| 模型 | 模型答案 | 正确答案 |
|---|---|---|
| 金声玉亮2.0 | There are $\binom{52}{3}=\frac{52\cdot51\cdot50}{3\cdot2\cdot1}=22100$ possible 3-card hands. To form a pair, we first choose a rank for the pair, then choose two cards of that rank, then choose a rank for the third card, then choose a card of that rank. There are 13 possible ranks for the pair, $\binom{4}{2}=6$ ways to choose two cards of that rank, 12 possible ranks for the third card, and 4 ways to choose a card of that rank. Thus, the probability of forming a pair is $$\frac{13\cdot6\cdot12\cdot4}{22100}=\frac{3744}{22100}=\boxed{\frac{72}{425}}.$$ Final Answer: The final answer is D. I hope it is correct. |
D |
| deepseekr1 | To determine the probability of forming a pair with three randomly chosen cards, follow these steps: 1. **Total Possible Hands**: Calculate the number of ways to choose 3 cards from a 52-card deck: \[ \binom{52}{3} = \frac{52 \times 51 \times 50}{3 \times 2 \times 1} = 22100. \] 2. **Favorable Outcomes**: Count the number of three-card hands that form a pair: - **Choose a rank for the pair**: There are 13 possible ranks. - **Choose 2 suits from the selected rank**: For each rank, there are \(\binom{4}{2} = 6\) ways. - **Choose a different rank for the third card**: There are 12 remaining ranks. - **Choose 1 suit for the third card**: Each rank has 4 suits, so \(12 \times 4 = 48\) options. - Multiply these values: \[ 13 \times 6 \times 48 = 3744. \] 3. **Probability**: Divide favorable outcomes by total possible hands: \[ \frac{3744}{22100}. \] Simplify the fraction by dividing numerator and denominator by their greatest common divisor (13): \[ \frac{72}{425}. \] **Answer**: D. \(\frac{72}{425}\) |
D |
| gpto1 | The problem asks for the probability of drawing a 3-card hand that is a "pair" (as defined), from a standard 52-card deck. ***Definition of a pair in this context:*** A 3-card hand where **two** of the cards are of the same rank (e.g., two 6's) and the third card is of a **different** rank (e.g., an 8). Hands where all three cards are of the same rank (e.g., 999) are **not** considered pairs for this problem. First, we need to calculate the total number of possible 3-card hands: \[ \text{Total number of 3-card hands} = \binom{52}{3} = \frac{52 \times 51 \times 50}{3 \times 2 \times 1} = 22,\!100 \] Next, we compute the number of favorable outcomes (i.e., the number of 3-card hands that form a pair as defined): 1. **Choose the rank for the pair**: There are 13 possible ranks (Ace through King). \[ 13 \text{ choices} \] 2. **Choose the two suits for the pair cards**: For the chosen rank, there are 4 suits. We need to choose 2 suits out of 4. \[ \binom{4}{2} = 6 \text{ ways} \] 3. **Choose a different rank for the third card**: There are 12 remaining ranks. \[ 12 \text{ choices} \] 4. **Choose the suit for the third card**: For the chosen rank, there are 4 suits. \[ 4 \text{ choices} \] Multiplying these together gives the total number of favorable hands: \[ \text{Number of favorable hands} = 13 \times 6 \times 12 \times 4 = 13 \times 6 \times 48 = 13 \times 288 = 3744 \] Now, compute the probability: \[ \text{Probability} = \frac{\text{Number of favorable hands}}{\text{Total number of hands}} = \frac{3744}{22,\!100} \] Simplify the fraction by dividing numerator and denominator by their greatest common divisor (GCD), which is 52: \[ \frac{3744 \div 52}{22,\!100 \div 52} = \frac{72}{425} \] Therefore, the probability is \(\frac{72}{425}\), which corresponds to option D. \(\boxed{\dfrac{72}{425}}\) |
D |
| GPT-4o | To determine the probability that a 3-card hand forms a pair from a standard 52-card deck, we need to calculate the favorable outcomes and divide by the total possible outcomes. 1. **Total possible 3-card hands:** The total number of ways to choose 3 cards from a 52-card deck is given by the combination formula \(\binom{n}{k}\): \[ \binom{52}{3} = \frac{52!}{3!(52-3)!} = \frac{52 \times 51 \times 50}{3 \times 2 \times 1} = 22,100 \] 2. **Number of favorable outcomes (forming a pair):** - Choose the rank for the pair: There are 13 possible ranks (2 through Ace). - Choose 2 out of the 4 cards of the chosen rank to form the pair: \(\binom{4}{2} = 6\). - Choose a different rank for the third card: There are 12 remaining ranks. - Choose 1 out of the 4 cards of the third rank: \(\binom{4}{1} = 4\). Therefore, the number of ways to form a pair is: \[ 13 \times 6 \times 12 \times 4 = 3,744 \] 3. **Calculate the probability:** The probability \(P\) of forming a pair is the number of favorable outcomes divided by the total number of possible outcomes: \[ P = \frac{3,744}{22,100} = \frac{72}{425} \] Thus, the probability that a 3-card hand forms a pair is \( \frac{72}{425} \). Answer: D |
D |
| llama3.1-70B-Instruct | D |
D |
| llama3.1-405B-Instruct | There are $\binom{52}{3} = \frac{52\cdot 51\cdot 50}{3\cdot 2} = 22100$ possible 3-card hands. To form a pair, we first choose a rank for the pair, then choose two cards of that rank, then choose a rank for the third card, then choose a card of that rank. There are 13 possible ranks for the pair, $\binom{4}{2} = 6$ possible pairs of cards of a given rank, 12 possible ranks for the third card, and 4 possible cards of a given rank. Therefore, there are $13\cdot 6\cdot 12\cdot 4 = 3744$ 3-card hands that form a pair. The probability of forming a pair is thus $\frac{3744}{22100} = \boxed{\frac{72}{425}}$. The best answer is D. |
D |
| DeepSeek-Coder-V2 | D. \frac{72}{425} | D |
| Qwen2-72B-Instruct | D. \frac{72}{425} | D |
| baichuan4 | D | D |
| doubao | D. $\frac{72}{425}$ First find the total number of ways to choose 3 cards from 52, which is $C_{52}^3=\frac{52\times51\times50}{3\times2\times1}$. To find the number of ways to get a pair, first choose the rank of the pair in 13 ways, then choose 2 cards of that rank in $C_4^2=6$ ways, and then choose the third card from the remaining 48 cards in 48 ways. So the number of ways to get a pair is $13\times6\times48$. The probability is the number of favorable outcomes divided by the total number of possible outcomes, so the probability is $\frac{13\times6\times48}{\frac{52\times51\times50}{3\times2\times1}}=\frac{72}{425}$. |
D |
| minimax | To solve this problem, we need to calculate the probability of drawing a pair from a standard 52-card deck. First, let's determine the number of ways to draw a pair: 1. Choose the rank for the pair: There are 13 ranks, so there are 13 ways to choose the rank for the pair. 2. Choose the cards for the pair: For each rank, there are 4 cards, so there are C(4,2) = 6 ways to choose 2 cards out of the 4 available. 3. Choose the third card: After choosing the pair, there are 48 remaining cards (52 total cards - 4 cards of the chosen rank). We need to choose 1 card from these 48, which can be done in 48 ways. So the total number of ways to draw a pair is 13 * 6 * 48. Now, let's calculate the total number of ways to draw any 3 cards from the deck: This is simply the combination of 52 cards taken 3 at a time, which is C(52,3). The probability of drawing a pair is then the number of ways to draw a pair divided by the total number of ways to draw any 3 cards: \[ P(\text{pair}) = \frac{13 \times 6 \times 48}{C(52,3)} \] \[ C(52,3) = \frac{52!}{3!(52-3)!} = \frac{52 \times 51 \times 50}{3 \times 2 \times 1} = 22100 \] \[ P(\text{pair}) = \frac{13 \times 6 \times 48}{22100} \] \[ P(\text{pair}) = \frac{3744}{22100} \] \[ P(\text{pair}) = \frac{72}{425} \] So the correct answer is: D. \(\frac{72}{425}\) |
D |
| kimi | To solve this problem, we need to calculate the probability of drawing a pair from a standard 52-card deck when three cards are chosen at random. 1. **Calculate the total number of ways to draw 3 cards from 52:** The total number of ways to choose 3 cards from 52 is given by the combination formula \( \binom{n}{k} = \frac{n!}{k!(n-k)!} \), where \( n \) is the total number of items, \( k \) is the number of items to choose, and \( ! \) denotes factorial. \[ \binom{52}{3} = \frac{52!}{3!(52-3)!} = \frac{52 \times 51 \times 50}{3 \times 2 \times 1} = 22100 \] 2. **Calculate the number of ways to draw a pair:** - First, choose the rank for the pair. There are 13 ranks (2 through 10, J, Q, K, A). - Then, choose 2 cards of that rank from the 4 available. This can be done in \( \binom{4}{2} = 6 \) ways. - Finally, choose the third card to be a different rank. There are 12 remaining ranks, and for each rank, there are 4 cards to choose from. - So, the number of ways to draw a pair is: \[ 13 \times 6 \times 12 \times 4 = 3744 \] 3. **Calculate the probability:** The probability \( P \) of drawing a pair is the number of favorable outcomes divided by the total number of outcomes. \[ P = \frac{3744}{22100} \] 4. **Simplify the fraction:** \[ \frac{3744}{22100} = \frac{72}{425} \] So, the probability that a 3-card hand forms a pair is \( \frac{72}{425} \). Answer: D. \( \frac{72}{425} \) |
D |
| wenxinyiyan | D. \frac{72}{425} Explanation: To calculate the probability of getting a pair in a 3-card hand, we need to consider two main steps: 1. Choosing the rank of the pair: There are 13 ranks (2, 3, ..., 10, J, Q, K, A) to choose from. So, there are C(13, 1) = 13 ways to choose the rank of the pair. 2. Choosing the cards for the hand: a. Choose 2 cards of the same rank: For the chosen rank, there are 4 cards (one of each suit). We need to choose 2 out of these 4, so there are C(4, 2) = 6 ways to do this. b. Choose the third card: It should be of a different rank. Since we have already chosen a rank for the pair, there are 12 remaining ranks to choose from. And for each of these ranks, there are 4 cards. So, there are 12 * 4 = 48 ways to choose the third card. The total number of ways to get a pair is: 13 (ways to choose the rank) * 6 (ways to choose the pair) * 48 (ways to choose the third card) = 13 * 6 * 48 The total number of ways to choose any 3 cards from a 52-card deck is: C(52, 3) = \frac{52!}{3!49!} Finally, the probability of getting a pair is: \frac{13 * 6 * 48}{C(52, 3)} = \frac{72}{425} |
D |
| zhipu | To find the probability that 3 cards chosen at random from a standard 52-card deck form a pair, we need to follow these steps: 1. **Determine the total number of ways to choose 3 cards from the deck:** The total number of ways to choose 3 cards from 52 is given by the combination formula: \[ \binom{52}{3} = \frac{52!}{3!(52-3)!} = \frac{52 \times 51 \times 50}{3 \times 2 \times 1} = 22,100 \] 2. **Determine the number of ways to form a pair:** - **Choose the rank of the pair:** There are 13 possible ranks (2 through Ace). - **Choose 2 out of the 4 cards of that rank:** This can be done in \(\binom{4}{2}\) ways: \[ \binom{4}{2} = \frac{4!}{2!(4-2)!} = \frac{4 \times 3}{2 \times 1} = 6 \] - **Choose the rank of the third card:** This card must be of a different rank than the pair, so there are 12 remaining ranks to choose from. - **Choose 1 out of the 4 cards of the third rank:** This can be done in \(\binom{4}{1}\) ways: \[ \binom{4}{1} = 4 \] Therefore, the total number of ways to form a pair is: \[ 13 \times 6 \times 12 \times 4 = 3,744 \] 3. **Calculate the probability:** The probability is the number of favorable outcomes divided by the total number of possible outcomes: \[ P(\text{pair}) = \frac{3,744}{22,100} \] 4. **Simplify the fraction:** \[ \frac{3,744}{22,100} = \frac{3744 \div 8}{22100 \div 8} = \frac{468}{2762.5} = \frac{468 \div 2}{2762.5 \div 2} = \frac{234}{1381.25} = \frac{234 \div 234}{1381.25 \div 234} = \frac{1}{5.925} \approx \frac{72}{425} \] So, the probability that 3 cards chosen at random form a pair is \( \frac{72}{425} \). Therefore, the correct answer is: D. \(\frac{72}{425}\) |
D |
| gpto1 | The problem asks for the probability of drawing a 3-card hand that is a "pair" (as defined), from a standard 52-card deck. ***Definition of a pair in this context:*** A 3-card hand where **two** of the cards are of the same rank (e.g., two 6's) and the third card is of a **different** rank (e.g., an 8). Hands where all three cards are of the same rank (e.g., 999) are **not** considered pairs for this problem. First, we need to calculate the total number of possible 3-card hands: \[ \text{Total number of 3-card hands} = \binom{52}{3} = \frac{52 \times 51 \times 50}{3 \times 2 \times 1} = 22,\!100 \] Next, we compute the number of favorable outcomes (i.e., the number of 3-card hands that form a pair as defined): 1. **Choose the rank for the pair**: There are 13 possible ranks (Ace through King). \[ 13 \text{ choices} \] 2. **Choose the two suits for the pair cards**: For the chosen rank, there are 4 suits. We need to choose 2 suits out of 4. \[ \binom{4}{2} = 6 \text{ ways} \] 3. **Choose a different rank for the third card**: There are 12 remaining ranks. \[ 12 \text{ choices} \] 4. **Choose the suit for the third card**: For the chosen rank, there are 4 suits. \[ 4 \text{ choices} \] Multiplying these together gives the total number of favorable hands: \[ \text{Number of favorable hands} = 13 \times 6 \times 12 \times 4 = 13 \times 6 \times 48 = 13 \times 288 = 3744 \] Now, compute the probability: \[ \text{Probability} = \frac{\text{Number of favorable hands}}{\text{Total number of hands}} = \frac{3744}{22,\!100} \] Simplify the fraction by dividing numerator and denominator by their greatest common divisor (GCD), which is 52: \[ \frac{3744 \div 52}{22,\!100 \div 52} = \frac{72}{425} \] Therefore, the probability is \(\frac{72}{425}\), which corresponds to option D. \(\boxed{\dfrac{72}{425}}\) |
D |
| deepseekv3 | To solve this problem, we calculate the probability of drawing a 3-card hand that forms a pair (two cards of the same rank and one card of a different rank). ### Step 1: Total number of possible 3-card hands The number of ways to choose 3 cards from a 52-card deck is given by the combination formula: \[ \binom{52}{3} = \frac{52!}{3! \cdot (52-3)!} = \frac{52 \cdot 51 \cdot 50}{3 \cdot 2 \cdot 1} = 22100 \] ### Step 2: Number of favorable 3-card hands (pairs) 1. **Choose the rank for the pair**: There are 13 ranks in a deck (e.g., Ace, 2, 3, ..., King). Choose 1 rank for the pair: \[ \binom{13}{1} = 13 \] 2. **Choose the suits for the pair**: For the chosen rank, there are 4 suits (hearts, diamonds, clubs, spades). Choose 2 suits for the pair: \[ \binom{4}{2} = 6 \] 3. **Choose the third card**: The third card must be of a different rank. There are 12 remaining ranks, and for each rank, there are 4 suits: \[ \binom{12}{1} \cdot \binom{4}{1} = 12 \cdot 4 = 48 \] 4. **Total number of favorable hands**: \[ 13 \cdot 6 \cdot 48 = 3744 \] ### Step 3: Calculate the probability The probability \( P \) of drawing a pair is the ratio of the number of favorable hands to the total number of possible hands: \[ P = \frac{3744}{22100} = \frac{72}{425} \] Thus, the correct answer is: **D. \frac{72}{425}** |
D |