113花蓮女中

計算第 6 題
n[f(n) - f(n - 1)] = - [f(n - 1) - f(n - 2)]
g(n) = f(n) - f(n - 1)
g(n)/g(n - 1) = -1/n
[g(3)/g(2)][g(4)/g(3)]……[g(n)/g(n - 1)] = (-1)^n * [1/(n!/2)] = (-1)^n * (2/n!)
g(2) = f(2) - f(1) = 1/2
g(n) = (-1)^n * (1/n!)

g(3) = f(3) - f(2) = -1/3!
g(4) = f(4) - f(3) = 1/4!
:
:
g(n) = f(n) - f(n - 1) = (-1)^n * (1/n!)
f(n) = -1/2 - 1/3! + 1/4! - …… + (-1)^n * (1/n!)

更正計算題第7題

設第\(k\)次抽到SSR的機率為\(p_k\)
\(p_1=p_2=...=p_{99}=p\)
\(p_{100}=p+(1-p)^{99}(1-p)=p+(1-p)^{100}\)
\(p_{101}=p+p_1(1-p)^{99}(1-p)=p+p_1(1-p)^{100}\)
\(p_{102}=p+p_2(1-p)^{99}(1-p)=p+p_2(1-p)^{100}\)
...
\(p_{250}=p+p_{150}(1-p)^{99}(1-p)=p+p_{150}(1-p)^{100}\)
因此250次可抽得SSR張數的期望值為
\(p_1+p_2+...+p_{250}\)
\(=250p+[1+p_1+p_2+...+p_{150}](1-p)^{100}\)
\(=250p+[1+p_1+p_2+...+p_{99}](1-p)^{100}+[p_{100}+p_{101}+...+p_{150}](1-p)^{100}\)
\(=250p+[1+99p](1-p)^{100}+[51p+(1+50p)p^{100}](1-p)^{100}\)
\(=250p+(1+150p)(1-p)^{100}+(1+50p)(1-p)^{200}\)

[ 本帖最後由 Jimmy92888 於 2024-6-20 23:16 編輯 ]

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