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統計検定 1級過去問解答/解答例と解説
2015年11月29日 (日) 試験

統計応用問1 [1]

設問の要約

$P(X = k)$ を求めよ．

解答例

\begin{aligned} P (X = 1) = \frac{1}{2} \end{aligned}

$\begin{align} P(X = 1) = \frac{1}{2} \end{align}$

\begin{aligned} P (X = 2) = \frac{1}{2} \times \frac{1}{3} = \frac{1}{2 (2 + 1)} \end{aligned}

$\begin{align} P(X = 2) = \frac{1}{2} \times \frac{1}{3} = \frac{1}{2(2 + 1)} \end{align}$

\begin{aligned} P (X = 3) = (\frac{1}{2} \times \frac{2}{3}) \times \frac{1}{4} = \frac{1}{3 (3 + 1)} \end{aligned}

$\begin{align} P(X = 3) = \left(\frac{1}{2} \times \frac{2}{3}\right) \times \frac{1}{4} = \frac{1}{3(3 + 1)} \end{align}$

\begin{aligned} P (X = 4) = (\frac{1}{2} \times \frac{2}{3} \times \frac{3}{4}) \times \frac{1}{5} = \frac{1}{4 (4 + 1)} \end{aligned}

$\begin{align} P(X = 4) = \left(\frac{1}{2} \times \frac{2}{3} \times \frac{3}{4}\right) \times \frac{1}{5} = \frac{1}{4(4 + 1)} \end{align}$

\begin{aligned} ⋮ \end{aligned}

$\begin{align} \quad\quad\vdots \end{align}$

\begin{aligned} P (X = k) = \frac{1}{k (k + 1)} \end{aligned}

$\begin{align} P(X = k) = \frac{1}{k(k+1)} \end{align}$

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