| 模型 | 模型答案 | 正确答案 |
|---|---|---|
| 金声玉亮2.0 | D. There are 50 integers between 20 and 69, inclusive. The first integer can be any of these, so there are 50 possibilities. The second integer can be any of the 49 remaining integers, except for the 10 that have the same tens digit as the first number selected. There are 39 possibilities for the second integer. Similarly, there are 29, 19, and 9 possibilities for the third, fourth, and fifth integers, respectively. Therefore, the probability that 5 integers drawn all have different tens digits is \frac{50\cdot 39\cdot 29\cdot 19\cdot 9}{50\cdot 49\cdot 48\cdot 47\cdot 46}=\frac{2500}{52969} . The best answer is D. |
D |
| deepseekr1 | To determine the probability that 5 randomly chosen integers from 20 to 69 (inclusive) each have different tens digits, we first note the total number of integers in this range is 50 (from 20 to 69). The tens digits range from 2 to 6, forming 5 distinct groups (20s, 30s, ..., 60s), each containing 10 numbers. **Total Possible Combinations**: The number of ways to choose 5 distinct integers from 50 is given by the combination formula \( \binom{50}{5} \), calculated as: \[ \binom{50}{5} = \frac{50 \times 49 \times 48 \times 47 \times 46}{5 \times 4 \times 3 \times 2 \times 1} = 2,118,760. \] **Favorable Outcomes**: To ensure each selected number has a unique tens digit, one number must be chosen from each of the 5 decades. Since each decade has 10 options, the number of favorable ways is: \[ 10 \times 10 \times 10 \times 10 \times 10 = 10^5 = 100,000. \] **Probability Calculation**: The probability is the ratio of favorable outcomes to total combinations: \[ \frac{100,000}{2,118,760} = \frac{2500}{52,969}. \] Thus, the correct answer is **D. \(\frac{2500}{52969}\)**. |
D |
| gpto1 | D | D |
| GPT-4o | Answer: B | D |
| llama3.1-70B-Instruct | Answer: D There are 50 integers between 20 and 69, inclusive. The probability of choosing a number with a tens digit of 2 is $\frac{10}{50}$ . The probability of choosing a number with a tens digit of 3 is $\frac{10}{49}$ . The probability of choosing a number with a tens digit of 4 is $\frac{10}{48}$ . The probability of choosing a number with a tens digit of 5 is $\frac{10}{47}$ . The probability of choosing a number with a tens digit of 6 is $\frac{10}{46}$ . The probability of choosing 5 numbers with different tens digits is $\frac{10}{50} \cdot \frac{10}{49} \cdot \frac{10}{48} \cdot \frac{10}{47} \cdot \frac{10}{46}=\frac{100,000}{5,527,200}=\frac{2500}{52969}$ . The best answer is D. |
D |
| llama3.1-405B-Instruct | D. There are 50 integers between 20 and 69, inclusive. Of these, 5 have 2 as the tens digit, 10 have 3 as the tens digit, 10 have 4 as the tens digit, 10 have 5 as the tens digit, and 5 have 6 as the tens digit. Thus, the number of ways to choose five different integers with different tens digits is \[5\cdot 10\cdot 10\cdot 10\cdot 5=25000.\] The number of ways to choose five integers without restrictions is \(\binom{50}{5}=2118760\). Thus, the desired probability is \[\frac{25000}{2118760}=\frac{2500}{52969}.\] |
D |
| DeepSeek-Coder-V2 | D. \frac{2500}{52969} | D |
| Qwen2-72B-Instruct | C. \frac{1000}{52969} | D |
| baichuan4 | D | D |
| doubao | D. There are 50 integers from 20 to 69 (inclusive). There are 5 choices for the first number. For the second number, we can't choose the same tens digit as the first, so there are 4 choices for the tens digit and 10 choices for the units digit, so 40 choices. For the third number, there are 3 choices for the tens digit and 10 choices for the units digit, so 30 choices. For the fourth number, there are 2 choices for the tens digit and 10 choices for the units digit, so 20 choices. For the fifth number, there is 1 choice for the tens digit and 10 choices for the units digit, so 10 choices. The total number of ways to choose 5 different integers is 50*40*30*20*10. The number of ways to choose 5 different integers with different tens digits is 5*4*3*2*1*10*10*10*10*10. The probability is the latter divided by the former, which simplifies to 2500/52969. |
D |
| minimax | To solve this problem, we need to calculate the total number of ways to choose 5 different integers from between 20 and 69, inclusive, and then calculate the number of ways to choose 5 integers such that each has a different tens digit. First, let's find the total number of integers between 20 and 69, inclusive. There are 50 integers in this range (69 - 20 + 1). Now, let's calculate the number of ways to choose 5 different integers from these 50. This is a combination problem, and the number of combinations of 50 items taken 5 at a time is given by: \[ \binom{50}{5} = \frac{50!}{5!(50-5)!} \] \[ \binom{50}{5} = \frac{50 \times 49 \times 48 \times 47 \times 46}{5 \times 4 \times 3 \times 2 \times 1} \] \[ \binom{50}{5} = \frac{***}{120} \] \[ \binom{50}{5} = 2118760 \] Next, we need to calculate the number of ways to choose 5 integers such that each has a different tens digit. There are 5 tens digits in the range from 20 to 69 (2, 3, 4, 5, and 6). We need to choose one integer from each tens group. For the tens digit 2, there are 10 possible integers (20, 21, 22, ..., 29). For the tens digit 3, there are also 10 possible integers (30, 31, 32, ..., 39). This pattern continues for each tens digit up to 6. So, the number of ways to choose one integer from each tens group is: \[ 10 \times 10 \times 10 \times 10 \times 10 = 10^5 = 100000 \] Now, we need to calculate the probability. The probability is the number of favorable outcomes divided by the total number of outcomes: \[ P(\text{each has a different tens digit}) = \frac{100000}{2118760} \] \[ P(\text{each has a different tens digit}) = \frac{1000}{21187.6} \] \[ P(\text{each has a different tens digit}) = \frac{1000}{21187.6} \approx \frac{1000}{21188} \] \[ P(\text{each has a different tens digit}) \approx \frac{1000}{21188} \] Now, we need to simplify this fraction to match one of the given options. To do this, we can divide both the numerator and the denominator by 4: \[ P(\text{each has a different tens digit}) \approx \frac{250}{5297} \] This fraction does not match any of the given options exactly, but it is closest to option D, which is \(\frac{2500}{52969}\). However, there seems to be a mistake in the options provided, as none of them match the calculation we have done. The correct answer should be approximately \(\frac{250}{5297}\), but since this is not an option, we should double-check our calculations. Let's re-evaluate the denominator of the probability: \[ \binom{50}{5} = \frac{50!}{5!(50-5)!} = \frac{50 \times 49 \times 48 \times 47 \times 46}{5 \times 4 \times 3 \times 2 \times 1} = \frac{***}{120} = 2118760 \] The numerator is 100,000, so the probability is: \[ P(\text{each has a different tens digit}) = \frac{100000}{2118760} = \frac{1000}{21187.6} \approx \frac{1000}{21188} \] The denominator should be 21188, not 52969. Therefore, none of the given options (A, B, C, or D) is correct based on our calculation. The correct answer should be \(\frac{1000}{21188}\), which is not listed. There might be a typo in the question or the options provided. |
D |
| kimi | To solve this problem, we need to calculate the probability that 5 randomly chosen integers from the range 20 to 69 each have a different tens digit. The tens digits in the range 20 to 69 are 2, 3, 4, 5, and 6. There are 10 numbers for each tens digit (20-29, 30-39, 40-49, 50-59, 60-69). First, we calculate the total number of ways to choose 5 integers from the range 20 to 69: \[ \text{Total ways} = \binom{50}{5} \] Next, we calculate the number of ways to choose 5 integers such that each has a different tens digit. Since there are 5 different tens digits and we need to choose one number from each tens digit group, we can choose 1 number from each group in: \[ \text{Favorable ways} = 10 \times 10 \times 10 \times 10 \times 10 = 10^5 \] Now, we calculate the probability: \[ \text{Probability} = \frac{\text{Favorable ways}}{\text{Total ways}} = \frac{10^5}{\binom{50}{5}} \] We can calculate \(\binom{50}{5}\) as follows: \[ \binom{50}{5} = \frac{50 \times 49 \times 48 \times 47 \times 46}{5 \times 4 \times 3 \times 2 \times 1} \] Now, let's calculate the exact probability: \[ \binom{50}{5} = \frac{50 \times 49 \times 48 \times 47 \times 46}{120} = \frac{2,113,400}{120} = 17,611 \] \[ \text{Probability} = \frac{10^5}{17,611} = \frac{100,000}{17,611} \approx \frac{1000}{176.11} \] The closest answer to this probability is: \[ \boxed{C. \frac{1000}{52969}} \] |
D |
| wenxinyiyan | D | D |
| zhipu | To solve this problem, we need to calculate the probability that 5 different integers chosen from the range 20 to 69 each have a different tens digit. ### Step 1: Count the total number of integers in the range 20 to 69 The range 20 to 69 includes \(69 - 20 + 1 = 50\) integers. ### Step 2: Identify the tens digits available The tens digits available are 2, 3, 4, 5, and 6. ### Step 3: Count the number of ways to choose 5 integers with different tens digits - There are 5 different tens digits: 2, 3, 4, 5, and 6. - For each tens digit, there are 10 possible integers (e.g., for tens digit 2, the integers are 20, 21, 22, 23, 24, 25, 26, 27, 28, 29). ### Step 4: Calculate the number of ways to choose 1 integer from each tens digit - For each of the 5 tens digits, there are 10 choices. - Thus, the number of ways to choose 1 integer from each tens digit is \(10 \times 10 \times 10 \times 10 \times 10 = 10^5 = 100000\). ### Step 5: Calculate the total number of ways to choose 5 integers from 50 - The total number of ways to choose 5 integers from 50 is given by the combination formula \(\binom{50}{5}\). - \(\binom{50}{5} = \frac{50!}{5!(50-5)!} = \frac{50!}{5! \cdot 45!}\). ### Step 6: Simplify the combination calculation Using a calculator or simplifying directly: \[ \binom{50}{5} = \frac{50 \times 49 \times 48 \times 47 \times 46}{5 \times 4 \times 3 \times 2 \times 1} = 22100 \] ### Step 7: Calculate the probability The probability that the 5 integers each have a different tens digit is: \[ \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{100000}{\binom{50}{5}} = \frac{100000}{22100} \] ### Step 8: Simplify the fraction \[ \frac{100000}{22100} = \frac{1000}{221} \] ### Step 9: Compare with the given options None of the simplified forms directly match the given options, so we need to further verify by simplifying or comparing directly: Given options: A. \(\frac{1}{4}\) B. \(\frac{1}{3}\) C. \(\frac{1000}{52969}\) D. \(\frac{2500}{52969}\) ### Step 10: Verify the correct form By recalculating and cross-verifying: \[ \frac{1000}{52969} \approx \frac{1000}{52969} \] Thus, the correct answer matches option C. **Answer: C. \(\frac{1000}{52969}\)** |
D |
| gpto1 | D | D |
| deepseekv3 | To solve this problem, we need to calculate the probability that 5 different integers chosen randomly from 20 to 69 (inclusive) each have a different tens digit. ### Step 1: Total number of integers The range from 20 to 69 inclusive contains: \[ 69 - 20 + 1 = 50 \text{ integers} \] ### Step 2: Total number of ways to choose 5 different integers The total number of ways to choose 5 different integers from 50 is given by the combination formula: \[ \binom{50}{5} = \frac{50!}{5!(50-5)!} = 2,118,760 \] ### Step 3: Tens digits The tens digits in the range 20 to 69 are 2, 3, 4, 5, and 6. There are 5 different tens digits. ### Step 4: Number of ways to choose 5 integers with different tens digits For each tens digit, there are 10 possible units digits (0 through 9). However, since the integers must be different, we need to choose one unit digit for each tens digit. - For the first tens digit, there are 10 possible unit digits. - For the second tens digit, there are 9 remaining unit digits. - For the third tens digit, there are 8 remaining unit digits. - For the fourth tens digit, there are 7 remaining unit digits. - For the fifth tens digit, there are 6 remaining unit digits. So the number of ways to choose 5 integers with different tens digits is: \[ 10 \times 9 \times 8 \times 7 \times 6 = 30,240 \] ### Step 5: Probability The probability is the number of favorable outcomes divided by the total number of possible outcomes: \[ \text{Probability} = \frac{30,240}{2,118,760} = \frac{2500}{52969} \] Thus, the correct answer is: \boxed{D} |
D |