| 模型 | 模型答案 | 正确答案 |
|---|---|---|
| 金声玉亮2.0 | To find the fraction of slots wasted due to multiple simultaneous transmission attempts, we need to calculate the probability that more than one station transmits at the same time. The probability that a station transmits in a given slot is 0.35, and the probability that it does not transmit is 1 - 0.35 = 0.65. Let's calculate the probability of each relevant scenario: 1. **No station transmits**: All 5 stations do not transmit. The probability of this happening is \(0.65^5\). 2. **Exactly one station transmits**: This can happen in 5 different ways (each of the 5 stations could be the one that transmits). The probability of one specific station transmitting while the others do not is \(0.35 \times 0.65^4\). Since there are 5 ways for this to happen, the total probability is \(5 \times 0.35 \times 0.65^4\). 3. **Two or more stations transmit**: This is the scenario we're interested in to find the fraction of slots wasted. Since we have the probabilities of 0 and 1 station transmitting, we can subtract these from 1 to find the probability of 2 or more stations transmitting simultaneously. Let's calculate these probabilities: 1. Probability no station transmits: \(0.65^5 = 0.116029\) 2. Probability exactly one station transmits: \(5 \times 0.35 \times 0.65^4 = 0.369645\) 3. Probability two or more stations transmit: \(1 - (0.116029 + 0.369645) = 0.514326\) Therefore, the fraction of slots wasted due to multiple simultaneous transmission attempts is approximately **0.514326**. |
0.572 |
| deepseekr1 | To determine the fraction of slots wasted due to multiple simultaneous transmission attempts in a LAN with 5 stations, each transmitting with a probability of 0.35 during each discrete time slot, we need to calculate the probability that a slot is wasted. A slot is wasted if two or more stations transmit simultaneously. First, we calculate the probability that no stations transmit (P(0 transmissions)) and the probability that exactly one station transmits (P(1 transmission)). 1. **Probability that no stations transmit (P(0 transmissions))**: \[ P(0 \text{ transmissions}) = (1 - p)^5 = 0.65^5 \approx 0.116029 \] 2. **Probability that exactly one station transmits (P(1 transmission))**: \[ P(1 \text{ transmission}) = 5 \cdot p \cdot (1 - p)^4 = 5 \cdot 0.35 \cdot 0.65^4 \approx 0.312386 \] 3. **Probability that a slot is wasted (P(wasted))**: \[ P(\text{wasted}) = 1 - P(0 \text{ transmissions}) - P(1 \text{ transmission}) \] Substituting the calculated values: \[ P(\text{wasted}) = 1 - 0.116029 - 0.312386 \approx 1 - 0.428415 \approx 0.571585 \] Thus, the fraction of slots wasted due to multiple simultaneous transmission attempts is approximately 0.5716, or 57.16%. \[ \boxed{0.5716} \] |
0.572 |
| gpto1 | To solve this problem, we need to calculate the probability that, in a given time slot, multiple stations attempt to transmit simultaneously, causing a collision and thus wasting the slot. **Given**: - Number of stations: \( n = 5 \) - Probability that a station attempts to transmit in a slot: \( p = 0.35 \) - Probability that a station does not attempt to transmit: \( q = 1 - p = 0.65 \) **Let \( X \) be the number of stations attempting to transmit in a slot.** We need to find \( P(X \geq 2) \), the probability that at least two stations attempt to transmit simultaneously. First, we'll compute \( P(X = 0) \) and \( P(X = 1) \), then use these to find \( P(X \geq 2) \): 1. **Probability that no stations attempt to transmit (\( X = 0 \))**: Since each station decides independently whether to transmit, the probability that all stations do not transmit is: \[ P(X = 0) = q^n = (0.65)^5 \] Let's compute \( (0.65)^5 \): \[ (0.65)^1 = 0.65 \\ (0.65)^2 = 0.65 \times 0.65 = 0.4225 \\ (0.65)^3 = 0.4225 \times 0.65 = 0.274625 \\ (0.65)^4 = 0.274625 \times 0.65 = 0.17850625 \\ (0.65)^5 = 0.17850625 \times 0.65 = 0.1160290625 \] So, \[ P(X = 0) = 0.1160290625 \] 2. **Probability that exactly one station attempts to transmit (\( X = 1 \))**: Any one of the 5 stations can be the one transmitting, while the others are not. The probability that a specific station attempts to transmit while the others do not is \( p \times q^{n-1} \). Since there are 5 such possibilities, we have: \[ P(X = 1) = n \times p \times q^{n-1} = 5 \times 0.35 \times (0.65)^4 \] We already have \( (0.65)^4 = 0.17850625 \), so: \[ P(X = 1) = 5 \times 0.35 \times 0.17850625 \] Compute \( 0.35 \times 0.17850625 \): \[ 0.35 \times 0.17850625 = 0.0624771875 \] Then multiply by 5: \[ P(X = 1) = 5 \times 0.0624771875 = 0.3123859375 \] 3. **Probability that multiple stations attempt to transmit (\( X \geq 2 \))**: \[ P(X \geq 2) = 1 - P(X = 0) - P(X = 1) \] Substitute the values we have: \[ P(X \geq 2) = 1 - 0.1160290625 - 0.3123859375 = 1 - 0.428415 = 0.571585 \] **Conclusion**: Approximately 57.16% of the slots are wasted due to multiple simultaneous transmission attempts. **Answer**: Approximately 57.16% of the slots are wasted due to simultaneous transmissions by multiple stations. |
0.572 |
| GPT-4o | To solve this problem, we need to determine the fraction of time slots that are wasted due to multiple stations attempting to transmit simultaneously. Let's break down the problem step by step. ### Step 1: Define the probability of a single station attempting to transmit Each station attempts to transmit with a probability \( p = 0.35 \) during each slot. ### Step 2: Determine the total number of stations The total number of stations is \( n = 5 \). ### Step 3: Calculate the probability of exactly \( k \) stations transmitting in a given slot The number of stations trying to transmit in a slot follows a binomial distribution. The probability of exactly \( k \) stations transmitting is given by: \[ P(k) = \binom{n}{k} p^k (1-p)^{n-k} \] Where: - \( \binom{n}{k} \) is the binomial coefficient, representing the number of ways to choose \( k \) stations out of \( n \). - \( p^k \) is the probability that \( k \) stations are transmitting. - \( (1-p)^{n-k} \) is the probability that the remaining \( n-k \) stations are not transmitting. ### Step 4: Calculate the probability of 0 stations transmitting (no transmission) \[ P(0) = \binom{5}{0} (0.35)^0 (0.65)^5 = 1 \cdot 1 \cdot (0.65)^5 \approx 0.116 \] ### Step 5: Calculate the probability of exactly 1 station transmitting (successful transmission) \[ P(1) = \binom{5}{1} (0.35)^1 (0.65)^4 = 5 \cdot 0.35 \cdot (0.65)^4 \approx 0.311 \] ### Step 6: Calculate the total probability of more than 1 station transmitting (collision) The fraction of slots wasted due to collisions is the probability that more than one station transmits. This is calculated as: \[ P(\text{collision}) = 1 - P(0) - P(1) \] Substitute the values from Step 4 and Step 5: \[ P(\text{collision}) = 1 - 0.116 - 0.311 = 0.573 \] ### Step 7: Conclusion The fraction of slots that is wasted due to multiple simultaneous transmission attempts is approximately \( 0.573 \) or 57.3%. So, the solution to the problem is: **The fraction of slots wasted due to multiple simultaneous transmission attempts is 57.3%.** |
0.572 |
| llama3.1-70B-Instruct | To find the fraction of slots wasted due to multiple simultaneous transmission attempts, we need to calculate the probability of two or more stations attempting to transmit during the same slot. Let's break it down step by step: 1. The probability of a single station attempting to transmit during a slot is 0.35. 2. The probability of a single station not attempting to transmit during a slot is 1 - 0.35 = 0.65. 3. Since there are 5 stations, the probability of all 5 stations not attempting to transmit during a slot is (0.65)^5 = 0.116. 4. The probability of exactly one station attempting to transmit during a slot is 5 × 0.35 × (0.65)^4 = 0.401 (using the binomial probability formula). 5. The probability of two or more stations attempting to transmit during a slot is 1 - (probability of no stations transmitting) - (probability of exactly one station transmitting) = 1 - 0.116 - 0.401 = 0.483. Therefore, the fraction of slots wasted due to multiple simultaneous transmission attempts is approximately 0.483 or 48.3%. |
0.572 |
| llama3.1-405B-Instruct | To find the fraction of slots wasted due to multiple simultaneous transmission attempts, we need to calculate the probability of two or more stations attempting to transmit in the same slot. Let's define the probability of a station attempting to transmit in a slot as p = 0.35. The probability of a station not attempting to transmit in a slot is q = 1 - p = 0.65. Since there are 5 stations, the probability of exactly k stations attempting to transmit in a slot is given by the binomial distribution: P(k) = (5 choose k) \* p^k \* q^(5-k) where (5 choose k) is the binomial coefficient. We want to find the probability of two or more stations attempting to transmit in the same slot, which is the sum of the probabilities of exactly 2, 3, 4, and 5 stations attempting to transmit: P(2 or more) = P(2) + P(3) + P(4) + P(5) Using the binomial distribution formula, we can calculate each of these probabilities: P(2) = (5 choose 2) \* (0.35)^2 \* (0.65)^3 = 0.216 P(3) = (5 choose 3) \* (0.35)^3 \* (0.65)^2 = 0.105 P(4) = (5 choose 4) \* (0.35)^4 \* (0.65)^1 = 0.031 P(5) = (5 choose 5) \* (0.35)^5 \* (0.65)^0 = 0.005 Adding these probabilities together, we get: P(2 or more) = 0.216 + 0.105 + 0.031 + 0.005 = 0.357 So, approximately 35.7% of slots are wasted due to multiple simultaneous transmission attempts. |
0.572 |
| DeepSeek-Coder-V2 | To solve the problem of determining the fraction of slots wasted due to multiple simultaneous transmission attempts in a LAN using the contention approach (e.g., CSMA/CD), we need to calculate the probability of having more than one station attempting to transmit in the same time slot. Given: - Number of stations, \( N = 5 \) - Probability of each station transmitting in a slot, \( p = 0.35 \) We need to find the probability that more than one station transmits in a given slot. This can be broken down into the following steps: 1. **Calculate the probability that exactly one station transmits:** \[ P(\text{exactly 1 station transmits}) = \binom{5}{1} \cdot (0.35)^1 \cdot (0.65)^4 \] \[ = 5 \cdot 0.35 \cdot 0.65^4 \] \[ = 5 \cdot 0.35 \cdot 0.17850625 \] \[ = 5 \cdot 0.0624771875 \] \[ = 0.3123859375 \] 2. **Calculate the probability that no station transmits:** \[ P(\text{no station transmits}) = (0.65)^5 \] \[ = 0.65^5 \] \[ = 0.1160290625 \] 3. **Calculate the probability that more than one station transmits:** \[ P(\text{more than 1 station transmits}) = 1 - P(\text{exactly 1 station transmits}) - P(\text{no station transmits}) \] \[ = 1 - 0.3123859375 - 0.1160290625 \] \[ = 1 - 0.428415 \] \[ = 0.571585 \] Thus, the fraction of slots wasted due to multiple simultaneous transmission attempts is approximately \( 0.571585 \). Therefore, the final answer is: \[ \boxed{0.571585} \] |
0.572 |
| Qwen2-72B-Instruct | To find the fraction of slots wasted due to multiple simultaneous transmission attempts, we need to calculate the probability that more than one station transmits at the same time. The probability that a station transmits in a given slot is 0.35, and the probability that it does not transmit is 1 - 0.35 = 0.65. Let's calculate the probability of each relevant scenario: 1. **No station transmits**: All 5 stations do not transmit. The probability of this happening is \(0.65^5\). 2. **Exactly one station transmits**: This can happen in 5 different ways (each of the 5 stations could be the one that transmits). The probability of one specific station transmitting while the others do not is \(0.35 \times 0.65^4\). Since there are 5 ways for this to happen, the total probability is \(5 \times 0.35 \times 0.65^4\). 3. **Two or more stations transmit**: This is the scenario we're interested in to find the fraction of slots wasted. Since we have the probabilities of 0 and 1 station transmitting, we can subtract these from 1 to find the probability of 2 or more stations transmitting simultaneously. Let's calculate these probabilities: 1. Probability no station transmits: \(0.65^5 = 0.116029\) 2. Probability exactly one station transmits: \(5 \times 0.35 \times 0.65^4 = 0.369645\) 3. Probability two or more stations transmit: \(1 - (0.116029 + 0.369645) = 0.514326\) Therefore, the fraction of slots wasted due to multiple simultaneous transmission attempts is approximately **0.514326**. |
0.572 |
| baichuan4 | To solve this problem, we can use the binomial distribution to calculate the probability of having more than one station attempting to transmit in the same slot. Given that there are 5 stations, each attempting to transmit with a probability of 0.35, we can calculate the probability of exactly k stations attempting to transmit simultaneously as follows: P(k stations attempt to transmit) = C(5, k) * (0.35)^k * (1 - 0.35)^(5 - k) where C(5, k) is the combination of 5 things taken k at a time, which can be calculated as: C(5, k) = 5! / (k! * (5 - k)!) We are interested in the cases where k is greater than 1 because we want to find the fraction of slots wasted due to multiple simultaneous transmission attempts. So, we need to sum the probabilities for k = 2 to k = 5: P(wasted slots) = Σ[C(5, k) * (0.35)^k * (1 - 0.35)^(5 - k)] for k = 2 to 5 Let's calculate these probabilities: P(2 stations attempt to transmit) = C(5, 2) * (0.35)^2 * (1 - 0.35)^(5 - 2) = 10 * 0.1225 * 0.3164 = 0.3841 P(3 stations attempt to transmit) = C(5, 3) * (0.35)^3 * (1 - 0.35)^(5 - 3) = 10 * 0.042875 * 0.441 = 0.1886 P(4 stations attempt to transmit) = C(5, 4) * (0.35)^4 * (1 - 0.35)^(5 - 4) = 5 * 0.01225 * 0.605 = 0.0370 P(5 stations attempt to transmit) = C(5, 5) * (0.35)^5 * (1 - 0.35)^(5 - 5) = 1 * 0.005625 * 1 = 0.0056 Now, we sum these probabilities to get the total probability of wasted slots: P(wasted slots) = 0.3841 + 0.1886 + 0.0370 + 0.0056 = 0.6153 Therefore, approximately 61.53% of the slots are wasted due to multiple simultaneous transmission attempts. |
0.572 |
| doubao | The probability that a specific station does not transmit is 1 - 0.35 = 0.65. The probability that all 5 stations do not transmit simultaneously is 0.65^5 ≈ 0.1160. So the probability that at least one station transmits (i.e., there is a multiple simultaneous transmission attempt) is 1 - 0.1160 = 0.8840. So the fraction of slots wasted due to multiple simultaneous transmission attempts is 0.8840. |
0.572 |
| minimax | To solve this problem, we need to calculate the probability of at least two stations attempting to transmit simultaneously and then determine the fraction of slots that are wasted due to these simultaneous transmission attempts. Let's denote the probability that a station attempts to transmit during a slot as \( p \), and the probability that it does not attempt to transmit as \( q \). Given that \( p = 0.35 \), it follows that \( q = 1 - p = 0.65 \). We are interested in the probability that no station attempts to transmit and the probability that exactly one station attempts to transmit, so that we can subtract these from 1 to find the probability of at least two stations transmitting simultaneously. The probability that no station attempts to transmit in a slot is \( q^5 \), since there are 5 independent stations and each has a probability \( q \) of not transmitting. The probability that exactly one station attempts to transmit is a bit more complex, as there are 5 different ways this can happen (any one of the 5 stations could be the one to transmit). The probability for each of these cases is \( p \cdot q^4 \), since one station transmits and the other four do not. So, the total probability for exactly one station transmitting is \( 5 \cdot p \cdot q^4 \). Now, let's calculate these probabilities: - Probability of no station transmitting: \( q^5 = 0.65^5 \) - Probability of exactly one station transmitting: \( 5 \cdot p \cdot q^4 = 5 \cdot 0.35 \cdot 0.65^4 \) The probability of at least two stations transmitting simultaneously is then: \[ 1 - (q^5 + 5 \cdot p \cdot q^4) \] Let's compute the values: \[ q^5 = 0.65^5 \approx 0.116025 \] \[ 5 \cdot p \cdot q^4 = 5 \cdot 0.35 \cdot 0.65^4 \approx 5 \cdot 0.35 \cdot 0.*** \approx 0.*** \] Now, sum these two probabilities: \[ q^5 + 5 \cdot p \cdot q^4 \approx 0.116025 + 0.*** \approx 0.*** \] Finally, subtract this sum from 1 to find the probability of at least two stations transmitting simultaneously: \[ 1 - (q^5 + 5 \cdot p \cdot q^4) \approx 1 - 0.*** \approx 0.*** \] So, approximately 57.17% of the slots are wasted due to multiple simultaneous transmission attempts. |
0.572 |
| kimi | To solve this problem, we can use the concept of the birthday paradox, which deals with the probability that at least two people share the same birthday in a group of people. In this case, we are looking at the probability that at least two stations attempt to transmit at the same time. Let's denote the probability that a station attempts to transmit as \( p = 0.35 \) and the probability that a station does not attempt to transmit as \( q = 1 - p = 0.65 \). The probability that exactly \( k \) stations attempt to transmit in a slot can be calculated using the binomial probability formula: \[ P(X = k) = \binom{n}{k} p^k q^{n-k} \] where \( n \) is the number of stations, \( k \) is the number of stations that attempt to transmit, and \( \binom{n}{k} \) is the binomial coefficient. We are interested in the probability that at least two stations attempt to transmit, which is the sum of the probabilities for 2, 3, 4, and 5 stations attempting to transmit: \[ P(\text{at least 2 transmit}) = P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) \] Let's calculate each term: 1. \( P(X = 2) = \binom{5}{2} (0.35)^2 (0.65)^3 \) 2. \( P(X = 3) = \binom{5}{3} (0.35)^3 (0.65)^2 \) 3. \( P(X = 4) = \binom{5}{4} (0.35)^4 (0.65)^1 \) 4. \( P(X = 5) = \binom{5}{5} (0.35)^5 (0.65)^0 \) Now, let's compute these probabilities: 1. \( P(X = 2) = 10 \times (0.35)^2 \times (0.65)^3 \) 2. \( P(X = 3) = 10 \times (0.35)^3 \times (0.65)^2 \) 3. \( P(X = 4) = 5 \times (0.35)^4 \times (0.65)^1 \) 4. \( P(X = 5) = 1 \times (0.35)^5 \times (0.65)^0 \) Calculating the values: 1. \( P(X = 2) = 10 \times 0.1225 \times 0.274625 = 0.3315375 \) 2. \( P(X = 3) = 10 \times 0.04225 \times 0.42225 = 0.1740125 \) 3. \( P(X = 4) = 5 \times 0.0150875 \times 0.65 = 0.04950625 \) 4. \( P(X = 5) = 1 \times 0.0050125 \times 1 = 0.0050125 \) Adding these probabilities together gives us the total probability of at least two stations attempting to transmit: \[ P(\text{at least 2 transmit}) = 0.3315375 + 0.1740125 + 0.04950625 + 0.0050125 = 0.56056125 \] Therefore, the fraction of slots wasted due to multiple simultaneous transmission attempts is approximately 0.56056125. |
0.572 |
| wenxinyiyan | 要解决这个问题，我们首先需要理解每个站点尝试传输的概率以及多个站点同时尝试传输时浪费的容量。 已知有5个站点，每个站点尝试传输的概率为0.35。我们需要计算的是由于多个站点同时尝试传输而浪费的时隙的比例。 首先，我们计算一个特定站点在一个时隙内尝试传输的概率，这是0.35。然后，我们计算一个时隙内没有站点尝试传输的概率，这是(1 - 0.35)^5，因为有5个站点，每个站点不尝试传输的概率是1 - 0.35。 接下来，我们计算只有一个站点尝试传输的概率。这是5 * 0.35 * (1 - 0.35)^4，因为有5个站点，任何一个站点都可以是尝试传输的那个，而其他4个站点则不尝试传输。 最后，我们计算有多于一个站点尝试传输的概率，这是1减去没有站点尝试传输的概率和只有一个站点尝试传输的概率。这个概率就是浪费的时隙的比例。 进行计算，我们得到： 没有站点尝试传输的概率 = (1 - 0.35)^5 = 0.180 只有一个站点尝试传输的概率 = 5 * 0.35 * (1 - 0.35)^4 = 0.368 有多于一个站点尝试传输的概率 = 1 - 0.180 - 0.368 = 0.452 所以，由于多个站点同时尝试传输而浪费的时隙的比例是0.452。 |
0.572 |
| zhipu | To solve this problem, we need to calculate the probability that more than one station attempts to transmit in the same slot, as this would result in a collision and thus a wasted slot. Here's a step-by-step approach: 1. **Define the probabilities:** - Probability that a single station attempts to transmit, \( p = 0.35 \) - Probability that a single station does not attempt to transmit, \( q = 1 - p = 0.65 \) - Number of stations, \( N = 5 \) 2. **Calculate the probability of exactly one station transmitting:** - For exactly one station to transmit, one station must attempt to transmit while the other four do not. - The probability of this happening is given by the binomial distribution: \[ P(\text{exactly one station transmits}) = \binom{N}{1} p^1 q^{N-1} = 5 \cdot 0.35 \cdot 0.65^4 \] 3. **Calculate the probability of no station transmitting:** - For no station to transmit, all five stations must not attempt to transmit. - The probability of this happening is: \[ P(\text{no station transmits}) = q^N = 0.65^5 \] 4. **Calculate the probability of multiple stations transmitting:** - The probability of multiple stations transmitting is the complement of the probabilities of exactly one station transmitting and no station transmitting. - Therefore: \[ P(\text{multiple stations transmit}) = 1 - P(\text{exactly one station transmits}) - P(\text{no station transmits}) \] Let's plug in the numbers: 1. **Probability of exactly one station transmitting:** \[ P(\text{exactly one station transmits}) = 5 \cdot 0.35 \cdot 0.65^4 \] \[ = 5 \cdot 0.35 \cdot (0.65 \times 0.65 \times 0.65 \times 0.65) \] \[ = 5 \cdot 0.35 \cdot 0.17850625 \] \[ = 5 \cdot 0.06247521875 \] \[ = 0.3123759375 \] 2. **Probability of no station transmitting:** \[ P(\text{no station transmits}) = 0.65^5 \] \[ = 0.65 \times 0.65 \times 0.65 \times 0.65 \times 0.65 \] \[ = 0.1160290625 \] 3. **Probability of multiple stations transmitting:** \[ P(\text{multiple stations transmit}) = 1 - 0.3123759375 - 0.1160290625 \] \[ = 1 - 0.428404999 \] \[ = 0.571595001 \] Therefore, the fraction of slots wasted due to multiple simultaneous transmission attempts is approximately \( 0.5716 \). So, the solution is: \[ \boxed{0.5716} \] |
0.572 |
| gpto1 | To solve this problem, we need to calculate the probability that, in a given time slot, multiple stations attempt to transmit simultaneously, causing a collision and thus wasting the slot. **Given**: - Number of stations: \( n = 5 \) - Probability that a station attempts to transmit in a slot: \( p = 0.35 \) - Probability that a station does not attempt to transmit: \( q = 1 - p = 0.65 \) **Let \( X \) be the number of stations attempting to transmit in a slot.** We need to find \( P(X \geq 2) \), the probability that at least two stations attempt to transmit simultaneously. First, we'll compute \( P(X = 0) \) and \( P(X = 1) \), then use these to find \( P(X \geq 2) \): 1. **Probability that no stations attempt to transmit (\( X = 0 \))**: Since each station decides independently whether to transmit, the probability that all stations do not transmit is: \[ P(X = 0) = q^n = (0.65)^5 \] Let's compute \( (0.65)^5 \): \[ (0.65)^1 = 0.65 \\ (0.65)^2 = 0.65 \times 0.65 = 0.4225 \\ (0.65)^3 = 0.4225 \times 0.65 = 0.274625 \\ (0.65)^4 = 0.274625 \times 0.65 = 0.17850625 \\ (0.65)^5 = 0.17850625 \times 0.65 = 0.1160290625 \] So, \[ P(X = 0) = 0.1160290625 \] 2. **Probability that exactly one station attempts to transmit (\( X = 1 \))**: Any one of the 5 stations can be the one transmitting, while the others are not. The probability that a specific station attempts to transmit while the others do not is \( p \times q^{n-1} \). Since there are 5 such possibilities, we have: \[ P(X = 1) = n \times p \times q^{n-1} = 5 \times 0.35 \times (0.65)^4 \] We already have \( (0.65)^4 = 0.17850625 \), so: \[ P(X = 1) = 5 \times 0.35 \times 0.17850625 \] Compute \( 0.35 \times 0.17850625 \): \[ 0.35 \times 0.17850625 = 0.0624771875 \] Then multiply by 5: \[ P(X = 1) = 5 \times 0.0624771875 = 0.3123859375 \] 3. **Probability that multiple stations attempt to transmit (\( X \geq 2 \))**: \[ P(X \geq 2) = 1 - P(X = 0) - P(X = 1) \] Substitute the values we have: \[ P(X \geq 2) = 1 - 0.1160290625 - 0.3123859375 = 1 - 0.428415 = 0.571585 \] **Conclusion**: Approximately 57.16% of the slots are wasted due to multiple simultaneous transmission attempts. **Answer**: Approximately 57.16% of the slots are wasted due to simultaneous transmissions by multiple stations. |
0.572 |
| deepseekv3 | To determine the fraction of slots wasted due to multiple simultaneous transmission attempts in a CSMA/CD LAN, we can model the situation using probability theory. ### Problem Setup: - **Number of stations (n):** 5 - **Probability of transmission (p):** 0.35 - **Time slots are discrete.** ### Definitions: - A **slot is wasted** if two or more stations attempt to transmit simultaneously, leading to a collision. ### Step 1: Calculate the probability of exactly one station transmitting. The probability that exactly one station transmits while the other four do not is given by the binomial probability formula: \[ P(\text{exactly one transmission}) = \binom{5}{1} \times p^1 \times (1-p)^{4} \] \[ P(\text{exactly one transmission}) = 5 \times 0.35 \times (0.65)^4 \] \[ P(\text{exactly one transmission}) \approx 5 \times 0.35 \times 0.17850625 \approx 0.312 \] ### Step 2: Calculate the probability of no stations transmitting. The probability that no station transmits is: \[ P(\text{no transmission}) = (1-p)^5 \] \[ P(\text{no transmission}) = (0.65)^5 \approx 0.116 \] ### Step 3: Calculate the probability of wasted slots. A slot is wasted if there are two or more transmissions. This is the complement of the events where either zero or one station transmits: \[ P(\text{wasted slot}) = 1 - P(\text{no transmission}) - P(\text{exactly one transmission}) \] \[ P(\text{wasted slot}) = 1 - 0.116 - 0.312 \approx 0.572 \] ### Conclusion: The fraction of slots wasted due to multiple simultaneous transmission attempts is approximately **0.572**, or **57.2%**. |
0.572 |