金声玉亮2.0模型评测报告

问题

There is a single choice question about high school statistics. Answer the question by replying A, B, C or D.
Question: In the jury pool available for this week, 30 percent of potential jurors are women. A particular trial requires that, out of a jury of 12, at least three are women. If a jury of 12 is to be selected at random from the pool, what is the probability it meets the requirements of this trial?
A. 0.168
B. 0.843
C. 0.915
D. The answer cannot be determined without knowing the size of the jury pool.
Answer:

答案

模型	模型答案	正确答案
金声玉亮2.0	B. Let X be the number of women on the jury. Then X is a binomial random variable with n = 12 and p = 0.3. The probability that the jury will have at least three women is P ( X ≥ 3) = 1 – P ( X ≤ 2) = 1 – [ P ( X = 0) + P ( X = 1) + P ( X = 2)] The probability that the jury will have at least three women is 0.843.	D
deepseekr1	The problem requires calculating the probability that a randomly selected jury of 12 has at least three women, given that 30% of the jury pool are women. The key consideration is whether the binomial or hypergeometric distribution applies. 1. Binomial Approach: Assumes independent trials with a fixed probability (30% chance each juror is a woman). Calculations for at least three women yield approximately 74.7% (1 - P(0) - P(1) - P(2)), but this result does not match any provided answer choices (A: 0.168, B: 0.843, C: 0.915). 2. Hypergeometric Consideration: Requires knowing the total jury pool size (N) to account for sampling without replacement. Without N, exact probabilities cannot be computed. For smaller pools, the hypergeometric distribution deviates significantly from the binomial approximation. Since the problem does not specify the pool size, the exact probability remains undeterminable. 3. Answer Analysis: Options A, B, and C do not align with the binomial result (~74.7%), suggesting the question tests awareness of distribution assumptions. Answer D highlights the necessity of pool size for accurate calculation, which is missing here. Thus, the correct choice is D. Answer: D	D
gpto1	B	D
GPT-4o	B. 0.843	D
llama3.1-70B-Instruct	B	D
llama3.1-405B-Instruct	B. Let X be the number of women on the jury. Then X is a binomial random variable with n = 12 and p = 0.3. The probability that the jury will have at least three women is P ( X ≥ 3) = 1 – P ( X ≤ 2) = 1 – [ P ( X = 0) + P ( X = 1) + P ( X = 2)] The probability that the jury will have at least three women is 0.843.	D
DeepSeek-Coder-V2	C. 0.915	D
Qwen2-72B-Instruct	C. 0.915	D
baichuan4	D	D
doubao	D. The answer cannot be determined without knowing the size of the jury pool.	D
minimax	To solve this problem, we need to calculate the probability that there are at least three women in a jury of 12. This can be done by finding the probability of having exactly 0, 1, or 2 women and subtracting that from 1. Let's denote the probability of selecting a woman as \( p \) and the probability of selecting a man as \( q \). Given that 30% of potential jurors are women, \( p = 0.3 \) and \( q = 0.7 \). The probability of having exactly \( k \) women in a jury of 12 can be calculated using the binomial distribution formula: \[ P(X = k) = \binom{n}{k} p^k q^{(n-k)} \] where \( n \) is the number of trials (in this case, 12), \( k \) is the number of successes (women in the jury), and \( \binom{n}{k} \) is the binomial coefficient. We need to calculate the probabilities for \( k = 0, 1, \) and \( 2 \), and then sum them up: \[ P(X \geq 3) = 1 - (P(X = 0) + P(X = 1) + P(X = 2)) \] Let's calculate each term: \[ P(X = 0) = \binom{12}{0} p^0 q^{12} = 1 \cdot 1 \cdot 0.7^{12} \] \[ P(X = 1) = \binom{12}{1} p^1 q^{11} = 12 \cdot 0.3 \cdot 0.7^{11} \] \[ P(X = 2) = \binom{12}{2} p^2 q^{10} = \frac{12 \cdot 11}{2} \cdot 0.3^2 \cdot 0.7^{10} \] Now we sum up the probabilities for \( k = 0, 1, \) and \( 2 \): \[ P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2) \] \[ P(X \geq 3) = 1 - P(X < 3) \] After calculating the exact values for \( P(X = 0) \), \( P(X = 1) \), and \( P(X = 2) \), we can find \( P(X \geq 3) \). However, since this is a multiple-choice question and the calculations involve binomial probabilities, which can be tedious without a calculator, I'll provide the correct answer based on the typical approach to such problems: The correct answer is C. 0.915. This answer is based on the assumption that the calculations have been done correctly and that the probability of having at least three women in a jury of 12 from a pool where 30% are women is approximately 0.915.	D
kimi	To solve this problem, we can use the binomial probability formula to calculate the probability of selecting at least three women in a jury of 12 when the probability of selecting a woman is 30% (or 0.3). The binomial probability formula is: \[ P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} \] where: - \( P(X = k) \) is the probability of exactly \( k \) successes (in this case, selecting a woman) in \( n \) trials (selecting 12 jurors). - \( \binom{n}{k} \) is the binomial coefficient, which represents the number of ways to choose \( k \) successes from \( n \) trials. - \( p \) is the probability of success on a single trial (selecting a woman). - \( (1-p) \) is the probability of failure on a single trial (not selecting a woman). Since we want the probability of selecting at least three women, we need to calculate the probability for selecting 3, 4, 5, ..., 12 women and sum these probabilities. However, calculating this by hand can be quite tedious, so we can use the complement rule, which states that the probability of an event occurring is 1 minus the probability of the event not occurring. In this case, we can calculate the probability of selecting fewer than three women (0, 1, or 2 women) and subtract that from 1 to get the probability of selecting at least three women. Let's calculate the probability of selecting fewer than three women: \[ P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2) \] \[ P(X = 0) = \binom{12}{0} (0.3)^0 (0.7)^{12} \] \[ P(X = 1) = \binom{12}{1} (0.3)^1 (0.7)^{11} \] \[ P(X = 2) = \binom{12}{2} (0.3)^2 (0.7)^{10} \] Now, let's calculate these probabilities: \[ P(X = 0) = 1 \cdot (1) \cdot (0.7)^{12} \] \[ P(X = 1) = 12 \cdot 0.3 \cdot (0.7)^{11} \] \[ P(X = 2) = \frac{12!}{2!(12-2)!} \cdot (0.3)^2 \cdot (0.7)^{10} \] \[ P(X = 0) = (0.7)^{12} \] \[ P(X = 1) = 12 \cdot 0.3 \cdot (0.7)^{11} \] \[ P(X = 2) = \frac{12 \cdot 11}{2 \cdot 1} \cdot 0.09 \cdot (0.7)^{10} \] \[ P(X = 0) \approx 0.0282 \] \[ P(X = 1) \approx 0.2187 \] \[ P(X = 2) \approx 0.2805 \] \[ P(X < 3) \approx 0.0282 + 0.2187 + 0.2805 = 0.5274 \] Now, the probability of selecting at least three women is: \[ P(X \geq 3) = 1 - P(X < 3) \] \[ P(X \geq 3) \approx 1 - 0.5274 = 0.4726 \] This probability does not match any of the given options exactly, but it is closest to option A, which is 0.168. However, this is a rough calculation and the actual probability should be calculated using a more precise method or a calculator that can handle binomial probabilities. If we were to choose the closest option, it would be: Answer: A. 0.168	D
wenxinyiyan	C. 0.915 To solve this problem, we can use the binomial distribution, which is a probability distribution that models the number of successes in a fixed number of independent trials with the same probability of success. Let's define the following: * n = 12 (number of jurors to be selected) * p = 0.3 (probability of selecting a woman) * q = 0.7 (probability of selecting a man) * X = number of women selected We want to find the probability that at least three women are selected, which can be expressed as P(X >= 3). To find this probability, we can calculate the probabilities of selecting 3, 4, 5, ..., 12 women and then sum them up. However, it's easier to calculate the probabilities of selecting 0, 1, or 2 women and then subtracting them from 1 to get the desired probability. The probability of selecting exactly k women is given by the binomial formula: P(X = k) = (n choose k) * p^k * q^(n-k) where (n choose k) is the binomial coefficient, which can be calculated as n! / (k! * (n-k)!). Let's calculate the probabilities of selecting 0, 1, and 2 women: P(X = 0) = (12 choose 0) * 0.3^0 * 0.7^12 = 1 * 1 * 0.7^12 P(X = 1) = (12 choose 1) * 0.3^1 * 0.7^11 = 12 * 0.3 * 0.7^11 P(X = 2) = (12 choose 2) * 0.3^2 * 0.7^10 = 66 * 0.3^2 * 0.7^10 Now, let's sum these probabilities and subtract them from 1 to get the desired probability: P(X >= 3) = 1 - (P(X = 0) + P(X = 1) + P(X = 2)) = 1 - (0.7^12 + 12 * 0.3 * 0.7^11 + 66 * 0.3^2 * 0.7^10) ≈ 1 - (0.000121 + 0.004876 + 0.080238) ≈ 1 - 0.085235 ≈ 0.914765 ≈ 0.915 (rounded to three decimal places)	D
zhipu	The key to solving this problem is understanding how to calculate the probability of having at least three women in a jury of 12, given that 30% of the potential jurors are women. To solve this, we can use the binomial probability formula: \[ P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} \] where: - \( n \) is the number of trials (juries selected), which is 12. - \( k \) is the number of successes (women in the jury). - \( p \) is the probability of success on a single trial, which is 0.30 (30%). - \( \binom{n}{k} \) is the binomial coefficient, which is the number of ways to choose \( k \) successes out of \( n \) trials. We need to find the probability that there are at least 3 women, which is: \[ P(X \geq 3) = 1 - P(X < 3) \] where \( P(X < 3) \) is the probability of having 0, 1, or 2 women. Let's calculate \( P(X < 3) \): \[ P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2) \] First, calculate each term: 1. \( P(X = 0) \): \[ P(X = 0) = \binom{12}{0} (0.30)^0 (0.70)^{12} = 1 \cdot 1 \cdot 0.70^{12} \approx 0.01384 \] 2. \( P(X = 1) \): \[ P(X = 1) = \binom{12}{1} (0.30)^1 (0.70)^{11} = 12 \cdot 0.30 \cdot 0.70^{11} \approx 0.06978 \] 3. \( P(X = 2) \): \[ P(X = 2) = \binom{12}{2} (0.30)^2 (0.70)^{10} = 66 \cdot 0.09 \cdot 0.70^{10} \approx 0.20667 \] Now sum these probabilities: \[ P(X < 3) \approx 0.01384 + 0.06978 + 0.20667 \approx 0.29029 \] Therefore, \[ P(X \geq 3) = 1 - P(X < 3) \approx 1 - 0.29029 \approx 0.70971 \] However, since this calculation is quite involved and the options provided are discrete probabilities, we can check the closest answer that aligns with our understanding. Based on the nature of the problem and typical binomial distributions, the closest match to our calculated probability is: B. 0.843 This option best fits the typical result when dealing with such probabilities in binomial distributions. Therefore, the answer is: B	D
gpto1	B	D
deepseekv3	To solve this problem, we can use the binomial distribution. The probability of selecting a woman is 0.3, and the probability of selecting a man is 0.7. We need to calculate the probability that at least 3 out of the 12 jurors are women. ### Step 1: Define the variables - Number of trials (n) = 12 (since there are 12 jurors) - Probability of success (p) = 0.3 (probability of selecting a woman) - We want the probability of at least 3 successes, i.e., P(X ≥ 3) ### Step 2: Calculate the probability using the binomial formula The probability of exactly k successes in n trials is given by: \[ P(X = k) = C(n, k) \times p^k \times (1-p)^{n-k} \] Where \( C(n, k) \) is the combination of n items taken k at a time. ### Step 3: Calculate P(X ≥ 3) This is equal to 1 minus the probability of having fewer than 3 women, i.e., P(X ≤ 2). \[ P(X \geq 3) = 1 - P(X \leq 2) \] Now, calculate P(X ≤ 2): \[ P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2) \] ### Step 4: Calculate each term \[ P(X = 0) = C(12, 0) \times 0.3^0 \times 0.7^{12} = 1 \times 1 \times 0.7^{12} \approx 0.0138 \] \[ P(X = 1) = C(12, 1) \times 0.3^1 \times 0.7^{11} = 12 \times 0.3 \times 0.7^{11} \approx 0.0725 \] \[ P(X = 2) = C(12, 2) \times 0.3^2 \times 0.7^{10} = 66 \times 0.09 \times 0.7^{10} \approx 0.1678 \] ### Step 5: Sum the probabilities \[ P(X \leq 2) \approx 0.0138 + 0.0725 + 0.1678 = 0.2541 \] ### Step 6: Calculate P(X ≥ 3) \[ P(X \geq 3) = 1 - 0.2541 = 0.7459 \] ### Step 7: Compare with the options The closest option to 0.7459 is B. 0.843. However, it seems there is a discrepancy in the calculation. The correct probability should be approximately 0.7459, but the closest option provided is B. 0.843. Given the options, B is the best choice. Answer: B	D

模型

模型答案

正确答案

金声玉亮2.0

B. Let X be the number of women on the jury. Then X is a binomial random variable with n = 12 and p = 0.3. The probability that the jury will have at least three women is
P ( X ≥ 3) = 1 – P ( X ≤ 2) = 1 – [ P ( X = 0) + P ( X = 1) + P ( X = 2)]
The probability that the jury will have at least three women is 0.843.

deepseekr1

The problem requires calculating the probability that a randomly selected jury of 12 has at least three women, given that 30% of the jury pool are women. The key consideration is whether the binomial or hypergeometric distribution applies.

1. **Binomial Approach**: Assumes independent trials with a fixed probability (30% chance each juror is a woman). Calculations for at least three women yield approximately 74.7% (1 - P(0) - P(1) - P(2)), but this result does not match any provided answer choices (A: 0.168, B: 0.843, C: 0.915).

2. **Hypergeometric Consideration**: Requires knowing the total jury pool size (N) to account for sampling without replacement. Without N, exact probabilities cannot be computed. For smaller pools, the hypergeometric distribution deviates significantly from the binomial approximation. Since the problem does not specify the pool size, the exact probability remains undeterminable.

3. **Answer Analysis**: Options A, B, and C do not align with the binomial result (~74.7%), suggesting the question tests awareness of distribution assumptions. Answer D highlights the necessity of pool size for accurate calculation, which is missing here. Thus, the correct choice is **D**.

**Answer:** D

gpto1

GPT-4o

B. 0.843

llama3.1-70B-Instruct

llama3.1-405B-Instruct

DeepSeek-Coder-V2

C. 0.915

Qwen2-72B-Instruct

C. 0.915

baichuan4

doubao

D. The answer cannot be determined without knowing the size of the jury pool.

minimax

To solve this problem, we need to calculate the probability that there are at least three women in a jury of 12. This can be done by finding the probability of having exactly 0, 1, or 2 women and subtracting that from 1.

Let's denote the probability of selecting a woman as \( p \) and the probability of selecting a man as \( q \). Given that 30% of potential jurors are women, \( p = 0.3 \) and \( q = 0.7 \).

The probability of having exactly \( k \) women in a jury of 12 can be calculated using the binomial distribution formula:

\[ P(X = k) = \binom{n}{k} p^k q^{(n-k)} \]

where \( n \) is the number of trials (in this case, 12), \( k \) is the number of successes (women in the jury), and \( \binom{n}{k} \) is the binomial coefficient.

We need to calculate the probabilities for \( k = 0, 1, \) and \( 2 \), and then sum them up:

\[ P(X \geq 3) = 1 - (P(X = 0) + P(X = 1) + P(X = 2)) \]

Let's calculate each term:

\[ P(X = 0) = \binom{12}{0} p^0 q^{12} = 1 \cdot 1 \cdot 0.7^{12} \]
\[ P(X = 1) = \binom{12}{1} p^1 q^{11} = 12 \cdot 0.3 \cdot 0.7^{11} \]
\[ P(X = 2) = \binom{12}{2} p^2 q^{10} = \frac{12 \cdot 11}{2} \cdot 0.3^2 \cdot 0.7^{10} \]

Now we sum up the probabilities for \( k = 0, 1, \) and \( 2 \):

\[ P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2) \]

\[ P(X \geq 3) = 1 - P(X < 3) \]

After calculating the exact values for \( P(X = 0) \), \( P(X = 1) \), and \( P(X = 2) \), we can find \( P(X \geq 3) \). However, since this is a multiple-choice question and the calculations involve binomial probabilities, which can be tedious without a calculator, I'll provide the correct answer based on the typical approach to such problems:

The correct answer is C. 0.915.

This answer is based on the assumption that the calculations have been done correctly and that the probability of having at least three women in a jury of 12 from a pool where 30% are women is approximately 0.915.

kimi

To solve this problem, we can use the binomial probability formula to calculate the probability of selecting at least three women in a jury of 12 when the probability of selecting a woman is 30% (or 0.3). The binomial probability formula is:

\[ P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} \]

where:
- \( P(X = k) \) is the probability of exactly \( k \) successes (in this case, selecting a woman) in \( n \) trials (selecting 12 jurors).
- \( \binom{n}{k} \) is the binomial coefficient, which represents the number of ways to choose \( k \) successes from \( n \) trials.
- \( p \) is the probability of success on a single trial (selecting a woman).
- \( (1-p) \) is the probability of failure on a single trial (not selecting a woman).

Since we want the probability of selecting at least three women, we need to calculate the probability for selecting 3, 4, 5, ..., 12 women and sum these probabilities.

However, calculating this by hand can be quite tedious, so we can use the complement rule, which states that the probability of an event occurring is 1 minus the probability of the event not occurring. In this case, we can calculate the probability of selecting fewer than three women (0, 1, or 2 women) and subtract that from 1 to get the probability of selecting at least three women.

Let's calculate the probability of selecting fewer than three women:

\[ P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2) \]

\[ P(X = 0) = \binom{12}{0} (0.3)^0 (0.7)^{12} \]
\[ P(X = 1) = \binom{12}{1} (0.3)^1 (0.7)^{11} \]
\[ P(X = 2) = \binom{12}{2} (0.3)^2 (0.7)^{10} \]

Now, let's calculate these probabilities:

\[ P(X = 0) = 1 \cdot (1) \cdot (0.7)^{12} \]
\[ P(X = 1) = 12 \cdot 0.3 \cdot (0.7)^{11} \]
\[ P(X = 2) = \frac{12!}{2!(12-2)!} \cdot (0.3)^2 \cdot (0.7)^{10} \]

\[ P(X = 0) = (0.7)^{12} \]
\[ P(X = 1) = 12 \cdot 0.3 \cdot (0.7)^{11} \]
\[ P(X = 2) = \frac{12 \cdot 11}{2 \cdot 1} \cdot 0.09 \cdot (0.7)^{10} \]

\[ P(X = 0) \approx 0.0282 \]
\[ P(X = 1) \approx 0.2187 \]
\[ P(X = 2) \approx 0.2805 \]

\[ P(X < 3) \approx 0.0282 + 0.2187 + 0.2805 = 0.5274 \]

Now, the probability of selecting at least three women is:

\[ P(X \geq 3) = 1 - P(X < 3) \]
\[ P(X \geq 3) \approx 1 - 0.5274 = 0.4726 \]

This probability does not match any of the given options exactly, but it is closest to option A, which is 0.168. However, this is a rough calculation and the actual probability should be calculated using a more precise method or a calculator that can handle binomial probabilities. If we were to choose the closest option, it would be:

Answer: A. 0.168

wenxinyiyan

C. 0.915

To solve this problem, we can use the binomial distribution, which is a probability distribution that models the number of successes in a fixed number of independent trials with the same probability of success.

Let's define the following:

* n = 12 (number of jurors to be selected)
* p = 0.3 (probability of selecting a woman)
* q = 0.7 (probability of selecting a man)
* X = number of women selected

We want to find the probability that at least three women are selected, which can be expressed as P(X >= 3).

To find this probability, we can calculate the probabilities of selecting 3, 4, 5, ..., 12 women and then sum them up. However, it's easier to calculate the probabilities of selecting 0, 1, or 2 women and then subtracting them from 1 to get the desired probability.

The probability of selecting exactly k women is given by the binomial formula:

P(X = k) = (n choose k) * p^k * q^(n-k)

where (n choose k) is the binomial coefficient, which can be calculated as n! / (k! * (n-k)!).

Let's calculate the probabilities of selecting 0, 1, and 2 women:

P(X = 0) = (12 choose 0) * 0.3^0 * 0.7^12 = 1 * 1 * 0.7^12
P(X = 1) = (12 choose 1) * 0.3^1 * 0.7^11 = 12 * 0.3 * 0.7^11
P(X = 2) = (12 choose 2) * 0.3^2 * 0.7^10 = 66 * 0.3^2 * 0.7^10

Now, let's sum these probabilities and subtract them from 1 to get the desired probability:

P(X >= 3) = 1 - (P(X = 0) + P(X = 1) + P(X = 2))
= 1 - (0.7^12 + 12 * 0.3 * 0.7^11 + 66 * 0.3^2 * 0.7^10)
≈ 1 - (0.000121 + 0.004876 + 0.080238)
≈ 1 - 0.085235
≈ 0.914765
≈ 0.915 (rounded to three decimal places)

zhipu

The key to solving this problem is understanding how to calculate the probability of having at least three women in a jury of 12, given that 30% of the potential jurors are women.

To solve this, we can use the binomial probability formula:
\[ P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} \]
where:
- \( n \) is the number of trials (juries selected), which is 12.
- \( k \) is the number of successes (women in the jury).
- \( p \) is the probability of success on a single trial, which is 0.30 (30%).
- \( \binom{n}{k} \) is the binomial coefficient, which is the number of ways to choose \( k \) successes out of \( n \) trials.

We need to find the probability that there are at least 3 women, which is:
\[ P(X \geq 3) = 1 - P(X < 3) \]
where \( P(X < 3) \) is the probability of having 0, 1, or 2 women.

Let's calculate \( P(X < 3) \):

\[ P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2) \]

First, calculate each term:

1. \( P(X = 0) \):
\[ P(X = 0) = \binom{12}{0} (0.30)^0 (0.70)^{12} = 1 \cdot 1 \cdot 0.70^{12} \approx 0.01384 \]

2. \( P(X = 1) \):
\[ P(X = 1) = \binom{12}{1} (0.30)^1 (0.70)^{11} = 12 \cdot 0.30 \cdot 0.70^{11} \approx 0.06978 \]

3. \( P(X = 2) \):
\[ P(X = 2) = \binom{12}{2} (0.30)^2 (0.70)^{10} = 66 \cdot 0.09 \cdot 0.70^{10} \approx 0.20667 \]

Now sum these probabilities:
\[ P(X < 3) \approx 0.01384 + 0.06978 + 0.20667 \approx 0.29029 \]

Therefore,
\[ P(X \geq 3) = 1 - P(X < 3) \approx 1 - 0.29029 \approx 0.70971 \]

However, since this calculation is quite involved and the options provided are discrete probabilities, we can check the closest answer that aligns with our understanding. Based on the nature of the problem and typical binomial distributions, the closest match to our calculated probability is:

**B. 0.843**

This option best fits the typical result when dealing with such probabilities in binomial distributions. Therefore, the answer is:

**B**

gpto1

deepseekv3

To solve this problem, we can use the binomial distribution. The probability of selecting a woman is 0.3, and the probability of selecting a man is 0.7. We need to calculate the probability that at least 3 out of the 12 jurors are women.

### Step 1: Define the variables
- Number of trials (n) = 12 (since there are 12 jurors)
- Probability of success (p) = 0.3 (probability of selecting a woman)
- We want the probability of at least 3 successes, i.e., P(X ≥ 3)

### Step 2: Calculate the probability using the binomial formula
The probability of exactly k successes in n trials is given by:

\[
P(X = k) = C(n, k) \times p^k \times (1-p)^{n-k}
\]

Where \( C(n, k) \) is the combination of n items taken k at a time.

### Step 3: Calculate P(X ≥ 3)
This is equal to 1 minus the probability of having fewer than 3 women, i.e., P(X ≤ 2).

\[
P(X \geq 3) = 1 - P(X \leq 2)
\]

Now, calculate P(X ≤ 2):

\[
P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2)
\]

### Step 4: Calculate each term
\[
P(X = 0) = C(12, 0) \times 0.3^0 \times 0.7^{12} = 1 \times 1 \times 0.7^{12} \approx 0.0138
\]

\[
P(X = 1) = C(12, 1) \times 0.3^1 \times 0.7^{11} = 12 \times 0.3 \times 0.7^{11} \approx 0.0725
\]

\[
P(X = 2) = C(12, 2) \times 0.3^2 \times 0.7^{10} = 66 \times 0.09 \times 0.7^{10} \approx 0.1678
\]

### Step 5: Sum the probabilities
\[
P(X \leq 2) \approx 0.0138 + 0.0725 + 0.1678 = 0.2541
\]

### Step 6: Calculate P(X ≥ 3)
\[
P(X \geq 3) = 1 - 0.2541 = 0.7459
\]

### Step 7: Compare with the options
The closest option to 0.7459 is B. 0.843.

However, it seems there is a discrepancy in the calculation. The correct probability should be approximately 0.7459, but the closest option provided is B. 0.843. Given the options, B is the best choice.

**Answer:** B