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]简答题
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Question

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]简答题

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BlackTom AI · Accepted Answer

The problem asks for the probability that at least 6 out of 10 patients recover, given a 30% recovery rate within two weeks. This is a classic binomial probability scenario with parameters n = 10 and p = 0.3.

First, define the random variable X as the number of recoveries among the 10 patients. X follows a Binomial(n=10, p=0.3) distribution. “At least 6 recoveries” translates to P(X ≥ 6).

To evaluate P(X ≥ 6) we sum the probabilities for X = 6, 7, 8, 9, and 10:
- P(X = 6) = C(10,6) * (0.3)^6 * (0.7)^4
- P(X = 7) = C(10,7) * (0.3)^7 * (0.7)^3
- P(X = 8) = C(10,8) * (0.3)^8 * (0.7)^2
- P(X = 9) = C(10,9) * (0.3)^9 * (0.7)^1
- P(X = 10) = C(10,10) * (0.3)^{10} * (0.7)^0

Compute each term step by step, keeping track of powers and combinatorial factors:
- For X = 6: C(10,6) = 210; (0.3)^6 ≈ 0.000729; (0.7)^4 ≈ 0.2401; product ≈ 210 * 0.000729 * 0.2401 ≈ 0.0368.
- For X = 7: C(10,7) = 120; (0.3)^7 ≈ 0.0002187; (0.7)^3 ≈ 0.343; product ≈ 120 * 0.0002187 * 0.343 ≈ 0.0090.
- For X = 8: C(10,8) = 45; (0.3)^8 ≈ 0.00006561; (0.7)^2 = 0.49; product ≈ 45 * 0.00006561 * 0.49 ≈ 0.00145.
- For X = 9: C(10,9) = 10; (0.3)^9 ≈ 0.000019683; (0.7)^1 = 0.7; product ≈ 10 * 0.000019683 * 0.7 ≈ 0.000138.
- For X = 10: C(10,10) = 1; (0.3)^{10} ≈ 0.000005905; (0.7)^0 = 1; product ≈ 0.0000059.

Add these contributions: 0.0368 + 0.0090 + 0.00145 + 0.00014 + 0.000006 ≈ 0.0473.

Thus, P(X ≥ 6) ≈ 0.0473, which, when recorded to three digits after the decimal, is 0.047.

If you want to cross-check, you can also compute the cumulative distribution up to 5 and subtract from 1: P(X ≥ 6) = 1 − P(X ≤ 5). Either route yields the same result to the requested precision.