金声玉亮2.0模型评测报告

问题

There is a single choice question about high school mathematics. Answer the question by replying A, B, C or D.
Question: Four packages are delivered to four houses, one to each house. If these packages are randomly delivered, what is the probability that exactly two of them are delivered to the correct houses? Express your answer as a common fraction.
A. \frac{1}{4}
B. \frac{27}{128}
C. \frac{1}{2}
D. \frac{1}{9}
Answer:

答案

模型	模型答案	正确答案
金声玉亮2.0	A derangement is a permutation of objects in which none of the objects appear in their original positions. The number of derangements of $n$ objects is given by the formula $$D_n=n!\left(1-\frac{1}{1!}+\frac{1}{2!}-\frac{1}{3!}+\cdots+(-1)^n\frac{1}{n!}\right).$$If we want to count the number of ways in which exactly two packages are delivered to the correct houses, we can do so by counting the number of derangements of two packages (the two that are delivered correctly) times the number of derangements of the remaining two packages (the two that are delivered incorrectly). This gives us $$D_2\cdot D_2=2!\left(1-\frac{1}{1!}+\frac{1}{2!}\right)\cdot 2!\left(1-\frac{1}{1!}+\frac{1}{2!}\right)=1\cdot1=1.$$However, this only accounts for the case in which the first two packages are delivered correctly. We must also consider the ${4\choose2}=6$ ways in which we can choose which two packages are delivered correctly. Therefore, the total number of ways in which exactly two packages are delivered to the correct houses is $6\cdot1=6.$ Since there are $4!=24$ total ways in which the packages can be delivered, the probability that exactly two of them are delivered to the correct houses is $\frac{6}{24}=\boxed{\frac{1}{4}}.$ The best answer is A.	A
deepseekr1	To determine the probability that exactly two out of four packages are delivered to the correct houses, follow these steps: 1. Total Permutations: There are \(4! = 24\) possible ways to deliver the packages. 2. Favorable Cases: Calculate permutations where exactly two packages are correct and the other two are swapped (deranged): - Choose 2 correct packages: \(\binom{4}{2} = 6\) ways. - Derange the remaining 2: For \(n=2\), derangements \(D(2) = 1\) (swap the two items). - Total favorable permutations: \(6 \times 1 = 6\). 3. Probability: Divide favorable cases by total permutations: \[ \frac{6}{24} = \frac{1}{4}. \] Verification using fixed-point counting confirms the result. Thus, the probability is \(\frac{1}{4}\). Answer: A	A
gpto1	A	A
GPT-4o	To solve this problem, we need to determine the probability that exactly two out of four packages are delivered to the correct houses when the packages are randomly delivered. First, let's calculate the total number of ways to deliver four packages to four houses. This is simply the number of permutations of four items, which is \(4!\): \[ 4! = 24 \] Next, we need to count the number of permutations where exactly two packages are delivered to the correct houses. This is a bit more complex, as we need to account for partial derangements (permutations where some but not all items are in their original position). For exactly two packages to be delivered correctly, the remaining two packages must not be delivered correctly. However, in a permutation of four elements, it is not possible to have exactly two elements in their original position and the remaining two not in their original position. This can be verified by considering the fixed points and derangements for smaller cases or using combinatorial arguments. Thus, there are zero ways to deliver exactly two packages correctly. The probability is therefore: \[ \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{0}{24} = 0 \] In common fraction form, this is: \[ 0 \] However, none of the provided answer choices include \(0\). This suggests a reevaluation of the problem is necessary to ensure there's no misunderstanding. Given the context of high school mathematics, this problem's intention is likely to highlight the impossibility of exactly two packages being correctly delivered in such a scenario. Answer: \(0\), but since this is not an option, the correct understanding demonstrates that the problem may have an error in its setup.	A
llama3.1-70B-Instruct	The correct answer is A.	A
llama3.1-405B-Instruct	A derangement is a permutation of objects in which none of the objects appear in their original positions. The number of derangements of $n$ objects is given by the formula $$!n = n!\left(1-\frac{1}{1!}+\frac{1}{2!}-\frac{1}{3!}+\cdots+(-1)^n\frac{1}{n!}\right).$$ If exactly two of the packages are delivered to the correct houses, then the other two packages must be swapped with each other. There are $\binom{4}{2}=6$ ways to choose which two packages are delivered correctly, and for each of these choices, there is only one way to derange the other two packages. Thus, the total number of ways to deliver exactly two packages correctly is $6\cdot 1=6.$ The total number of ways to deliver the packages is $4!=24$, so the probability that exactly two of them are delivered to the correct houses is $\frac{6}{24}=\boxed{\frac{1}{4}}.$ The best answer is A.	A
DeepSeek-Coder-V2	D. \frac{1}{9}	A
Qwen2-72B-Instruct	B. \frac{27}{128}	A
baichuan4	B. \frac{27}{128}	A
doubao	D. $\frac{1}{9}$. The total number of ways to deliver 4 packages to 4 houses is 4! = 24. To have exactly 2 packages delivered to the correct houses, we can choose 2 of the 4 packages in $\binom{4}{2}$ = 6 ways. The other 2 packages are delivered to the wrong houses in 1 way. So the number of favorable outcomes is 6. The probability is $\frac{6}{24}$ = $\frac{1}{4}$. But we have to divide by 2 because we counted each pair of correct packages twice. So the final probability is $\frac{1}{9}$.	A
minimax	To solve this problem, we need to calculate the probability that exactly two packages are delivered to the correct houses. This is a classic example of a derangement problem, but with a twist because we want exactly two correct deliveries, not zero. First, let's consider the total number of ways to deliver the four packages to the four houses. Since each package can go to any of the four houses, there are \(4! = 4 \times 3 \times 2 \times 1 = 24\) possible arrangements. Now, we need to find the number of arrangements where exactly two packages are delivered to the correct houses. To do this, we can choose two houses to receive the correct packages in \(\binom{4}{2} = 6\) ways. For each of these ways, we have to arrange the remaining two packages such that neither of them goes to the correct house. This is a derangement problem for two items, and there is only one derangement for two items: they must be swapped. So for each of the 6 ways to choose the two correct houses, there is only 1 way to arrange the remaining two packages incorrectly. This gives us a total of \(6 \times 1 = 6\) favorable outcomes. The probability that exactly two packages are delivered to the correct houses is then the number of favorable outcomes divided by the total number of possible arrangements: \[ P(\text{exactly two correct}) = \frac{6}{24} = \frac{1}{4} \] So the correct answer is: A. \(\frac{1}{4}\)	A
kimi	To solve this problem, we need to consider the concept of derangements, which are permutations where none of the elements appear in their original positions. In this case, we are looking for the probability that exactly two packages are delivered to the correct houses, which means two packages are correct and two are deranged. First, let's calculate the total number of ways to deliver the packages without any restrictions. There are 4 houses and 4 packages, so there are 4! (4 factorial) ways to deliver the packages, which is 4 × 3 × 2 × 1 = 24. Now, let's calculate the number of ways to deliver exactly two packages correctly. We can choose 2 out of the 4 houses to receive the correct packages in 4C2 ways, which is the number of combinations of 4 items taken 2 at a time: 4C2 = 4! / (2! * (4-2)!) = (4 × 3) / (2 × 1) = 6 ways. For the remaining two packages, we want them to be deranged. There is only 1 way to derange 2 items (i.e., they are swapped with each other). So, the number of favorable outcomes is 6 (ways to deliver two packages correctly) times 1 (way to derange the other two), which is 6. The probability is then the number of favorable outcomes divided by the total number of outcomes: Probability = Number of favorable outcomes / Total number of outcomes Probability = 6 / 24 Probability = 1 / 4 The correct answer is A. \frac{1}{4}.	A
wenxinyiyan	B. \frac{27}{128} To find the probability that exactly two packages are delivered to the correct houses, we need to consider all possible ways this can happen and divide by the total number of delivery options. Total Delivery Options: Since each package can be delivered to any of the four houses, there are 4! = 4 × 3 × 2 × 1 = 24 total delivery options. Ways to Deliver Exactly Two Packages Correctly: 1. Choose two packages to deliver correctly: C(4, 2) = 4! / (2! × (4-2)!) = 6 ways. 2. The remaining two packages must be delivered incorrectly. For these two packages, there is only one way each can be delivered incorrectly (i.e., the first incorrect package can only go to the house of the second incorrect package, and vice versa). So, there are 6 × 1 × 1 = 6 ways to choose and deliver two packages correctly and the other two incorrectly. However, for the incorrectly delivered packages, there are 2! = 2 ways to arrange them (since they are both going to the wrong house). Therefore, the total number of ways to deliver exactly two packages correctly is 6 × 2 = 12. Probability: The probability is the number of ways to deliver exactly two packages correctly divided by the total number of delivery options: 12 / 24 = 1/2. Wait! We forgot to consider that the two incorrectly delivered packages cannot be delivered to each other's houses in any order, as that would make it three or four correct deliveries. So, we need to adjust our calculation. Adjusted Calculation for Incorrect Deliveries: Consider the two remaining packages after choosing two for correct delivery. Let's call them A and B. Package A cannot go to B's house, and package B cannot go to A's house. So, A can go to one of the two remaining houses, and B will then go to the other. There are 2! = 2 ways to arrange this, but we need to exclude the case where A goes to the correct house of B and B goes to the correct house of A. Therefore, there is only 1 way for the two remaining packages to be delivered incorrectly (i.e., they are swapped). Adjusted Total Ways: With this adjustment, there are 6 ways to choose two packages for correct delivery and 1 way for the other two to be delivered incorrectly, totaling 6 ways. Adjusted Probability: The adjusted probability is 6 / 24 = 1/4 if we only consider the order of the first two correct packages. However, since the order of delivery for all four packages matters, we need to consider all permutations. Permutations of Correct and Incorrect Packages: For each of the 6 ways to choose two correct packages, there are 2! = 2 ways to permute the remaining two incorrect packages (since they are distinct). Additionally, there are 2! = 2 ways to permute the two correct packages themselves. So, the total number of permutations for exactly two correct deliveries is 6 × 2 × 2 = 24. But we need to divide by the total number of permutations for all four packages, which is 4! = 24. Final Probability: The final probability is the number of permutations for exactly two correct deliveries divided by the total number of permutations: 24 / 24 × 1/4 = 1/4 × 1/4 = 1/16. But since there are two distinct ways (out of four) for the remaining two packages to be incorrect, we multiply by 2: 1/16 × 2 = 1/8. Oops! We made a mistake in our initial calculation. Let's redo this using a different approach. Alternative Approach: 1. Choose two houses to receive the correct packages: C(4, 2) = 6 ways. 2. Arrange the remaining two packages to go to the incorrect houses: Since each package has only one incorrect house option (the other package's correct house), there is only 1 way to arrange them. But we need to consider that there are 2! = 2 ways to permute these two incorrect deliveries. So, the total number of ways is 6 × 1 × 2 = 12. The total number of possible delivery options is 4!. So, the probability is: Probability = 12 / 4! = 12 / 24 = 1/2 (this is still incorrect because we overcounted).	A
zhipu	To solve this problem, we need to use combinatorial probability. We are given four packages and four houses, and we want to find the probability that exactly two of the packages are delivered to the correct houses. 1. Total number of ways to deliver the packages: There are \(4!\) (4 factorial) ways to deliver 4 packages to 4 houses, which is: \[ 4! = 24 \] 2. Number of ways to choose which two houses receive the correct packages: We need to choose 2 out of the 4 houses to receive the correct packages. The number of ways to choose 2 houses from 4 is given by the combination formula \(\binom{4}{2}\): \[ \binom{4}{2} = \frac{4!}{2!(4-2)!} = 6 \] 3. Number of ways to arrange the remaining two packages incorrectly: For the remaining two houses, we need to arrange the two remaining packages such that neither of them is delivered to the correct house. This is a derangement problem for 2 items, and there is exactly 1 way to derange 2 items (i.e., each package goes to the wrong house). 4. Calculate the number of favorable outcomes: The total number of favorable outcomes is the product of the number of ways to choose the 2 correct houses and the number of ways to derange the remaining 2 packages: \[ 6 \times 1 = 6 \] 5. Calculate the probability: The probability that exactly two packages are delivered to the correct houses is the ratio of the number of favorable outcomes to the total number of possible outcomes: \[ \frac{6}{24} = \frac{1}{4} \] Therefore, the correct answer is: A. \(\frac{1}{4}\)	A
gpto1	A	A
deepseekv3	To solve this problem, we need to calculate the probability that exactly two out of four packages are delivered to the correct houses. ### Step 1: Total number of possible delivery arrangements There are 4 packages and 4 houses, so the total number of possible delivery arrangements is \(4! = 24\). ### Step 2: Number of favorable arrangements where exactly two packages are delivered correctly - Choose the two packages that are delivered correctly: The number of ways to choose 2 packages out of 4 is given by the combination formula \(C(4, 2) = 6\). - Deliver the remaining two packages incorrectly: For the remaining two packages, they must both be delivered incorrectly. There is only 1 way for both packages to be delivered incorrectly (this is known as a derangement). Thus, the number of favorable arrangements is \(6 \times 1 = 6\). ### Step 3: Calculate the probability The probability is the number of favorable arrangements divided by the total number of possible arrangements: \[ \text{Probability} = \frac{6}{24} = \frac{1}{4} \] ### Final Answer The probability that exactly two packages are delivered to the correct houses is \(\boxed{A}\).	A

模型

模型答案

正确答案

金声玉亮2.0

A derangement is a permutation of objects in which none of the objects appear in their original positions. The number of derangements of $n$ objects is given by the formula

$$D_n=n!\left(1-\frac{1}{1!}+\frac{1}{2!}-\frac{1}{3!}+\cdots+(-1)^n\frac{1}{n!}\right).$$If we want to count the number of ways in which exactly two packages are delivered to the correct houses, we can do so by counting the number of derangements of two packages (the two that are delivered correctly) times the number of derangements of the remaining two packages (the two that are delivered incorrectly). This gives us

$$D_2\cdot D_2=2!\left(1-\frac{1}{1!}+\frac{1}{2!}\right)\cdot 2!\left(1-\frac{1}{1!}+\frac{1}{2!}\right)=1\cdot1=1.$$However, this only accounts for the case in which the first two packages are delivered correctly. We must also consider the ${4\choose2}=6$ ways in which we can choose which two packages are delivered correctly. Therefore, the total number of ways in which exactly two packages are delivered to the correct houses is $6\cdot1=6.$ Since there are $4!=24$ total ways in which the packages can be delivered, the probability that exactly two of them are delivered to the correct houses is $\frac{6}{24}=\boxed{\frac{1}{4}}.$
The best answer is A.

deepseekr1

To determine the probability that exactly two out of four packages are delivered to the correct houses, follow these steps:

1. **Total Permutations**: There are $4! = 24$ possible ways to deliver the packages.

2. **Favorable Cases**: Calculate permutations where exactly two packages are correct and the other two are swapped (deranged):
- Choose 2 correct packages: $\binom{4}{2} = 6$ ways.
- Derange the remaining 2: For $n=2$, derangements $D(2) = 1$ (swap the two items).
- Total favorable permutations: $6 \times 1 = 6$.

3. **Probability**: Divide favorable cases by total permutations:
\[
\frac{6}{24} = \frac{1}{4}.
\]

Verification using fixed-point counting confirms the result. Thus, the probability is $\frac{1}{4}$.

**Answer: A**

gpto1

GPT-4o

To solve this problem, we need to determine the probability that exactly two out of four packages are delivered to the correct houses when the packages are randomly delivered.

First, let's calculate the total number of ways to deliver four packages to four houses. This is simply the number of permutations of four items, which is $4!$:

\[ 4! = 24 \]

Next, we need to count the number of permutations where exactly two packages are delivered to the correct houses. This is a bit more complex, as we need to account for partial derangements (permutations where some but not all items are in their original position).

For exactly two packages to be delivered correctly, the remaining two packages must not be delivered correctly. However, in a permutation of four elements, it is not possible to have exactly two elements in their original position and the remaining two not in their original position. This can be verified by considering the fixed points and derangements for smaller cases or using combinatorial arguments.

Thus, there are zero ways to deliver exactly two packages correctly.

The probability is therefore:

\[ \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{0}{24} = 0 \]

In common fraction form, this is:

\[ 0 \]

However, none of the provided answer choices include $0$. This suggests a reevaluation of the problem is necessary to ensure there's no misunderstanding. Given the context of high school mathematics, this problem's intention is likely to highlight the impossibility of exactly two packages being correctly delivered in such a scenario.

Answer: $0$, but since this is not an option, the correct understanding demonstrates that the problem may have an error in its setup.

llama3.1-70B-Instruct

The correct answer is A.

llama3.1-405B-Instruct

A derangement is a permutation of objects in which none of the objects appear in their original positions. The number of derangements of $n$ objects is given by the formula $$!n = n!\left(1-\frac{1}{1!}+\frac{1}{2!}-\frac{1}{3!}+\cdots+(-1)^n\frac{1}{n!}\right).$$ If exactly two of the packages are delivered to the correct houses, then the other two packages must be swapped with each other. There are $\binom{4}{2}=6$ ways to choose which two packages are delivered correctly, and for each of these choices, there is only one way to derange the other two packages. Thus, the total number of ways to deliver exactly two packages correctly is $6\cdot 1=6.$ The total number of ways to deliver the packages is $4!=24$, so the probability that exactly two of them are delivered to the correct houses is $\frac{6}{24}=\boxed{\frac{1}{4}}.$
The best answer is A.

DeepSeek-Coder-V2

D. \frac{1}{9}

Qwen2-72B-Instruct

B. \frac{27}{128}

baichuan4

B. \frac{27}{128}

doubao

D. $\frac{1}{9}$.

The total number of ways to deliver 4 packages to 4 houses is 4! = 24. To have exactly 2 packages delivered to the correct houses, we can choose 2 of the 4 packages in $\binom{4}{2}$ = 6 ways. The other 2 packages are delivered to the wrong houses in 1 way. So the number of favorable outcomes is 6. The probability is $\frac{6}{24}$ = $\frac{1}{4}$. But we have to divide by 2 because we counted each pair of correct packages twice. So the final probability is $\frac{1}{9}$.

minimax

To solve this problem, we need to calculate the probability that exactly two packages are delivered to the correct houses. This is a classic example of a derangement problem, but with a twist because we want exactly two correct deliveries, not zero.

First, let's consider the total number of ways to deliver the four packages to the four houses. Since each package can go to any of the four houses, there are $4! = 4 \times 3 \times 2 \times 1 = 24$ possible arrangements.

Now, we need to find the number of arrangements where exactly two packages are delivered to the correct houses. To do this, we can choose two houses to receive the correct packages in $\binom{4}{2} = 6$ ways. For each of these ways, we have to arrange the remaining two packages such that neither of them goes to the correct house. This is a derangement problem for two items, and there is only one derangement for two items: they must be swapped.

So for each of the 6 ways to choose the two correct houses, there is only 1 way to arrange the remaining two packages incorrectly. This gives us a total of $6 \times 1 = 6$ favorable outcomes.

The probability that exactly two packages are delivered to the correct houses is then the number of favorable outcomes divided by the total number of possible arrangements:

\[ P(\text{exactly two correct}) = \frac{6}{24} = \frac{1}{4} \]

So the correct answer is:

A. $\frac{1}{4}$

kimi

To solve this problem, we need to consider the concept of derangements, which are permutations where none of the elements appear in their original positions. In this case, we are looking for the probability that exactly two packages are delivered to the correct houses, which means two packages are correct and two are deranged.

First, let's calculate the total number of ways to deliver the packages without any restrictions. There are 4 houses and 4 packages, so there are 4! (4 factorial) ways to deliver the packages, which is 4 × 3 × 2 × 1 = 24.

Now, let's calculate the number of ways to deliver exactly two packages correctly. We can choose 2 out of the 4 houses to receive the correct packages in 4C2 ways, which is the number of combinations of 4 items taken 2 at a time:

4C2 = 4! / (2! * (4-2)!) = (4 × 3) / (2 × 1) = 6 ways.

For the remaining two packages, we want them to be deranged. There is only 1 way to derange 2 items (i.e., they are swapped with each other).

So, the number of favorable outcomes is 6 (ways to deliver two packages correctly) times 1 (way to derange the other two), which is 6.

The probability is then the number of favorable outcomes divided by the total number of outcomes:

Probability = Number of favorable outcomes / Total number of outcomes
Probability = 6 / 24
Probability = 1 / 4

The correct answer is A. \frac{1}{4}.

wenxinyiyan

B. \frac{27}{128}

To find the probability that exactly two packages are delivered to the correct houses, we need to consider all possible ways this can happen and divide by the total number of delivery options.

Total Delivery Options:
Since each package can be delivered to any of the four houses, there are 4! = 4 × 3 × 2 × 1 = 24 total delivery options.

Ways to Deliver Exactly Two Packages Correctly:

1. Choose two packages to deliver correctly: C(4, 2) = 4! / (2! × (4-2)!) = 6 ways.
2. The remaining two packages must be delivered incorrectly. For these two packages, there is only one way each can be delivered incorrectly (i.e., the first incorrect package can only go to the house of the second incorrect package, and vice versa).

So, there are 6 × 1 × 1 = 6 ways to choose and deliver two packages correctly and the other two incorrectly. However, for the incorrectly delivered packages, there are 2! = 2 ways to arrange them (since they are both going to the wrong house).

Therefore, the total number of ways to deliver exactly two packages correctly is 6 × 2 = 12.

Probability:
The probability is the number of ways to deliver exactly two packages correctly divided by the total number of delivery options: 12 / 24 = 1/2.

Wait! We forgot to consider that the two incorrectly delivered packages cannot be delivered to each other's houses in any order, as that would make it three or four correct deliveries. So, we need to adjust our calculation.

Adjusted Calculation for Incorrect Deliveries:
Consider the two remaining packages after choosing two for correct delivery. Let's call them A and B. Package A cannot go to B's house, and package B cannot go to A's house. So, A can go to one of the two remaining houses, and B will then go to the other. There are 2! = 2 ways to arrange this, but we need to exclude the case where A goes to the correct house of B and B goes to the correct house of A.

Therefore, there is only 1 way for the two remaining packages to be delivered incorrectly (i.e., they are swapped).

Adjusted Total Ways:
With this adjustment, there are 6 ways to choose two packages for correct delivery and 1 way for the other two to be delivered incorrectly, totaling 6 ways.

Adjusted Probability:
The adjusted probability is 6 / 24 = 1/4 if we only consider the order of the first two correct packages. However, since the order of delivery for all four packages matters, we need to consider all permutations.

Permutations of Correct and Incorrect Packages:
For each of the 6 ways to choose two correct packages, there are 2! = 2 ways to permute the remaining two incorrect packages (since they are distinct). Additionally, there are 2! = 2 ways to permute the two correct packages themselves.

So, the total number of permutations for exactly two correct deliveries is 6 × 2 × 2 = 24. But we need to divide by the total number of permutations for all four packages, which is 4! = 24.

Final Probability:
The final probability is the number of permutations for exactly two correct deliveries divided by the total number of permutations: 24 / 24 × 1/4 = 1/4 × 1/4 = 1/16. But since there are two distinct ways (out of four) for the remaining two packages to be incorrect, we multiply by 2: 1/16 × 2 = 1/8.

Oops! We made a mistake in our initial calculation. Let's redo this using a different approach.

Alternative Approach:

1. Choose two houses to receive the correct packages: C(4, 2) = 6 ways.
2. Arrange the remaining two packages to go to the incorrect houses: Since each package has only one incorrect house option (the other package's correct house), there is only 1 way to arrange them. But we need to consider that there are 2! = 2 ways to permute these two incorrect deliveries.

So, the total number of ways is 6 × 1 × 2 = 12.

The total number of possible delivery options is 4!. So, the probability is:

Probability = 12 / 4! = 12 / 24 = 1/2 (this is still incorrect because we overcounted).

zhipu

To solve this problem, we need to use combinatorial probability. We are given four packages and four houses, and we want to find the probability that exactly two of the packages are delivered to the correct houses.

1. **Total number of ways to deliver the packages:**
There are $4!$ (4 factorial) ways to deliver 4 packages to 4 houses, which is:
\[
4! = 24
\]

2. **Number of ways to choose which two houses receive the correct packages:**
We need to choose 2 out of the 4 houses to receive the correct packages. The number of ways to choose 2 houses from 4 is given by the combination formula $\binom{4}{2}$:
\[
\binom{4}{2} = \frac{4!}{2!(4-2)!} = 6
\]

3. **Number of ways to arrange the remaining two packages incorrectly:**
For the remaining two houses, we need to arrange the two remaining packages such that neither of them is delivered to the correct house. This is a derangement problem for 2 items, and there is exactly 1 way to derange 2 items (i.e., each package goes to the wrong house).

4. **Calculate the number of favorable outcomes:**
The total number of favorable outcomes is the product of the number of ways to choose the 2 correct houses and the number of ways to derange the remaining 2 packages:
\[
6 \times 1 = 6
\]

5. **Calculate the probability:**
The probability that exactly two packages are delivered to the correct houses is the ratio of the number of favorable outcomes to the total number of possible outcomes:
\[
\frac{6}{24} = \frac{1}{4}
\]

Therefore, the correct answer is:
A. $\frac{1}{4}$

gpto1

deepseekv3

To solve this problem, we need to calculate the probability that exactly two out of four packages are delivered to the correct houses.

### Step 1: Total number of possible delivery arrangements
There are 4 packages and 4 houses, so the total number of possible delivery arrangements is $4! = 24$.

### Step 2: Number of favorable arrangements where exactly two packages are delivered correctly
- **Choose the two packages that are delivered correctly**: The number of ways to choose 2 packages out of 4 is given by the combination formula $C(4, 2) = 6$.
- **Deliver the remaining two packages incorrectly**: For the remaining two packages, they must both be delivered incorrectly. There is only 1 way for both packages to be delivered incorrectly (this is known as a derangement).

Thus, the number of favorable arrangements is $6 \times 1 = 6$.

### Step 3: Calculate the probability
The probability is the number of favorable arrangements divided by the total number of possible arrangements:

\[
\text{Probability} = \frac{6}{24} = \frac{1}{4}
\]

### Final Answer
The probability that exactly two packages are delivered to the correct houses is $\boxed{A}$.