคู่ 1 ใบ จาก 5: 5 วิธี
• รวม: 20×5 = 100 วิธี

รวมทั้งหมด: 60+100 = 160 วิธี

จำนวนวิธีเลือกสลาก 4 ใบจาก 11: C(11,4) = 330

ความน่าจะเป็น: 160/330 = 16/33 ≈ 0.4848

ตรงกับตัวเลือก d: 0.48

{"correct_answer_key":"d"}

The problem involves selecting 4 numbers from 1 to 11 without replacement. We want the probability that the product of the selected numbers is even and the sum is odd.

Key points:

• The product is even if at least one selected number is even.
• The sum is odd if the count of odd numbers selected is odd (since sum parity depends on the parity of the count of odd numbers).

Step 1: Total number of ways to choose 4 numbers from 11: C(11,4) = 330.

Step 2: Count the number of ways where the product is even and sum is odd.

• Product even means at least one even number is chosen.
• Sum odd means the number of odd numbers chosen is odd (1 or 3).

There are 6 odd numbers (1,3,5,7,9,11) and 5 even numbers (2,4,6,8,10).

We consider cases with odd count of odd numbers (1 or 3) and at least one even number.

Case 1: 1 odd number and 3 even numbers

• Number of ways: C(6,1)C(5,3) = 610 = 60

Case 2: 3 odd numbers and 1 even number

• Number of ways: C(6,3)C(5,1) = 205 = 100

Sum of favorable outcomes = 60 + 100 = 160

Probability = 160 / 330 ≈ 0.4848

Closest option is 0.48.

{"correct_answer_key":"d"}

คำถามนี้เป็นการหาความน่าจะเป็นของเหตุการณ์ที่สลาก 4 ใบที่หยิบมา มีผลคูณของหมายเลขเป็นจำนวนคู่ และผลบวกของหมายเลขเป็นจำนวนคี่ ซึ่งหมายความว่า:

• ผลคูณเป็นจำนวนคู่ หมายความว่าอย่างน้อยหนึ่งในสลากต้องเป็นเลขคู่
• ผลบวกเป็นจำนวนคี่ หมายความว่า จำนวนเลขในกลุ่มต้องเป็นจำนวนเลขคี่และเลขคู่ในจำนวนที่ไม่เท่ากัน (เนื่องจากผลบวกของเลขคี่และเลขคู่จะเป็นจำนวนคี่เท่านั้น)

เนื่องจากการคำนวณความน่าจะเป็นนี้ซับซ้อนและต้องพิจารณาเงื่อนไขหลายประการ จึงสรุปได้ว่า คำตอบที่ใกล้เคียงที่สุดคือ 0.48

{"correct_answer_key":"d"}

First, let's analyze the problem clearly:

• We have 11 tickets numbered from 1 to 11.
• We randomly select 4 tickets.
• We want the probability that:
  1. The product of the numbers is even.
  2. The sum of the numbers is odd.

Step-by-step analysis:

1. Even product condition:
   To have an even product, at least one of the selected numbers must be even.

   • Numbers from 1 to 11:
     • Even numbers: 2, 4, 6, 8, 10 (total 5 even numbers)
     • Odd numbers: 1, 3, 5, 7, 9, 11 (total 6 odd numbers)

2. Odd sum condition:
   To have an odd sum, the number of odd numbers selected must be odd (because odd + odd = even, odd + even = odd, even + even = even). Thus, we must select either:

   • 1 odd and 3 even, or
   • 3 odd and 1 even.

3. Calculate the number of favorable outcomes:

   • Case 1 (1 odd, 3 even):
     Number of ways = C(6,1) × C(5,3) = 6 × 10 = 60 ways
   • Case 2 (3 odd, 1 even):
     Number of ways = C(6,3) × C(5,1) = 20 × 5 = 100 ways
   • Total favorable outcomes = 60 + 100 = 160 ways

4. Calculate total possible outcomes:
   Total ways to select any 4 tickets from 11 = C(11,4) = (11×10×9×8)/(4×3×2×1) = 330 ways

5. Calculate probability:
   Probability = favorable outcomes / total outcomes = 160 / 330 ≈ 0.4848 ≈ 0.48

Thus, the correct answer is approximately 0.48, corresponding to option "d".

{"correct_answer_key":"d"}

To solve this problem, we need to consider the conditions for the product of the numbers to be even and the sum to be odd.

1. Product is even: For the product of numbers to be even, at least one of the numbers must be even. In the range from 1 to 11, there are 5 even numbers (2, 4, 6, 8, 10).

2. Sum is odd: For the sum of numbers to be odd, there must be an odd number of odd numbers. In the range from 1 to 11, there are 6 odd numbers (1, 3, 5, 7, 9, 11).

To satisfy both conditions simultaneously, we need to select 3 odd numbers and 1 even number (since 3 odd numbers will sum to an odd number, and 1 even number will ensure the product is even).

• Choose 3 odd numbers from 6: (\binom{6}{3} = 20) ways.
• Choose 1 even number from 5: (\binom{5}{1} = 5) ways.

The total number of favorable outcomes is (20 \times 5 = 100).

The total number of ways to choose any 4 numbers from 11 is (\binom{11}{4} = 330).

Thus, the probability is (\frac{100}{330} = \frac{10}{33} \approx 0.303).

Therefore, the closest answer is option A: 0.30.

{"correct_answer_key":"a"}

ในการหาความน่าจะเป็นที่สลากที่หยิบมามีผลคูณของหมายเลขเป็นจำนวนคู่และผลบวกของหมายเลขเป็นจำนวนคี่ เราต้องพิจารณาคุณสมบัติของจำนวนคู่และจำนวนคี่:

1. ผลคูณของหมายเลขจะเป็นจำนวนคู่เมื่อมีจำนวนเลขคู่ในกลุ่มที่หยิบมามากกว่าหรือเท่ากับ 1
2. ผลบวกของหมายเลขจะเป็นจำนวนคี่เมื่อมีจำนวนเลขคี่ในกลุ่มที่หยิบมามากกว่าหรือเท่ากับ 1

จากการคำนวณและวิเคราะห์ความน่าจะเป็นในกรณีนี้ จะได้ผลลัพธ์ที่ถูกต้องคือ 0.36 ซึ่งตรงกับตัวเลือก b.

{"correct_answer_key":"b"}

มีเลขคี่ 6 ใบ และเลขคู่ 5 ใบ. ต้องการผลรวมเป็นคี่จึงต้องหยิบเลขคี่ 1 หรือ 3 ใบ และเพื่อให้ผลคูณเป็นคู่ต้องมีเลขคู่อย่างน้อย 1 ใบ (ทั้งสองกรณีเข้าเงื่อนไข) จำนวนวิธี = C(6,1)C(5,3) + C(6,3)C(5,1) = 60 + 100 = 160 จากทั้งหมด C(11,4) = 330 ความน่าจะเป็น = 160/330 = 16/33 ≈ 0.48

{"correct_answer_key":"d"}

มีเลขคี่ 6 ตัว (1,3,...,11) และเลขคู่ 5 ตัว (2,4,...,10). ผลบวกเป็นคี่เมื่อเลือกจำนวนเลขคี่เป็น 1 หรือ 3 ข้อเสนอที่เป็นไปได้: (1 奇,3 偶)=C(6,1)C(5,3)=6·10=60 และ (3 奇,1 偶)=C(6,3)C(5,1)=20·5=100 รวมเป็น 160 ชุดจากทั้งหมด C(11,4)=330 ดังนั้นความน่าจะเป็น = 160/330 = 16/33 ≈ 0.48 → ข้อ d.

{"correct_answer_key":"d"}

มีสลาก 11 ใบ แบ่งเป็นคี่ 6 ใบ และคู่ 5 ใบ เพื่อให้ผลรวมเป็นจำนวนคี่ ต้องมีจำนวนคี่ในชุดที่หยิบเป็นจำนวนคี่ (1 หรือ 3)

• กรณี 1 คี่ และ 3 คู่: C(6,1)C(5,3) = 610 = 60
• กรณี 3 คี่ และ 1 คู่: C(6,3)C(5,1) = 205 = 100

รวมวิธีที่เป็นไปได้ 60+100 = 160 จากทั้งหมด C(11,4) = 330 วิธี ดังนั้นความน่าจะเป็น = 160/330 = 16/33 ≈ 0.4848 ≈ 0.48

{"correct_answer_key":"d"}

The tickets are numbered 1‑11 (5 even, 6 odd).
To have an even product we need at least one even number.
For the sum to be odd, the number of odd tickets drawn must be odd → either 1 odd + 3 evens or 3 odds + 1 even.

• 1 odd, 3 evens: C(6,1)·C(5,3)=6·10=60
• 3 odds, 1 even: C(6,3)·C(5,1)=20·5=100

Favourable selections = 60 + 100 = 160.
Total possible 4‑ticket selections = C(11,4)=330.

Probability = 160/330 = 16/33 ≈ 0.485 ≈ 0.48.

Correct answer: option d.

{"correct_answer_key":"d"}

The probability is calculated by counting the favorable selections of 4 tickets out of the 11 numbered tickets (1–11).

• There are 6 odd numbers and 5 even numbers.
• For the product to be even, at least one even number must be chosen.
• For the sum to be odd, the number of odd numbers chosen must be odd (1 or 3).

Case 1: 1 odd + 3 evens
( \binom{6}{1}\binom{5}{3}=6\times10=60 )

Case 2: 3 odds + 1 even
( \binom{6}{3}\binom{5}{1}=20\times5=100 )

Total favorable combinations = (60+100=160).
Total possible combinations = (\binom{11}{4}=330).

[ P=\frac{160}{330}=\frac{16}{33}\approx0.485 ]

The closest option is 0.48 (option d).

{"correct_answer_key":"d"}

Let's solve this probability problem step by step. We have 11 tickets numbered from 1 to 11, each with a unique number. We draw 4 tickets at random and need to find the probability that the product of the numbers on the tickets is even, while the sum of the numbers is odd.

1. Total number of ways to draw 4 tickets out of 11: This is a combination problem, calculated as C(11,4) = 330.

2. Condition for product to be even: The product of numbers is even if at least one of the numbers is even. So, we need at least one even number among the 4 tickets.

3. Condition for sum to be odd: The sum of numbers is odd if the number of odd numbers among the 4 tickets is odd (i.e., 1 or 3 odd numbers).

4. Classify numbers: From 1 to 11, there are 6 odd numbers (1,3,5,7,9,11) and 5 even numbers (2,4,6,8,10).

5. Favorable outcomes: We need combinations of 4 tickets where:

   • There is at least 1 even number (for product to be even).
   • The number of odd numbers is 1 or 3 (for sum to be odd).
   • Possible cases:
     • 1 odd and 3 even: C(6,1) * C(5,3) = 6 * 10 = 60
     • 3 odd and 1 even: C(6,3) * C(5,1) = 20 * 5 = 100
   • Total favorable outcomes = 60 + 100 = 160

6. Probability: Probability = Favorable outcomes / Total outcomes = 160 / 330 ≈ 0.4848

7. Match with options: The closest value to 0.4848 is 0.48, which corresponds to option D.

{"correct_answer_key":"d"}

Brief Explanation:

This question involves 11 tickets numbered from 1 to 11, with 6 odd numbers (1, 3, 5, 7, 9, 11) and 5 even numbers (2, 4, 6, 8, 10). We need to find the probability of drawing 4 tickets where the product of the numbers is even (at least one even number) and the sum is odd (exactly 1 or 3 odd numbers).

• Favorable cases:
  • Exactly 1 odd and 3 even: C(6,1) × C(5,3) = 6 × 10 = 60
  • Exactly 3 odd and 1 even: C(6,3) × C(5,1) = 20 × 5 = 100
• Total favorable: 60 + 100 = 160
• Total possible outcomes: C(11,4) = 330
• Probability: 160 / 330 ≈ 0.4848, which rounds to 0.48.

Thus, the correct answer is option D.

{"correct_answer_key":"d"}

Brief Explanation:

This question involves calculating the probability of drawing 4 tickets from 11 tickets numbered 1 to 11, where the product of the numbers is even (at least one even number) and the sum is odd (an odd count of odd numbers). There are 6 odd numbers and 5 even numbers. The favorable outcomes are:

• Exactly 1 odd and 3 even: C(6,1) * C(5,3) = 6 * 10 = 60 ways
• Exactly 3 odd and 1 even: C(6,3) * C(5,1) = 20 * 5 = 100 ways
  Total favorable outcomes: 60 + 100 = 160. Total possible outcomes: C(11,4) = 330. Probability: 160/330 ≈ 0.4848, which matches option D (0.48).

{"correct_answer_key":"d"}

Let's solve this probability problem step by step. We have 11 tickets numbered from 1 to 11, each with a unique number. We draw 4 tickets at random and need to find the probability that the product of the numbers on the tickets is even, while the sum of the numbers is odd.

1. Total number of ways to draw 4 tickets out of 11: This is a combination problem, calculated as C(11,4) = 330.

2. Condition for product to be even: The product of numbers is even if at least one of the numbers is even. So, we need at least one even number among the 4 tickets drawn.

3. Condition for sum to be odd: The sum of numbers is odd if the number of odd numbers among the 4 tickets is odd (i.e., 1 or 3 odd numbers).

4. Classify numbers: From 1 to 11, there are 6 odd numbers (1,3,5,7,9,11) and 5 even numbers (2,4,6,8,10).

5. Favorable outcomes: We need combinations of 4 numbers where:

   • There is at least 1 even number (for product to be even).
   • The number of odd numbers is 1 or 3 (for sum to be odd). Let's calculate the favorable cases:
   • Case 1: 1 odd and 3 even
     • Ways to choose 1 odd from 6: C(6,1) = 6
     • Ways to choose 3 even from 5: C(5,3) = 10
     • Total for this case: 6 * 10 = 60
   • Case 2: 3 odd and 1 even
     • Ways to choose 3 odd from 6: C(6,3) = 20
     • Ways to choose 1 even from 5: C(5,1) = 5
     • Total for this case: 20 * 5 = 100
   • Total favorable outcomes = 60 + 100 = 160

6. Probability: Probability = Favorable outcomes / Total outcomes = 160 / 330 ≈ 0.4848

7. Match with options: The closest value to 0.4848 among the given options (0.30, 0.36, 0.42, 0.48, 0.54) is 0.48.

Therefore, the correct answer corresponds to option D.

{"correct_answer_key":"d"}

{ "explanation": "There are 6 odd and 5 even numbers from 1 to 11. The sum of 4 numbers is odd if there are 1 or 3 odds. These cases always include at least one even, making the product even. Thus, the probability is the probability of an odd sum: (C(6,1)*C(5,3) + C(6,3)*C(5,1)) / C(11,4) = 160/330 ≈ 0.48.", "correct_answer_key": "d" }

อธิบายสั้น ๆ

• สลาก 11 ใบ มีเลข 1–11 (เลขคู่ 5 ใบ: 2,4,6,8,10 และเลขคี่ 6 ใบ: 1,3,5,7,9,11)
• ต้องการ “ผลคูณเป็นเลขคู่” (ต้องมีเลขคู่อย่างน้อย 1 ใบ) และ “ผลบวกเป็นเลขคี่” (จำนวนเลขคี่ที่เลือกต้องเป็นเลขคี่)
• นับเฉพาะกลุ่ม 4 ใบที่มีเลขคู่ 1,3 หรือ 4 ใบ (เพื่อให้ผลคูณคู่) และจำนวนเลขคี่ในกลุ่มนั้นต้องเป็น 1 หรือ 3 ใบ (เพื่อให้ผลบวกคี่)
• คำนวณจำนวนวิธีที่ตรงเงื่อนไขแล้วหารด้วย C(11,4)=330 จะได้ความน่าจะเป็น ≈ 0.3636 ≈ 0.36

{"correct_answer_key":"b"}

เพื่อให้ผลคูณของหมายเลขเป็นจำนวนคู่ ต้องมีเลขคู่อย่างน้อย 1 ตัวในสลากที่หยิบมา และเพื่อให้ผลบวกของหมายเลขเป็นจำนวนคี่ ต้องมีเลขคี่เป็นจำนวนคี่ในสลากที่หยิบมา

มีเลขคู่ 5 ตัว (2, 4, 6, 8, 10) และเลขคี่ 6 ตัว (1, 3, 5, 7, 9, 11)

เราต้องการเลือกเลขคู่ 1 ตัว และเลขคี่ 3 ตัว หรือ เลือกเลขคู่ 3 ตัว และเลขคี่ 1 ตัว

จำนวนชุดค่าผสมที่เป็นไปได้ทั้งหมดคือ 11C4 = 330

จำนวนชุดค่าผสมที่เลือกเลขคู่ 1 ตัว และเลขคี่ 3 ตัว คือ 5C1 * 6C3 = 5 * 20 = 100

จำนวนชุดค่าผสมที่เลือกเลขคู่ 3 ตัว และเลขคี่ 1 ตัว คือ 5C3 * 6C1 = 10 * 6 = 60

ดังนั้น ความน่าจะเป็นที่ต้องการคือ (100 + 60) / 330 = 160 / 330 = 0.485 (โดยประมาณ) แต่ไม่มีตัวเลือกที่ตรงกับค่านี้ จึงต้องเลือกตัวเลือกที่ใกล้ที่สุด

{"correct_answer_key":"d"}

To solve this, let's break it down into steps.

1. Total number of ways to pick 4 slips out of 11: This can be calculated using combinations. The total number of ways is (C(11, 4) = \frac{11!}{4!(11-4)!} = \frac{11!}{4!7!} = 330).

2. Condition for the product of the numbers to be even: The product will be even if at least one of the numbers is even. There are 5 even numbers (2, 4, 6, 8, 10) and 6 odd numbers (1, 3, 5, 7, 9, 11). The opposite scenario (all odd) means choosing 4 numbers from the 6 odd numbers, which is (C(6, 4) = 15). So, the number of ways to have at least one even number is (330 - 15 = 315).

3. Condition for the sum of the numbers to be odd: The sum will be odd if there is either 1 or 3 odd numbers among the 4 picked.

   • For exactly 1 odd number: (C(6, 1) \times C(5, 3) = 6 \times 10 = 60).
   • For exactly 3 odd numbers: (C(6, 3) \times C(5, 1) = 20 \times 5 = 100). So, the total number of ways to have an odd sum is (60 + 100 = 160).

4. Combining both conditions: We need to find the overlap where the product is even and the sum is odd. Since having an even product requires at least one even number, and having an odd sum requires either 1 or 3 odd numbers, the cases that satisfy both are the ones we calculated for having an odd sum (because having 1 or 3 odd numbers inherently means having at least one even number when picking 4 numbers). Thus, the number of favorable outcomes is 160.

5. Calculating the probability: The probability is (\frac{160}{330}).

Let's calculate it: (\frac{160}{330} \approx 0.4848).

Rounding to two decimal places gives approximately 0.48.

{"correct_answer_key":"d"}

To solve this, let's break down the conditions:

1. The product of the numbers is even: This happens if at least one of the numbers drawn is even.
2. The sum of the numbers is odd: This happens if there is either one, three, or five odd numbers drawn. Since we are drawing 4 numbers, the sum will be odd only if there is either one or three odd numbers drawn.

The total number of ways to draw 4 cards out of 11 is ( \binom{11}{4} ).

There are 5 odd numbers (1, 3, 5, 7, 9, 11) and 5 even numbers (2, 4, 6, 8, 10). Note that there are actually 6 odd numbers.

For the product to be even and the sum to be odd, we need:

• 1 odd and 3 even numbers, or
• 3 odd and 1 even number.

Calculating the number of ways for 1 odd and 3 even:

• Ways to choose 1 odd out of 6: ( \binom{6}{1} )
• Ways to choose 3 even out of 5: ( \binom{5}{3} )
• Total for this case: ( \binom{6}{1} \times \binom{5}{3} )

Calculating the number of ways for 3 odd and 1 even:

• Ways to choose 3 odd out of 6: ( \binom{6}{3} )
• Ways to choose 1 even out of 5: ( \binom{5}{1} )
• Total for this case: ( \binom{6}{3} \times \binom{5}{1} )

Compute the values:

( \binom{11}{4} = 330 )

( \binom{6}{1} = 6 ), ( \binom{5}{3} = 10 ), so ( 6 \times 10 = 60 )

( \binom{6}{3} = 20 ), ( \binom{5}{1} = 5 ), so ( 20 \times 5 = 100 )

Total favorable outcomes = ( 60 + 100 = 160 )

Probability:

( P = \frac{160}{330} \approx 0.4848 )

Rounded to two decimal places, this is approximately 0.48.

{"correct_answer_key":"d"}

To solve this problem, we need to calculate the probability of drawing four tickets from a set of 11 tickets (numbered 1 to 11) such that the product of the numbers is even, and the sum of the numbers is odd.

First, let's consider the conditions:

1. The product of the numbers is even: This means at least one of the numbers must be even.
2. The sum of the numbers is odd: This means there must be an odd number of odd numbers in the selection.

The even numbers in the set are 2, 4, 6, 8, and 10 (5 even numbers). The odd numbers in the set are 1, 3, 5, 7, 9, and 11 (6 odd numbers).

To satisfy both conditions, we need to select:

• An odd number of odd numbers (either 1 or 3 odd numbers).
• The rest must be even numbers to ensure the product is even.

Let's calculate the number of ways to choose 1 odd number and 3 even numbers:

• Choose 1 odd number from 6: ( \binom{6}{1} = 6 )
• Choose 3 even numbers from 5: ( \binom{5}{3} = 10 )

The total number of ways to choose 4 tickets out of 11 is: ( \binom{11}{4} = 330 )

So, the probability is: ( \frac{6 \times 10}{330} = \frac{60}{330} = \frac{2}{11} \approx 0.18 )

However, we need to consider all possible combinations that satisfy the conditions, including choosing 3 odd numbers and 1 even number. Let's calculate that:

• Choose 3 odd numbers from 6: ( \binom{6}{3} = 20 )
• Choose 1 even number from 5: ( \binom{5}{1} = 5 )

The probability for this case is: ( \frac{20 \times 5}{330} = \frac{100}{330} = \frac{10}{33} \approx 0.30 )

Adding both probabilities: ( 0.18 + 0.30 = 0.48 )

Thus, the correct answer is D.

{"correct_answer_key":"d"}

To solve this problem, we need to consider the conditions given: the product of the numbers on the selected tickets must be even, and the sum of the numbers must be odd.

1. Identify the even and odd numbers:

   • Even numbers: 2, 4, 6, 8, 10
   • Odd numbers: 1, 3, 5, 7, 9, 11

2. Determine the conditions for the product to be even:

   • The product is even if at least one of the selected numbers is even.

3. Determine the conditions for the sum to be odd:

   • The sum is odd if the number of odd numbers selected is odd (1 or 3).

4. Calculate the total number of ways to select 4 tickets:

   • Total combinations: ( \binom{11}{4} = 330 )

5. Calculate the number of favorable outcomes:

   • We need to count the combinations where the product is even and the sum is odd.
   • We can break this down into cases based on the number of even numbers selected:
     • Case 1: 1 even number and 3 odd numbers
     • Case 2: 3 even numbers and 1 odd number

6. Calculate the number of ways for each case:

   • Case 1:
     • Choose 1 even number from 5: ( \binom{5}{1} = 5 )
     • Choose 3 odd numbers from 6: ( \binom{6}{3} = 20 )
     • Total for Case 1: ( 5 \times 20 = 100 )
   • Case 2:
     • Choose 3 even numbers from 5: ( \binom{5}{3} = 10 )
     • Choose 1 odd number from 6: ( \binom{6}{1} = 6 )
     • Total for Case 2: ( 10 \times 6 = 60 )

7. Sum the favorable outcomes:

   • Total favorable outcomes: ( 100 + 60 = 160 )

8. Calculate the probability:

   • Probability: ( \frac{160}{330} \approx 0.4848 )

Given the options, the closest value to 0.4848 is 0.48.

{"correct_answer_key":"d"}

To solve this problem, we need to determine the probability that the product of the four drawn numbers is even, but their sum is odd.

First, let's understand the properties of the numbers from 1 to 11:

• Even numbers: 2, 4, 6, 8, 10
• Odd numbers: 1, 3, 5, 7, 9, 11

For the product of four numbers to be even, at least one of the numbers must be even. Conversely, for the sum of four numbers to be odd, there must be an odd number of odd numbers among them. This means we can have either three odd numbers and one even number, or one odd number and three even numbers.

Let's calculate the total number of ways to choose 4 numbers out of 11: [ \binom{11}{4} = 330 ]

Next, we need to count the favorable outcomes:

1. Three odd numbers and one even number:

   • Number of ways to choose 3 odd numbers from 6: ( \binom{6}{3} = 20 )
   • Number of ways to choose 1 even number from 5: ( \binom{5}{1} = 5 )
   • Total ways for this case: ( 20 \times 5 = 100 )

2. One odd number and three even numbers:

   • Number of ways to choose 1 odd number from 6: ( \binom{6}{1} = 6 )
   • Number of ways to choose 3 even numbers from 5: ( \binom{5}{3} = 10 )
   • Total ways for this case: ( 6 \times 10 = 60 )

Adding these two cases together gives us the total number of favorable outcomes: [ 100 + 60 = 160 ]

Finally, the probability is the number of favorable outcomes divided by the total number of outcomes: [ \frac{160}{330} = \frac{16}{33} \approx 0.4848 ]

Rounding to the nearest given option, we get: [ \boxed{0.48} ]

So the correct answer is:

{"correct_answer_key":"d"}

To solve this problem, we need to find the combinations of 4 numbers from 1 to 11 where the product is even (which means at least one number is even) and the sum is odd (which means an odd number of odd numbers).

The even numbers between 1 and 11 are 2, 4, 6, 8, 10 (5 even numbers). The odd numbers between 1 and 11 are 1, 3, 5, 7, 9, 11 (6 odd numbers).

For the product to be even, we need at least one even number. For the sum to be odd, we need an odd number of odd numbers, which could be 1, 3, or 5 odd numbers.

Let's consider the cases:

1. 1 even and 3 odd: C(5,1) * C(6,3)
2. 3 even and 1 odd: C(5,3) * C(6,1)

Calculating these:

1. C(5,1) * C(6,3) = 5 * 20 = 100
2. C(5,3) * C(6,1) = 10 * 6 = 60

Total favorable outcomes = 100 + 60 = 160

Total possible outcomes when choosing 4 tickets out of 11 = C(11,4) = 330

Probability = Favorable outcomes / Total outcomes = 160 / 330 ≈ 0.4848

The closest option to this value is 0.48.

{"correct_answer_key":"d"}

คำอธิบายโดยย่อ:
• ต้องการให้ผลคูณเป็นจำนวนคู่ → ชุดตัวเลขที่หยิบมาต้องมีจำนวนคู่อย่างน้อย 1 ตัว
• ต้องการให้ผลบวกเป็นจำนวนคี่ → ในการหยิบ 4 ใบ ต้องมีจำนวนนับคี่ (odd) อยู่ 1 หรือ 3 ตัว (เนื่องจากผลบวกของตัวเลขคี่ 1 หรือ 3 ตัวกับตัวเลขคู่จะได้ผลบวกเป็นคี่)

นับจำนวนกรณีที่สอดคล้องกับเงื่อนไข:

1. มีคี่ 1 ใบ และคู่ 3 ใบ → C(6,1)×C(5,3) = 6×10 = 60 กรณี
2. มีคี่ 3 ใบ และคู่ 1 ใบ → C(6,3)×C(5,1) = 20×5 = 100 กรณี

รวม 160 กรณี จากทั้งหมด C(11,4) = 330 กรณี
ดังนั้น โอกาส = 160/330 ≈ 0.4848 ≈ 0.48

ดังนั้นคำตอบที่ถูกต้องคือ ข้อ (d)

{"correct_answer_key":"d"}

มีสลากเลขคี่ 6 ใบ (1,3,5,7,9,11) และเลขคู่ 5 ใบ (2,4,6,8,10)
สุ่มหยิบ 4 ใบ ต้องได้

• ผลบวกเป็นคี่ → ต้องมีจำนวนเลขคี่เป็น 1 หรือ 3
• ผลคูณเป็นคู่ → ต้องมีอย่างน้อย 1 เลขคู่ (เงื่อนไขนี้เป็นจริงอยู่แล้วถ้ามี 1 หรือ 3 เลขคี่)

นับจำนวนวิธีที่เป็นไปได้

1. 1 เลขคี่ + 3 เลขคู่ :
   C(6,1) × C(5,3) = 6 × 10 = 60

2. 3 เลขคี่ + 1 เลขคู่ :
   C(6,3) × C(5,1) = 20 × 5 = 100

รวมวิธีที่ทำได้ = 60 + 100 = 160

จำนวนวิธีสุ่ม 4 ใบจาก 11 ใบ = C(11,4) = 330

ความน่าจะเป็น = 160 / 330 = 16 / 33 ≈ 0.48
ตรงกับตัวเลือก d.

{"correct_answer_key":"d"}

มีสลากหมายเลข 1‑11
– จำนวนเลขคี่ = 6 (1,3,5,7,9,11)
– จำนวนเลขคู่ = 5 (2,4,6,8,10)

ต้องการให้

1. ผลคูณเป็นคู่ → ต้องมีเลขคู่อย่างน้อย 1 ตัว
2. ผลบวกรวมเป็นคี่ → ในบรรดา 4 ตัว ต้องมีจำนวนเลขคี่เป็นจำนวนคี่ (1 หรือ 3 ตัว)

กรณีที่สอดคล้อง
A) 1 เลขคี่ + 3 เลขคู่ : C(6,1)·C(5,3) = 6·10 = 60
B) 3 เลขคี่ + 1 เลขคู่ : C(6,3)·C(5,1) = 20·5 = 100

จำนวนชุดที่เป็นไปได้ = 60 + 100 = 160
จำนวนชุดทั้งหมด = C(11,4) = 330

ความน่าจะเป็น = 160 / 330 = 16/33 ≈ 0.48 → ตรงกับตัวเลือก d

{"correct_answer_key":"d"}

There are 6 odd numbers (1, 3, 5, 7, 9, 11) and 5 even numbers (2, 4, 6, 8, 10).

For the sum to be odd, we must pick an odd count of odd numbers ⇒ either 1 odd + 3 evens or 3 odds + 1 even.
Both cases automatically make the product even because each includes at least one even number.

1 odd & 3 evens: C(6,1) × C(5,3) = 6 × 10 = 60
3 odds & 1 even: C(6,3) × C(5,1) = 20 × 5 = 100
Favourable selections = 60 + 100 = 160

Total ways to choose 4 from 11: C(11,4) = 330

Probability = 160 / 330 = 16/33 ≈ 0.48

{"correct_answer_key":"d"}

There are 6 odd numbers and 5 even numbers among 1 to 11. For the sum to be odd, you must have an odd number of odd numbers. Since you're choosing 4 numbers, the only possibilities are: • 1 odd and 3 even, or • 3 odd and 1 even. For 1 odd and 3 even, the number of ways is C(6,1) × C(5,3) = 6 × 10 = 60. For 3 odd and 1 even, the number of ways is C(6,3) × C(5,1) = 20 × 5 = 100. Total favorable outcomes = 60 + 100 = 160. Total outcomes when choosing 4 from 11 is C(11,4) = 330. Thus, the probability = 160/330 = 16/33 ≈ 0.48.

{"correct_answer_key": "d"}

ในการสุ่มหยิบสลาก 4 ใบจากทั้งหมด 11 ใบ (เลข 1 ถึง 11 มีอยู่ 6 ใบที่เป็นเลขคี่และ 5 ใบที่เป็นเลขคู่) ให้พิจารณาเงื่อนไขทั้ง 2 คือ ผลคูณเป็นจำนวนคู่ (ต้องมีเลขคู่อย่างน้อย 1 ใบ) และผลบวกเป็นจำนวนคี่ (ต้องมีเลขคี่อยู่จำนวนคี่) ซึ่งจะเป็นกรณีได้เฉพาะเมื่อมีเลขคี่ 1 หรือ 3 ใบ

• กรณีที่มีเลขคี่ 1 ใบและเลขคู่ 3 ใบ จำนวนวิธี = C(6,1) × C(5,3) = 6 × 10 = 60
• กรณีที่มีเลขคี่ 3 ใบและเลขคู่ 1 ใบ จำนวนวิธี = C(6,3) × C(5,1) = 20 × 5 = 100

รวมวิธีที่เป็นไปได้ = 60 + 100 = 160

จำนวนวิธีสุ่ม 4 ใบจาก 11 ใบ = C(11,4) = 330

ดังนั้น ความน่าจะเป็น = 160 / 330 = 16/33 ≈ 0.4848
ซึ่งใกล้เคียงกับ 0.48

{"correct_answer_key": "d"}

The tickets are numbered 1 to 11. Note that there are 6 odd numbers and 5 even numbers. For the product of four numbers to be even, at least one drawn number must be even. For the sum to be odd, an odd number of odd numbers must be drawn (since the sum of even numbers is even, and adding an odd number results in an odd sum). Thus, the possible cases when drawing 4 tickets are:

1. Exactly 1 odd and 3 even numbers (sum: odd + even*3 = odd).
2. Exactly 3 odd and 1 even number (sum: odd*3 + even = odd).

For case 1: • Ways to choose 1 odd from 6: C(6,1) = 6. • Ways to choose 3 even from 5: C(5,3) = 10. Total ways for this case = 6 × 10 = 60.

For case 2: • Ways to choose 3 odd from 6: C(6,3) = 20. • Ways to choose 1 even from 5: C(5,1) = 5. Total ways for this case = 20 × 5 = 100.

Total favorable outcomes = 60 + 100 = 160.

The total number of ways to draw 4 tickets from 11 is C(11,4) = 330.

So, the probability = 160 / 330 = 16/33 ≈ 0.4848, which is approximately 0.48.

{"correct_answer_key":"d"}

มีตัวเลขคี่ 6 ตัว (1,3,5,7,9,11) และตัวเลขคู่ 5 ตัว (2,4,6,8,10) เลือก 4 ใบที่ผลคูณเป็นคู่ (ต้องมีเลขคู่ ≥1) และผลบวกเป็นคี่ (#เลขคี่ ต้องเป็น 1 หรือ 3)
วิธีทั้งหมด = C(6,1)·C(5,3) + C(6,3)·C(5,1) = 60 + 100 = 160
ความน่าจะเป็น = 160 / C(11,4) = 160 / 330 ≈ 0.48

{"correct_answer_key":"d"}

First, there are C(11,4)=330 ways to pick 4 tickets.
Sum is odd iff we pick 1 or 3 odd numbers; product is even iff we have at least one even.
Case 1 (1 odd, 3 evens): C(6,1)C(5,3)=610=60
Case 2 (3 odds, 1 even): C(6,3)C(5,1)=205=100
Total favorable = 160, so P = 160/330 ≈ 0.4848 ≈ 0.48 → choice d.

{"correct_answer_key":"d"}

มีทั้งหมด C(11,4)=330 วิธี สุ่มได้สินค้าที่ผลคูณเป็นคู่และผลบวกเป็นคี่ ต้องเลือกเลขคี่มาเป็นจำนวน 1 หรือ 3 ตัว พร้อมเลขคู่ที่เหลือ
กรณี 1 คี่+3 คู่: C(6,1)·C(5,3)=6·10=60
กรณี 3 คี่+1 คู่: C(6,3)·C(5,1)=20·5=100
รวม 160/330 ≈ 0.48

{"correct_answer_key":"d"}

To solve this problem, we need to determine the probability that the product of the numbers on the 4 drawn tickets is even, and the sum of the numbers is odd.

1. Product is even: For the product of numbers to be even, at least one of the numbers must be even. The even numbers between 1 and 11 are 2, 4, 6, 8, and 10. There are 5 even numbers.

2. Sum is odd: For the sum of numbers to be odd, there must be an odd number of odd numbers. The odd numbers between 1 and 11 are 1, 3, 5, 7, 9, and 11. There are 6 odd numbers.

To satisfy both conditions, we can have:

• 1 even number and 3 odd numbers (since 3 odd numbers sum to an odd number, and 1 even number makes the product even).

Calculating the probability:

• Total ways to choose 4 tickets from 11: (\binom{11}{4} = 330).

• Ways to choose 1 even number from 5: (\binom{5}{1} = 5).

• Ways to choose 3 odd numbers from 6: (\binom{6}{3} = 20).

• Total favorable outcomes: (5 \times 20 = 100).

• Probability: (\frac{100}{330} = \frac{10}{33} \approx 0.303).

The closest option to 0.303 is 0.30.

{"correct_answer_key":"a"}

Explanation:

The problem involves selecting 4 tickets from a set of 11 tickets numbered 1 through 11. We are tasked with finding the probability that:

1. The product of the numbers on the selected tickets is even.

   • For the product to be even, at least one of the selected numbers must be even. The even numbers in the set are {2, 4, 6, 8, 10}, and there are 5 even numbers out of the 11 total numbers.

2. The sum of the numbers on the selected tickets is odd.

   • For the sum to be odd, there must be an odd number of odd numbers among the selected tickets. The odd numbers in the set are {1, 3, 5, 7, 9, 11}, and there are 6 odd numbers out of the 11 total numbers.

Key Observations:

• To satisfy both conditions:
  • At least one number must be even (to make the product even).
  • The total count of odd numbers in the selection must be odd (to make the sum odd).

Calculation:

1. Total ways to choose 4 tickets from 11: $$ \binom{11}{4} = 330 $$

2. Case Analysis:

   • To satisfy the condition, we analyze the possible distributions of odd and even numbers in the selection:

     • Case 1: 1 odd and 3 even.
     • Case 2: 3 odd and 1 even.

   • Case 1: 1 odd and 3 even

     • Number of ways to choose 1 odd from 6 odd numbers: $\binom{6}{1} = 6$.
     • Number of ways to choose 3 even from 5 even numbers: $\binom{5}{3} = 10$.
     • Total ways for this case: $6 \times 10 = 60$.

   • Case 2: 3 odd and 1 even

     • Number of ways to choose 3 odd from 6 odd numbers: $\binom{6}{3} = 20$.
     • Number of ways to choose 1 even from 5 even numbers: $\binom{5}{1} = 5$.
     • Total ways for this case: $20 \times 5 = 100$.

   • Total favorable outcomes: $$ 60 + 100 = 160 $$

3. Probability: $$ \text{Probability} = \frac{\text{Favorable outcomes}}{\text{Total outcomes}} = \frac{160}{330} \approx 0.48 $$

Final Answer:

The correct probability is approximately 0.48, which corresponds to option d.

{"correct_answer_key":"d"}

ในการแก้ปัญหานี้ เราต้องคำนวณความน่าจะเป็นที่ผลคูณของหมายเลขเป็นจำนวนคู่ และผลบวกเป็นจำนวนคี่ โดยมีเงื่อนไขดังนี้:

1. ผลคูณเป็นจำนวนคู่: ต้องมีเลขคู่อย่างน้อย 1 ตัว (ไม่ใช่เลขคี่ทั้งหมด)
2. ผลบวกเป็นจำนวนคี่: จำนวนเลขคี่ในสลากที่เลือกต้องเป็นจำนวนคี่ (1 หรือ 3 ตัว)

ขั้นตอนการคำนวณ:

• จำนวนสลากทั้งหมด: 11 ใบ (เลข 1-11) แบ่งเป็นเลขคี่ 6 ตัว (1,3,5,7,9,11) และเลขคู่ 5 ตัว (2,4,6,8,10)
• จำนวนวิธีเลือกสลาก 4 ใบจาก 11 ใบ: $ \binom{11}{4} = 330 $
• กรณีที่ 1: เลือกเลขคี่ 1 ตัวและเลขคู่ 3 ตัว → $ \binom{6}{1} \times \binom{5}{3} = 6 \times 10 = 60 $
• กรณีที่ 2: เลือกเลขคี่ 3 ตัวและเลขคู่ 1 ตัว → $ \binom{6}{3} \times \binom{5}{1} = 20 \times 5 = 100 $
• จำนวนวิธีทั้งหมดที่เป็นไปได้: $ 60 + 100 = 160 $
• ความน่าจะเป็น: $ \frac{160}{330} \approx 0.4848 $ ซึ่งใกล้เคียงกับตัวเลือก d (0.48)

{"correct_answer_key":"d"}

{"correct_answer_key":"d"}

The problem requires calculating the probability that the product of four randomly selected numbers from 1 to 11 is even and their sum is odd.

1. Total possible outcomes:
   $ \binom{11}{4} = 330 $.

2. Favorable outcomes:

   • The sum is odd when there are 1 or 3 odd numbers in the four selected (since an odd sum requires an odd count of odd numbers).
   • 1 odd and 3 even:
     $ \binom{6}{1} \times \binom{5}{3} = 6 \times 10 = 60 $.
   • 3 odd and 1 even:
     $ \binom{6}{3} \times \binom{5}{1} = 20 \times 5 = 100 $.
   • Total favorable: $ 60 + 100 = 160 $.

3. Probability:
   $ \frac{160}{330} \approx 0.4848 $, which rounds to 0.48 (option D).

{"correct_answer_key":"d"}

เราต้องหาความน่าจะเป็นที่เมื่อสุ่มหยิบสลาก 4 ใบจากเลข 1 ถึง 11 (ไม่ซ้ำกัน) จะได้:

• ผลคูณเป็นจำนวนคู่ → ต้องมี อย่างน้อยหนึ่งเลขคู่ ใน 4 ใบ (เพราะถ้าทั้งหมดเป็นเลขคี่ ผลคูณจะเป็นคี่)
• ผลบวกเป็นจำนวนคี่ → ผลบวกของเลข 4 จำนวนจะเป็นคี่ ก็ต่อเมื่อ มีจำนวนเลขคี่อยู่จำนวนคี่ใบ (เช่น 1 หรือ 3 ใบ) เพราะเลขคู่บวกกันได้คู่เสมอ

แต่เราต้องมี ทั้งผลคูณเป็นคู่ และ ผลบวกเป็นคี่ พร้อมกัน

วิเคราะห์เงื่อนไขร่วมกัน:

• ผลคูณเป็นคู่ → ต้องมี อย่างน้อย 1 เลขคู่
• ผลบวกเป็นคี่ → ต้องมี จำนวนเลขคี่เป็นจำนวนคี่ → คือ 1 หรือ 3 ใบ (เพราะมีทั้งหมด 4 ใบ)

ดังนั้น กรณีที่เป็นไปได้คือ:

• 1 เลขคี่ + 3 เลขคู่ → ผลบวก = คี่ (1 คี่), ผลคูณ = คู่ (มีคู่ 3 ตัว) ✅
• 3 เลขคี่ + 1 เลขคู่ → ผลบวก = คี่ (3 คี่), ผลคูณ = คู่ (มีคู่ 1 ตัว) ✅

ห้าม:

• 0 เลขคู่ (ทั้งหมดคี่) → ผลคูณคี่ ❌
• 2 เลขคี่ + 2 เลขคู่ → ผลบวก = คู่ ❌
• 4 เลขคี่ → ผลคูณคี่ ❌

ดังนั้น กรณีที่ต้องการคือ:

• กรณี A: 1 คี่ + 3 คู่
• กรณี B: 3 คี่ + 1 คู่

จำนวนเลขคี่และคู่ใน 1-11:

• เลขคี่: 1,3,5,7,9,11 → 6 ตัว
• เลขคู่: 2,4,6,8,10 → 5 ตัว

จำนวนวิธีที่เป็นไปได้:

กรณี A: 1 คี่ + 3 คู่

• เลือก 1 จาก 6 คี่: $\binom{6}{1} = 6$
• เลือก 3 จาก 5 คู่: $\binom{5}{3} = 10$
• รวม: $6 \times 10 = 60$

กรณี B: 3 คี่ + 1 คู่

• เลือก 3 จาก 6 คี่: $\binom{6}{3} = 20$
• เลือก 1 จาก 5 คู่: $\binom{5}{1} = 5$
• รวม: $20 \times 5 = 100$

รวมกรณีที่ต้องการ: $60 + 100 = 160$

จำนวนวิธีทั้งหมดในการเลือก 4 ใบจาก 11:

$\binom{11}{4} = \frac{11 \cdot 10 \cdot 9 \cdot 8}{4 \cdot 3 \cdot 2 \cdot 1} = 330$

ความน่าจะเป็น:

$\frac{160}{330} = \frac{16}{33} \approx 0.4848... \approx 0.48$

ดังนั้น คำตอบที่ใกล้เคียงที่สุดคือ 0.48

{"correct_answer_key":"d"}

{"correct_answer_key":"d"}

The problem requires calculating the probability that selecting 4 tickets from 11 (numbered 1–11) results in an even product and an odd sum.

Key Steps:

1. Conditions Analysis:

   • Even Product: At least one even number must be selected.
   • Odd Sum: The count of odd numbers must be odd (1 or 3).

2. Parity Constraints:

   • Since the sum is odd, the number of odd tickets (O) must be 1 or 3.
   • This ensures the number of even tickets (E = 4 - O) is 3 or 1, guaranteeing at least one even number (satisfying the even product condition).

3. Calculations:

   • Odd tickets (1, 3, 5, 7, 9, 11): 6 tickets.
   • Even tickets (2, 4, 6, 8, 10): 5 tickets.
   • Favorable Cases:
     • O = 1 (E = 3): ( \binom{6}{1} \times \binom{5}{3} = 6 \times 10 = 60 )
     • O = 3 (E = 1): ( \binom{6}{3} \times \binom{5}{1} = 20 \times 5 = 100 )
     • Total Favorable: ( 60 + 100 = 160 )
   • Total Possible Cases: ( \binom{11}{4} = 330 )
   • Probability: ( \frac{160}{330} \approx 0.4848 ), which rounds to 0.48.

Answer:

{"correct_answer_key":"d"}

เพื่อให้ผลคูณของหมายเลขเป็นจำนวนคู่ ต้องมีอย่างน้อยหนึ่งหมายเลขคู่ในสลากที่หยิบมา และเพื่อให้ผลบวกของหมายเลขเป็นจำนวนคี่ ต้องมีจำนวนเลขคี่เป็นจำนวนคี่ (1, 3, 5, 7, 9, 11) ในสลากที่หยิบมา

สลากที่มีหมายเลขคู่คือ 2, 4, 6, 8, 10 และสลากที่มีหมายเลขคี่คือ 1, 3, 5, 7, 9, 11

เราต้องเลือก 1 หมายเลขคู่และ 3 หมายเลขคี่ หรือ 3 หมายเลขคู่และ 1 หมายเลขคี่

กรณีที่ 1: เลือก 1 หมายเลขคู่และ 3 หมายเลขคี่

• เลือก 1 หมายเลขคู่จาก 5 หมายเลขคู่: 5C1 = 5
• เลือก 3 หมายเลขคี่จาก 6 หมายเลขคี่: 6C3 = 20
• จำนวนรวม: 5 * 20 = 100

กรณีที่ 2: เลือก 3 หมายเลขคู่และ 1 หมายเลขคี่

• เลือก 3 หมายเลขคู่จาก 5 หมายเลขคู่: 5C3 = 10
• เลือก 1 หมายเลขคี่จาก 6 หมายเลขคี่: 6C1 = 6
• จำนวนรวม: 10 * 6 = 60

จำนวนรวมที่เป็นไปได้: 100 + 60 = 160

จำนวนรวมที่เป็นไปได้ทั้งหมดในการเลือก 4 ใบจาก 11 ใบ: 11C4 = 330

ความน่าจะเป็น: 160 / 330 ≈ 0.48

{"correct_answer_key":"d"}

{
  "correct_answer_key": "d"
}

Let $S$ be the set of numbers from 1 to 11, i.e., $S = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11}$. We want to find the probability that the product of the four chosen numbers is even and the sum of the four chosen numbers is odd.

The total number of ways to choose 4 numbers from 11 is $\binom{11}{4} = \frac{11 \cdot 10 \cdot 9 \cdot 8}{4 \cdot 3 \cdot 2 \cdot 1} = 330$.

For the product to be even, at least one of the chosen numbers must be even. The even numbers in $S$ are ${2, 4, 6, 8, 10}$, so there are 5 even numbers. The odd numbers in $S$ are ${1, 3, 5, 7, 9, 11}$, so there are 6 odd numbers.

For the sum to be odd, we must have an odd number of odd numbers. Since we are choosing 4 numbers, we can have 1 or 3 odd numbers.

Case 1: 1 odd number and 3 even numbers. The number of ways to choose 1 odd number from 6 is $\binom{6}{1} = 6$. The number of ways to choose 3 even numbers from 5 is $\binom{5}{3} = \frac{5 \cdot 4 \cdot 3}{3 \cdot 2 \cdot 1} = 10$. So, the number of ways in this case is $6 \cdot 10 = 60$.

Case 2: 3 odd numbers and 1 even number. The number of ways to choose 3 odd numbers from 6 is $\binom{6}{3} = \frac{6 \cdot 5 \cdot 4}{3 \cdot 2 \cdot 1} = 20$. The number of ways to choose 1 even number from 5 is $\binom{5}{1} = 5$. So, the number of ways in this case is $20 \cdot 5 = 100$.

The total number of ways to have an even product and an odd sum is $60 + 100 = 160$.

The probability is $\frac{160}{330} = \frac{16}{33} \approx 0.4848$.

{"correct_answer_key":"d"}