回復 3# Latte 的帖子
第\(n\)次取出紅球的機率為\({{a}_{n}}\)
\({{a}_{1}}=\frac{4}{7}\)成立
設\({{a}_{k}}=\frac{4}{7}\)成立，可設第\(k\)次取球前，袋中有\(4x\)個紅球，\(3x\)個白球
則
\(\begin{align}
& {{a}_{k+1}}={{a}_{k}}\times \frac{4x+1}{7x+1}+\left( 1-{{a}_{k}} \right)\times \frac{4x}{7x+1} \\
& =\frac{4}{7}\times \frac{4x+1}{7x+1}+\frac{3}{7}\times \frac{4x}{7x+1} \\
& =\frac{4}{7} \\
\end{align}\)
