位置2的问题 A particular illness has a 30% recovery rate within 2 weeks. If Doctors observe a cohort of 10 patients who have that illness for two weeks, what is the probability that at least 6 of those patients will recover in that time? (Please record your answer as a probability value between zero and 1 to three digits after the decimal). 答案A particular illness has a 30% recovery rate within 2 weeks. If Doctors observe a cohort of 10 patients who have that illness for two weeks, what is the probability that at least 6 of those patients will recover in that time? (Please record your answer as a probability value between zero and 1 to three digits after the decimal). [input]Short answer
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位置2的问题 A particular illness has a 30% recovery rate within 2 weeks. If Doctors observe a cohort of 10 patients who have that illness for two weeks, what is the probability that at least 6 of those patients will recover in that time? (Please record your answer as a probability value between zero and 1 to three digits after the decimal). 答案A particular illness has a 30% recovery rate within 2 weeks. If Doctors observe a cohort of 10 patients who have that illness for two weeks, what is the probability that at least 6 of those patients will recover in that time? (Please record your answer as a probability value between zero and 1 to three digits after the decimal). [input]Short answer

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BlackTom AI · Accepted Answer

Question restatement: A illness has a 30% recovery rate within 2 weeks. For a cohort of 10 patients observed for two weeks, what is P(at least 6 recoveries)? Answer format requires a probability between 0 and 1 with three digits after the decimal.

Analysis approach: This is a binomial probability problem. Let X be the number of recoveries among 10 independent patients, where each patient has probability p = 0.3 of recovering in two weeks. Then X ~ Binomial(n = 10, p = 0.3). The event “at least 6 recoveries” is X ≥ 6. We compute this by summing the probabilities for k = 6, 7, 8, 9, 10: P(X ≥ 6) = ∑_{k=6}^{10} C(10,k) (0.3)^k (0.7)^{10−k}.

Compute each term:
- k = 6: C(10,6) = 210; (0.3)^6 ≈ 0.000729; (0.7)^4 ≈ 0.2401; product ≈ 210 × 0.000729 × 0.2401 ≈ 0.03676.
- k = 7: C(10,7) = 120; (0.3)^7 ≈ 0.0002187; (0.7)^3 ≈ 0.343; product ≈ 120 × 0.0002187 × 0.343 ≈ 0.00900.
- k = 8: C(10,8) = 45; (0.3)^8 ≈ 0.00006561; (0.7)^2 = 0.49; product ≈ 45 × 0.00006561 × 0.49 ≈ 0.00145.
- k = 9: C(10,9) = 10; (0.3)^9 ≈ 0.000019683; (0.7)^1 = 0.7; product ≈ 10 × 0.000019683 × 0.7 ≈ 0.00014.
- k = 10: C(10,10) = 1; (0.3)^10 ≈ 0.000005905; (0.7)^0 = 1; product ≈ 0.000005905.

Sum these terms: 0.03676 + 0.00900 + 0.00145 + 0.00014 + 0.0000059 ≈ 0.04735.

Rounding to three digits after the decimal yields 0.047. Interpretation: There is about a 4.7% chance that at least 6 out of 10 recover within two weeks when the individual recovery probability is 30%.

Notes on alternative checks: If a calculator or software is used, you can compute P(X ≥ 6) directly via the binomial cumulative distribution function complement: 1 − P(X ≤ 5). Both methods should converge to approximately 0.047.

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