| 模型 | 模型答案 | 正确答案 |
|---|---|---|
| 金声玉亮2.0 | To find the probability that exactly 2 out of 3 selected suitcases are damaged, we can use the concept of combinations and the formula for probability. First, let's find the total number of ways to select 3 suitcases out of 25: Total combinations = C(25, 3) = 25! / (3! * (25-3)!) = 25! / (3! * 22!) = (25 * 24 * 23) / (3 * 2 * 1) = 125 * 24 * 23 / 6 = 2,300 Next, let's find the number of ways to select exactly 2 damaged suitcases out of 5 and 1 undamaged suitcase out of 20: Combinations of 2 damaged suitcases = C(5, 2) = 5! / (2! * (5-2)!) = 5! / (2! * 3!) = (5 * 4) / (2 * 1) = 10 Combinations of 1 undamaged suitcase = C(20, 1) = 20! / (1! * (20-1)!) = 20 Now, we multiply the combinations of 2 damaged suitcases and 1 undamaged suitcase to get the total number of favorable outcomes: Favorable outcomes = Combinations of 2 damaged suitcases * Combinations of 1 undamaged suitcase = 10 * 20 = 200 Finally, we can find the probability by dividing the number of favorable outcomes by the total number of combinations: Probability = Favorable outcomes / Total combinations = 200 / 2,300 To simplify the fraction, we can divide both the numerator and the denominator by their greatest common divisor, which is 100: Probability = (200 / 100) / (2,300 / 100) = 2 / 23 So, the probability that exactly 2 out of 3 selected suitcases are damaged is 2/23. The correct answer is C. |
C |
| deepseekr1 | The problem involves calculating the probability of selecting exactly 2 damaged suitcases out of 3 chosen from 25 total (5 damaged, 20 undamaged). 1. **Total Combinations**: The number of ways to choose 3 suitcases from 25 is \( \binom{25}{3} \): \[ \binom{25}{3} = \frac{25 \times 24 \times 23}{3 \times 2 \times 1} = 2300. \] 2. **Favorable Combinations**: To have exactly 2 damaged and 1 undamaged suitcase: - Choose 2 damaged from 5: \( \binom{5}{2} = \frac{5 \times 4}{2 \times 1} = 10 \). - Choose 1 undamaged from 20: \( \binom{20}{1} = 20 \). - Total favorable combinations: \( 10 \times 20 = 200 \). 3. **Probability**: Divide favorable by total combinations: \[ \frac{200}{2300} = \frac{2}{23}. \] 4. **Verification**: Using probability multiplication for specific orders (e.g., D-D-ND) and accounting for \( \binom{3}{2} = 3 \) arrangements: \[ 3 \times \left( \frac{5}{25} \times \frac{4}{24} \times \frac{20}{23} \right) = 3 \times \frac{2}{69} = \frac{6}{69} = \frac{2}{23}. \] Both methods confirm the answer is **C. 2/23**. Answer: C |
C |
| gpto1 | C | C |
| GPT-4o | To determine the probability that exactly 2 out of the 3 selected suitcases are damaged, we can use combinations and the concept of probability. First, we calculate the total number of ways to select 3 suitcases out of 25: \[ \binom{25}{3} = \frac{25!}{3!(25-3)!} = \frac{25 \times 24 \times 23}{3 \times 2 \times 1} = 2300 \] Next, we calculate the number of ways to select exactly 2 damaged suitcases out of the 5 damaged ones: \[ \binom{5}{2} = \frac{5!}{2!(5-2)!} = \frac{5 \times 4}{2 \times 1} = 10 \] We also need to select the remaining 1 suitcase from the 20 undamaged suitcases: \[ \binom{20}{1} = \frac{20!}{1!(20-1)!} = 20 \] Now, we multiply the number of ways to select the damaged suitcases by the number of ways to select the undamaged suitcase: \[ 10 \times 20 = 200 \] The probability is then the ratio of the favorable outcomes to the total outcomes: \[ \frac{200}{2300} = \frac{2}{23} \] Therefore, the probability that exactly 2 out of the 3 selected suitcases are damaged is: \[ \boxed{\text{C}} \] Answer: C |
C |
| llama3.1-70B-Instruct | A |
C |
| llama3.1-405B-Instruct | To find the probability that exactly 2 out of 3 selected suitcases are damaged, we can use the concept of combinations and the formula for probability. First, let's find the total number of ways to select 3 suitcases out of 25: Total combinations = C(25, 3) = 25! / (3! * (25-3)!) = 25! / (3! * 22!) = (25 * 24 * 23) / (3 * 2 * 1) = 125 * 24 * 23 / 6 = 2,300 Next, let's find the number of ways to select exactly 2 damaged suitcases out of 5 and 1 undamaged suitcase out of 20: Combinations of 2 damaged suitcases = C(5, 2) = 5! / (2! * (5-2)!) = 5! / (2! * 3!) = (5 * 4) / (2 * 1) = 10 Combinations of 1 undamaged suitcase = C(20, 1) = 20! / (1! * (20-1)!) = 20 Now, we multiply the combinations of damaged and undamaged suitcases to get the total number of favorable outcomes: Favorable outcomes = Combinations of 2 damaged suitcases * Combinations of 1 undamaged suitcase = 10 * 20 = 200 Finally, we can calculate the probability: Probability = Favorable outcomes / Total combinations = 200 / 2,300 To simplify the fraction, we can divide both numerator and denominator by their greatest common divisor, which is 100: Probability = (200 / 100) / (2,300 / 100) = 2 / 23 So, the probability that exactly 2 out of 3 selected suitcases are damaged is 2/23. The correct answer is C. |
C |
| DeepSeek-Coder-V2 | C. 2/23 | C |
| Qwen2-72B-Instruct | The probability that exactly 2 out of 3 selected suitcases are damaged can be calculated using combinations. There are 5 damaged suitcases and 20 undamaged suitcases. The number of ways to choose 2 damaged suitcases out of 5 is given by the combination formula \(C(n, k) = \frac{n!}{k!(n-k)!}\), where \(n\) is the total number of items, \(k\) is the number of items to choose, and \(!\) denotes factorial. So, the number of ways to choose 2 damaged suitcases out of 5 is \(C(5, 2) = \frac{5!}{2!(5-2)!} = 10\). The number of ways to choose 1 undamaged suitcase out of 20 is \(C(20, 1) = \frac{20!}{1!(20-1)!} = 20\). The total number of ways to choose 3 suitcases out of 25 is \(C(25, 3) = \frac{25!}{3!(25-3)!} = 2300\). Therefore, the probability that exactly 2 out of the 3 selected suitcases are damaged is \(\frac{10 \times 20}{2300} = \frac{200}{2300} = \frac{20}{230} = \frac{2}{23}\). Answer: C. 2/23 |
C |
| baichuan4 | D | C |
| doubao | D. 12/125 The total number of ways to choose 3 suitcases out of 25 is C(25,3). The number of ways to choose 2 damaged suitcases out of 5 is C(5,2), and the number of ways to choose 1 undamaged suitcase out of 20 is C(20,1). The probability of exactly 2 damaged suitcases is [C(5,2) * C(20,1)] / C(25,3) = 12/125. |
C |
| minimax | To solve this problem, we can use combinations to calculate the probability of selecting exactly 2 damaged suitcases out of 3 selected. First, we calculate the number of ways to choose 2 damaged suitcases from the 5 available. This is given by the combination formula: C(5, 2) = 5! / (2! * (5 - 2)!) = (5 * 4) / (2 * 1) = 10 ways. Next, we calculate the number of ways to choose 1 undamaged suitcase from the 20 available (since there are 25 - 5 = 20 undamaged suitcases). This is given by: C(20, 1) = 20! / (1! * (20 - 1)!) = 20 ways. Now, we multiply these two results to find the total number of ways to choose exactly 2 damaged suitcases and 1 undamaged suitcase: 10 ways (damaged) * 20 ways (undamaged) = 200 ways. Finally, we calculate the total number of ways to choose any 3 suitcases from the 25: C(25, 3) = 25! / (3! * (25 - 3)!) = (25 * 24 * 23) / (3 * 2 * 1) = 2300 ways. The probability of selecting exactly 2 damaged suitcases is the number of favorable outcomes divided by the total number of outcomes: Probability = 200 / 2300 = 20 / 230 = 2 / 23. So the correct answer is: C. 2/23 |
C |
| kimi | To solve this problem, we need to calculate the probability of selecting exactly 2 damaged suitcases out of 5 and 1 undamaged suitcase out of 20 (since there are 25 suitcases in total and 5 are damaged). First, let's calculate the total number of ways to select 3 suitcases out of 25. This is a combination problem, which can be solved using the formula for combinations: \[ C(n, k) = \frac{n!}{k!(n-k)!} \] where \( n \) is the total number of items, \( k \) is the number of items to choose, and \( ! \) denotes factorial. The total number of ways to choose 3 suitcases from 25 is: \[ C(25, 3) = \frac{25!}{3!(25-3)!} = \frac{25!}{3! \cdot 22!} \] Now, let's calculate the number of ways to choose exactly 2 damaged suitcases from 5 and 1 undamaged suitcase from 20: \[ C(5, 2) \times C(20, 1) = \frac{5!}{2!(5-2)!} \times \frac{20!}{1!(20-1)!} \] \[ = \frac{5!}{2! \cdot 3!} \times \frac{20!}{1! \cdot 19!} \] \[ = \frac{5 \times 4}{2 \times 1} \times 20 \] \[ = 10 \times 20 \] \[ = 200 \] Now, we calculate the probability by dividing the number of favorable outcomes by the total number of outcomes: \[ P(\text{exactly 2 damaged}) = \frac{C(5, 2) \times C(20, 1)}{C(25, 3)} \] \[ = \frac{200}{\frac{25 \times 24 \times 23}{3 \times 2 \times 1}} \] \[ = \frac{200}{2300} \] \[ = \frac{4}{23} \] However, none of the options match this result. Let's re-evaluate the calculation: \[ C(25, 3) = \frac{25 \times 24 \times 23}{3 \times 2 \times 1} = 2300 \] \[ C(5, 2) \times C(20, 1) = 10 \times 20 = 200 \] \[ P(\text{exactly 2 damaged}) = \frac{200}{2300} = \frac{200}{2300} \times \frac{100}{100} = \frac{20000}{23000} = \frac{100}{115} \] \[ = \frac{20}{23} \] This result is still not matching any of the options, so there must be a mistake in the calculation. Let's try again: \[ C(25, 3) = \frac{25 \times 24 \times 23}{3 \times 2 \times 1} = 2300 \] \[ C(5, 2) \times C(20, 1) = \frac{5 \times 4}{2 \times 1} \times 20 = 200 \] \[ P(\text{exactly 2 damaged}) = \frac{200}{2300} \] \[ = \frac{200}{2300} \times \frac{100}{100} = \frac{20000}{23000} \] \[ = \frac{100}{115} \] \[ = \frac{20}{23} \] This is incorrect. Let's correct the calculation: \[ C(25, 3) = \frac{25 \times 24 \times 23}{3 \times 2 \times 1} = 2300 \] \[ C(5, 2) \times C(20, 1) = \frac{5 \times 4}{2 \times 1} \times 20 = 200 \] \[ P(\text{exactly 2 damaged}) = \frac{200}{2300} = \frac{20}{230} = \frac{2}{23} \] So the correct probability is \( \frac{2}{23} \), which corresponds to option C. Answer: C. |
C |
| wenxinyiyan | C Explanation: To solve this problem, we can use the combination formula to calculate the probabilities. The combination formula is defined as: C(n, k) = n! / (k!(n-k)!) where n is the total number of items, k is the number of items we want to select, and ! represents the factorial operation. Let's define the following events: D: selecting a damaged suitcase N: selecting an undamaged suitcase We want to find the probability of selecting exactly 2 damaged suitcases out of 3 selected suitcases. This can be represented as: P(D, D, N) or P(D, N, D) or P(N, D, D) Since the order of selection doesn't matter, we can combine these probabilities into one formula: P(exactly 2 damaged) = (C(5, 2) * C(20, 1)) / C(25, 3) where: C(5, 2) is the number of ways to select 2 damaged suitcases from 5 damaged suitcases. C(20, 1) is the number of ways to select 1 undamaged suitcase from 20 undamaged suitcases. C(25, 3) is the total number of ways to select 3 suitcases from 25 suitcases. Calculating these combinations: C(5, 2) = 5! / (2! * (5-2)!) = 10 C(20, 1) = 20! / (1! * (20-1)!) = 20 C(25, 3) = 25! / (3! * (25-3)!) = 2300 Plugging these values into our probability formula: P(exactly 2 damaged) = (10 * 20) / 2300 = 200 / 2300 = 2 / 23 So the answer is C: 2/23. |
C |
| zhipu | To solve this problem, we need to calculate the probability of selecting exactly 2 damaged suitcases out of the 3 selected from the 25 suitcases, where 5 are damaged and 20 are not damaged. First, let's define the events: - Total number of suitcases: \( n = 25 \) - Number of damaged suitcases: \( d = 5 \) - Number of non-damaged suitcases: \( n - d = 20 \) - Number of suitcases selected: \( k = 3 \) We need to find the probability of selecting exactly 2 damaged suitcases and 1 non-damaged suitcase. The number of ways to choose 2 damaged suitcases from 5 is given by the combination formula \( \binom{d}{2} \): \[ \binom{5}{2} = \frac{5!}{2!(5-2)!} = \frac{5 \times 4}{2 \times 1} = 10 \] The number of ways to choose 1 non-damaged suitcase from 20 is given by the combination formula \( \binom{20}{1} \): \[ \binom{20}{1} = 20 \] So, the number of favorable ways to choose exactly 2 damaged and 1 non-damaged suitcase is: \[ \binom{5}{2} \times \binom{20}{1} = 10 \times 20 = 200 \] Next, we need to find the total number of ways to choose 3 suitcases from 25, which is given by the combination formula \( \binom{25}{3} \): \[ \binom{25}{3} = \frac{25!}{3!(25-3)!} = \frac{25 \times 24 \times 23}{3 \times 2 \times 1} = 2300 \] Therefore, the probability of selecting exactly 2 damaged suitcases out of the 3 selected is: \[ P(\text{exactly 2 damaged}) = \frac{\text{Number of favorable ways}}{\text{Total number of ways}} = \frac{200}{2300} = \frac{2}{23} \] So, the correct answer is: C. 2/23 |
C |
| gpto1 | C | C |
| deepseekv3 | To solve this problem, we can use the concept of **combinations** in probability. **Given:** - Total suitcases: 25 - Damaged suitcases: 5 - Non-damaged suitcases: 20 - Suitcases selected: 3 **Goal:** Find the probability that exactly 2 of the selected suitcases are damaged. --- ### Step 1: Total number of ways to select 3 suitcases The total number of ways to select 3 suitcases out of 25 is given by the combination formula: \[ \text{Total ways} = \binom{25}{3} = \frac{25!}{3!(25-3)!} = \frac{25 \times 24 \times 23}{3 \times 2 \times 1} = 2300 \] --- ### Step 2: Number of favorable outcomes (exactly 2 damaged) To have exactly 2 damaged suitcases, we need to: 1. Choose 2 damaged suitcases out of 5. 2. Choose 1 non-damaged suitcase out of 20. The number of ways to do this is: \[ \text{Favorable ways} = \binom{5}{2} \times \binom{20}{1} = \frac{5!}{2!(5-2)!} \times 20 = \frac{5 \times 4}{2 \times 1} \times 20 = 10 \times 20 = 200 \] --- ### Step 3: Calculate the probability The probability is the ratio of favorable outcomes to total outcomes: \[ \text{Probability} = \frac{\text{Favorable ways}}{\text{Total ways}} = \frac{200}{2300} = \frac{2}{23} \] --- ### Final Answer: **C. 2/23** |
C |