David mengerjakan 10 soal pilihan ganda dengan 5 opsi jawaban. Jika David hanya menebak jawaban, hitunglah peluang bahwa dia menjawab benar paling banyak 5 soal. A. 0.967 B. 0.879 C. 0.623 D. 0.377 E. 0.121

Alright, I need to determine the probability that David correctly answers at most 5 out of 10 multiple-choice questions, each with 5 options, by guessing. Let's break this down step by step. First, since David is guessing, the probability of getting any single question correct is 1 out of 5, which is 0.2, and the probability of getting it wrong is 0.8. This scenario can be modeled using the binomial probability formula, which is suitable for situations with a fixed number of independent trials, two possible outcomes, and a constant probability of success. The formula is: \[ P(X = k) = C(n, k) \times p^k \times (1-p)^{n-k} \] Where: - \( n = 10 \) (total number of questions) - \( k \) is the number of correct answers (from 0 to 5) - \( p = 0.2 \) (probability of getting a question correct) - \( C(n, k) \) is the combination of \( n \) items taken \( k \) at a time We need to calculate the probability that David gets at most 5 correct answers, which means we need to sum the probabilities from \( k = 0 \) to \( k = 5 \). So, the probability \( P(X \leq 5) \) is: \[ P(X \leq 5) = \sum_{k=0}^{5} C(10, k) \times (0.2)^k \times (0.8)^{10-k} \] Calculating each term individually: 1. **For \( k = 0 \):** \[ C(10, 0) = 1 \] \[ P(X = 0) = 1 \times (0.2)^0 \times (0.8)^{10} = 1 \times 1 \times 0.1073741824 = 0.1073741824 \] 2. **For \( k = 1 \):** \[ C(10, 1) = 10 \] \[ P(X = 1) = 10 \times (0.2)^1 \times (0.8)^9 = 10 \times 0.2 \times 0.134217728 = 0.268435456 \] 3. **For \( k = 2 \):** \[ C(10, 2) = 45 \] \[ P(X = 2) = 45 \times (0.2)^2 \times (0.8)^8 = 45 \times 0.04 \times 0.16777216 = 0.301989888 \] 4. **For \( k = 3 \):** \[ C(10, 3) = 120 \] \[ P(X = 3) = 120 \times (0.2)^3 \times (0.8)^7 = 120 \times 0.008 \times 0.2097152 = 0.201326592 \] 5. **For \( k = 4 \):** \[ C(10, 4) = 210 \] \[ P(X = 4) = 210 \times (0.2)^4 \times (0.8)^6 = 210 \times 0.0016 \times 0.262144 = 0.088080384 \] 6. **For \( k = 5 \):** \[ C(10, 5) = 252 \] \[ P(X = 5) = 252 \times (0.2)^5 \times (0.8)^5 = 252 \times 0.00032 \times 0.32768 = 0.026424112 \] Now, summing these probabilities: \[ P(X \leq 5) = 0.1073741824 + 0.268435456 + 0.301989888 + 0.201326592 + 0.088080384 + 0.026424112 = 0.99373052 \] Rounding this to three decimal places gives approximately 0.994. However, looking at the provided options: A. 0.967 B. 0.879 C. 0.623 D. 0.377 E. 0.121 None of these options match the calculated probability of approximately 0.994. This discrepancy suggests that there might be an error in the calculation or the options provided. It's possible that the options are rounded differently or there's a different interpretation of the problem. Alternatively, if we consider that David might have some knowledge or strategy, the probability could be different. But based on pure guessing, the calculated probability is approximately 0.994, which is not among the given options.