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"}

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"}

{"correct_answer_key":"d"}

Explanation:
To satisfy both conditions (product even and sum odd), the selected numbers must include exactly 1 or 3 odd numbers (to make the sum odd) and at least 1 even number (to make the product even).

1. Odd/Even Distribution:

   • Total odd numbers (1-11): 6 (1,3,5,7,9,11).
   • Total even numbers (1-11): 5 (2,4,6,8,10).

2. Valid Combinations:

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

3. Total Possible Outcomes:
   ( \binom{11}{4} = 330 ).

4. Probability:
   ( \frac{160}{330} \approx 0.4848 \approx 0.48 ).

Thus, the answer is d (0.48).

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"}

{
  "explanation": "To have a product of numbers be even, at least one number must be even. To have a sum of 4 numbers be odd, there must be an odd number of odd numbers in the selection. This means we can have either 1 odd and 3 even numbers, or 3 odd and 1 even numbers. In both cases, there is at least one even number, so the product will be even.

Number of odd numbers from 1 to 11 is 6 (1, 3, 5, 7, 9, 11).
Number of even numbers from 1 to 11 is 5 (2, 4, 6, 8, 10).
Total number of tickets is 11.

Total ways to choose 4 tickets from 11 is C(11, 4) = 330.

Case 1: 1 odd and 3 even numbers.
Number of ways = C(6, 1) * C(5, 3) = 6 * 10 = 60.

Case 2: 3 odd and 1 even numbers.
Number of ways = C(6, 3) * C(5, 1) = 20 * 5 = 100.

Total favorable outcomes = 60 + 100 = 160.

Probability = Favorable outcomes / Total outcomes = 160 / 330 = 16 / 33 ≈ 0.4848.

The closest option is 0.48.",
  "correct_answer_key": "d"
}

โจทย์ถามความน่าจะเป็นของการสุ่มหยิบสลาก 4 ใบจาก 11 ใบ (หมายเลข 1 ถึง 11) โดยที่ผลคูณของหมายเลขทั้งสี่เป็นจำนวนคู่ แต่ผลบวกของหมายเลขทั้งสี่เป็นจำนวนคี่

สลาก 11 ใบ ประกอบด้วยเลขคี่ 6 ใบ (1, 3, 5, 7, 9, 11) และเลขคู่ 5 ใบ (2, 4, 6, 8, 10) จำนวนวิธีทั้งหมดในการหยิบสลาก 4 ใบจาก 11 ใบ คือ C(11, 4) = 11! / (4! * 7!) = (11 * 10 * 9 * 8) / (4 * 3 * 2 * 1) = 330 วิธี

เงื่อนไขที่ต้องการ:

1. ผลคูณเป็นจำนวนคู่: ผลคูณจะเป็นจำนวนคู่ก็ต่อเมื่อมีเลขคู่อย่างน้อย 1 ใบ
2. ผลบวกเป็นจำนวนคี่: ผลบวกจะเป็นจำนวนคี่ก็ต่อเมื่อจำนวนเลขคี่ที่หยิบมาเป็นจำนวนคี่ (คือ 1 ใบ หรือ 3 ใบ)

พิจารณาการผสมของเลขคี่ (O) และเลขคู่ (E) ในสลาก 4 ใบที่หยิบมา:

• หยิบได้เลขคี่ 4 ใบ (OOOO): C(6,4) * C(5,0) = 15 วิธี. ผลคูณเป็นคี่, ผลบวกเป็นคู่ (ไม่ตรงเงื่อนไข)
• หยิบได้เลขคี่ 3 ใบ, เลขคู่ 1 ใบ (OOOE): C(6,3) * C(5,1) = 20 * 5 = 100 วิธี. ผลคูณเป็นคู่ (OOO*E = E), ผลบวกเป็นคี่ (O+O+O+E = O) (ตรงตามเงื่อนไขทั้งสอง)
• หยิบได้เลขคี่ 2 ใบ, เลขคู่ 2 ใบ (OOEE): C(6,2) * C(5,2) = 15 * 10 = 150 วิธี. ผลคูณเป็นคู่ (OOE*E = E), ผลบวกเป็นคู่ (O+O+E+E = E) (ไม่ตรงเงื่อนไขผลบวก)
• หยิบได้เลขคี่ 1 ใบ, เลขคู่ 3 ใบ (OEEE): C(6,1) * C(5,3) = 6 * 10 = 60 วิธี. ผลคูณเป็นคู่ (OEE*E = E), ผลบวกเป็นคี่ (O+E+E+E = O) (ตรงตามเงื่อนไขทั้งสอง)
• หยิบได้เลขคู่ 4 ใบ (EEEE): C(6,0) * C(5,4) = 1 * 5 = 5 วิธี. ผลคูณเป็นคู่ (EEE*E = E), ผลบวกเป็นคู่ (E+E+E+E = E) (ไม่ตรงเงื่อนไขผลบวก)

จำนวนวิธีที่ตรงตามเงื่อนไขทั้งสอง (ผลคูณคู่ และ ผลบวกคี่) คือกรณี OOOE และ OEEE จำนวนวิธีที่ตรงตามเงื่อนไข = 100 + 60 = 160 วิธี

ความน่าจะเป็น = (จำนวนวิธีที่ตรงตามเงื่อนไข) / (จำนวนวิธีทั้งหมด) = 160 / 330 = 16 / 33

ค่า 16/33 เมื่อแปลงเป็นทศนิยมจะได้ประมาณ 0.4848... ซึ่งใกล้เคียงกับ 0.48 ในตัวเลือก

{"correct_answer_key":"d"}

The problem asks for the probability of two conditions being met when selecting 4 tickets out of 11 (numbered 1 to 11): the product of the numbers is even, and the sum of the numbers is odd.

First, let's determine the total number of ways to choose 4 tickets from 11. This is a combination problem, calculated as C(11, 4) = 11! / (4! * 7!) = (11 * 10 * 9 * 8) / (4 * 3 * 2 * 1) = 11 * 10 * 3 = 330.

Next, let's analyze the conditions:

1. The product of the numbers is even. The product of a set of integers is even if at least one of the integers is even. The product is odd only if all the integers are odd.
2. The sum of the numbers is odd. The sum of a set of integers is odd if there is an odd number of odd integers in the set.

The numbers from 1 to 11 consist of 6 odd numbers (1, 3, 5, 7, 9, 11) and 5 even numbers (2, 4, 6, 8, 10).

Let's consider the possible combinations of odd (O) and even (E) numbers when selecting 4 tickets and check if they satisfy the conditions:

Case 1: 4 Odd (OOOO)

• Product: Odd * Odd * Odd * Odd = Odd. This does not satisfy condition 1.
• Sum: Odd + Odd + Odd + Odd = Even. This does not satisfy condition 2.

Case 2: 3 Odd, 1 Even (OOOE)

• Product: Odd * Odd * Odd * Even = Even. This satisfies condition 1.
• Sum: Odd + Odd + Odd + Even = Odd + Even = Odd. This satisfies condition 2. Number of ways to choose 3 odd from 6 and 1 even from 5: C(6, 3) * C(5, 1) = (6! / (3! * 3!)) * (5! / (1! * 4!)) = (20) * (5) = 100.

Case 3: 2 Odd, 2 Even (OOEE)

• Product: Odd * Odd * Even * Even = Even. This satisfies condition 1.
• Sum: Odd + Odd + Even + Even = Even + Even = Even. This does not satisfy condition 2.

Case 4: 1 Odd, 3 Even (OEEE)

• Product: Odd * Even * Even * Even = Even. This satisfies condition 1.
• Sum: Odd + Even + Even + Even = Odd + Even = Odd. This satisfies condition 2. Number of ways to choose 1 odd from 6 and 3 even from 5: C(6, 1) * C(5, 3) = (6! / (1! * 5!)) * (5! / (3! * 2!)) = (6) * (10) = 60.

Case 5: 4 Even (EEEE)

• Product: Even * Even * Even * Even = Even. This satisfies condition 1.
• Sum: Even + Even + Even + Even = Even. This does not satisfy condition 2.

The cases that satisfy both conditions are Case 2 (3 Odd, 1 Even) and Case 4 (1 Odd, 3 Even). The total number of favorable outcomes is the sum of the ways in these two cases: 100 + 60 = 160.

The probability is the number of favorable outcomes divided by the total number of possible outcomes: 160 / 330. 160 / 330 = 16 / 33.

Now, let's convert this fraction to a decimal: 16 / 33 ≈ 0.4848...

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

The closest option to 0.4848... is 0.48.

{"correct_answer_key":"d"}

Explanation:

1. Identify Numbers: There are 11 tickets numbered 1 to 11.
   • Even numbers (E): 2, 4, 6, 8, 10 (5 total)
   • Odd numbers (O): 1, 3, 5, 7, 9, 11 (6 total)
2. Total Outcomes: The total number of ways to choose 4 tickets from 11 is C(11, 4) = (11 * 10 * 9 * 8) / (4 * 3 * 2 * 1) = 330.
3. Conditions for Favorable Outcomes:
   • The product of the numbers must be EVEN. This means at least one even number must be drawn.
   • The sum of the numbers must be ODD. This requires an odd number of odd numbers to be drawn (either 1 odd or 3 odds).
4. Combining Conditions: To satisfy both, the chosen set of 4 tickets must have either (1 Even, 3 Odd) or (3 Even, 1 Odd).
   • Case 1 (1E, 3O): Choose 1 Even from 5 [C(5, 1)] and 3 Odd from 6 [C(6, 3)]. Ways = C(5, 1) * C(6, 3) = 5 * 20 = 100.
   • Case 2 (3E, 1O): Choose 3 Even from 5 [C(5, 3)] and 1 Odd from 6 [C(6, 1)]. Ways = C(5, 3) * C(6, 1) = 10 * 6 = 60.
5. Total Favorable Outcomes: 100 + 60 = 160.
6. Calculate Probability: Probability = (Favorable Outcomes) / (Total Outcomes) = 160 / 330 = 16 / 33.
7. Convert to Decimal: 16 / 33 ≈ 0.4848... which is approximately 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"}

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"}

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"}

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"}

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"}

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"
}