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東京大学 新領域創成科学研究科 メディカル情報生命専攻 2015年8月実施 問題11

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Author​

zephyr, 祭音Myyura

Description​

Suppose that there is an urn that contains mm black balls and (l−m)(l - m) white balls (0<m<l)(0 < m < l). You randomly draw nn balls with replacement (n>0)(n > 0). Answer the following questions with explanation.

(1) Find the probability that you draw a black ball for the first time at the kk-th draw (1<k<n)(1 < k < n).

(2) Suppose that you have drawn a black ball for the first time at the kk-th draw. Find the probability that you draw one or more black balls in the remaining (n−k)(n - k) draws.

Let XjX_j be a random variable the value of which is 1 if the jj-th ball is black, and 0 otherwise (j=1,…,n)(j = 1, \dots, n). If necessary, you can use the equalities ∑j=1nj=n(n+1)/2\sum_{j=1}^{n} j = n(n + 1)/2, and ∑j=1nj2=n(n+1)(2n+1)/6\sum_{j=1}^{n} j^2 = n(n + 1)(2n + 1)/6.

(3) Find the expected value E[Xj]\mathrm{E}[X_j] of XjX_j.

(4) Let R=∑j=1njXjR = \sum_{j=1}^{n} j X_j. Find the expected value E[R]\mathrm{E}[R] of RR.

(5) Find the variance Var[R]=E[R2]−(E[R])2\mathrm{Var}[R] = \mathrm{E}[R^2] - (\mathrm{E}[R])^2 of RR.

假设有一个包含 mm 个黑球和 (l−m)(l - m) 个白球的罐子 (0<m<l)(0 < m < l)。你随机有放回地抽取 nn 个球 (n>0)(n > 0)。回答以下问题并解释。

(1) 计算第一次在第 kk 次抽到黑球的概率 (1<k<n)(1 < k < n)。

(2) 假设你第一次在第 kk 次抽到黑球。计算在接下来的 (n−k)(n - k) 次抽中至少再抽到一个黑球的概率。

令 XjX_j 是一个随机变量，如果第 jj 个球是黑色的，则其值为 1，否则为 0 (j=1,…,n)(j = 1, \dots, n)。如果需要，你可以使用以下等式：∑j=1nj=n(n+1)/2\sum_{j=1}^{n} j = n(n + 1)/2，∑j=1nj2=n(n+1)(2n+1)/6\sum_{j=1}^{n} j^2 = n(n + 1)(2n + 1)/6。

(3) 计算 XjX_j 的期望值 E[Xj]\mathrm{E}[X_j]。

(4) 令 R=∑j=1njXjR = \sum_{j=1}^{n} j X_j。计算 RR 的期望值 E[R]\mathrm{E}[R]。

(5) 计算 RR 的方差 Var[R]=E[R2]−(E[R])2\mathrm{Var}[R] = \mathrm{E}[R^2] - (\mathrm{E}[R])^2。

题目描述​

一个罐中有 mm 个黑球与 l−ml-m 个白球，其中 0<m<l0<m<l。独立地进行 n>0n>0 次有放回抽取，并对答案说明理由：

  1. 对 1<k<n1<k<n，求第 kk 次抽取时首次抽到黑球的概率。

  2. 已知首次黑球出现在第 kk 次，求余下 n−kn-k 次中至少再出现一个黑球的概率。

  3. 定义

    Xj={1,第 j 次抽到黑球,0,否则,j=1,…,n,X_j= \begin{cases} 1,&\text{第 }j\text{ 次抽到黑球},\\ 0,&\text{否则}, \end{cases} \qquad j=1,\ldots,n,

    求 E[Xj]E[X_j]。

  4. 令加权随机和

    R=∑j=1njXj,R=\sum_{j=1}^{n}jX_j,

    求 E[R]E[R]。

  5. 求

    Var⁡(R)=E[R2]−E[R]2.\operatorname{Var}(R)=E[R^2]-E[R]^2.

计算时可使用

∑j=1nj=n(n+1)2,∑j=1nj2=n(n+1)(2n+1)6.\sum_{j=1}^{n}j=\frac{n(n+1)}2,\qquad \sum_{j=1}^{n}j^2=\frac{n(n+1)(2n+1)}6.

Kai​

(1)​

To find the probability of drawing a black ball for the first time at the kk-th draw, we need to calculate the probability of drawing white balls in the first (k−1)(k-1) draws and a black ball on the kk-th draw.

Let p=mlp = \frac{m}{l} be the probability of drawing a black ball in any given draw, and q=1−p=l−mlq = 1 - p = \frac{l - m}{l} be the probability of drawing a white ball.

The probability of the first k−1k-1 draws being white is qk−1q^{k-1}, and the probability of the kk-th draw being black is pp. Hence, the probability of this event is:

P(Xk=1 for the first time at k)=qk−1p\mathrm{P}(X_k = 1 \text{ for the first time at } k) = q^{k-1}p

(2)​

Given that the first black ball is drawn at the kk-th draw, we want to find the probability of drawing at least one more black ball in the remaining (n−k)(n - k) draws.

The probability of not drawing a black ball in any of the remaining (n−k)(n - k) draws is qn−kq^{n-k}. Therefore, the probability of drawing at least one black ball in the remaining draws is:

1−qn−k1 - q^{n-k}

(3)​

The random variable XjX_j indicates whether the jj-th ball is black. Since the draws are independent, the expected value of XjX_j is simply the probability of drawing a black ball in a single draw:

E[Xj]=p=ml\mathrm{E}[X_j] = p = \frac{m}{l}

(4)​

To find E[R]\mathrm{E}[R], we first compute E[jXj]\mathrm{E}[j X_j] for each jj and then sum over all jj:

E[R]=E[∑j=1njXj]=∑j=1njE[Xj]\mathrm{E}[R] = \mathrm{E}\left[\sum_{j=1}^{n} j X_j\right] = \sum_{j=1}^{n} j \mathrm{E}[X_j]
E[R]=∑j=1nj⋅ml\mathrm{E}[R] = \sum_{j=1}^{n} j \cdot \frac{m}{l}

Using the given summation formula:

∑j=1nj=n(n+1)2\sum_{j=1}^{n} j = \frac{n(n+1)}{2}
E[R]=ml⋅n(n+1)2\mathrm{E}[R] = \frac{m}{l} \cdot \frac{n(n+1)}{2}

(5)​

To find Var[R]\mathrm{Var}[R], we first compute E[R2]\mathrm{E}[R^2] and then subtract (E[R])2(\mathrm{E}[R])^2.

E[R2]=E[(∑j=1njXj)2]\mathrm{E}[R^2] = \mathrm{E}\left[\left(\sum_{j=1}^{n} j X_j\right)^2\right]

We need to compute E[R2]\mathrm{E}[R^2]. Let's express it as:

E[R2]=∑j=1nj2E[Xj2]+2∑1≤j<k≤njkE[XjXk]\mathrm{E}[R^2] = \sum_{j=1}^{n} j^2 \mathrm{E}[X_j^2] + 2 \sum_{1 \leq j < k \leq n} jk \mathrm{E}[X_j X_k]

Since XjX_j is a Bernoulli random variable:

E[Xj2]=E[Xj]=p\mathrm{E}[X_j^2] = \mathrm{E}[X_j] = p

The covariance term E[XjXk]\mathrm{E}[X_j X_k] for j≠kj \neq k is p2p^2 due to independence. Thus,

E[R2]=∑j=1nj2p+2∑1≤j<k≤njkp2\mathrm{E}[R^2] = \sum_{j=1}^{n} j^2 p + 2 \sum_{1 \leq j < k \leq n} jk p^2

Using the given formula for ∑j=1nj2\sum_{j=1}^{n} j^2:

∑j=1nj2=n(n+1)(2n+1)6\sum_{j=1}^{n} j^2 = \frac{n(n + 1)(2n + 1)}{6}
E[R2]=p∑j=1nj2+2p2∑1≤j<k≤njk\mathrm{E}[R^2] = p \sum_{j=1}^{n} j^2 + 2p^2 \sum_{1 \leq j < k \leq n} jk
E[R2]=ml⋅n(n+1)(2n+1)6+2(ml)2⋅∑1≤j<k≤njk\mathrm{E}[R^2] = \frac{m}{l} \cdot \frac{n(n + 1)(2n + 1)}{6} + 2\left(\frac{m}{l}\right)^2 \cdot \sum_{1 \leq j < k \leq n} jk

The variance is

Var[R]=E[R2]−(E[R])2\mathrm{Var}[R] = \mathrm{E}[R^2] - (\mathrm{E}[R])^2

Using

2∑1≤j<k≤njk=(∑j=1nj)2−∑j=1nj2,2\sum_{1\le j<k\le n}jk =\left(\sum_{j=1}^{n}j\right)^2-\sum_{j=1}^{n}j^2,

the terms containing p2(∑j)2p^2\left(\sum j\right)^2 cancel. Hence

Var[R]=(p−p2)∑j=1nj2=p(1−p)n(n+1)(2n+1)6=m(l−m)l2n(n+1)(2n+1)6.\begin{aligned} \mathrm{Var}[R] &=(p-p^2)\sum_{j=1}^{n}j^2\\ &=p(1-p)\frac{n(n+1)(2n+1)}{6}\\ &=\boxed{\frac{m(l-m)}{l^2}\frac{n(n+1)(2n+1)}{6}}. \end{aligned}

Knowledge​

期望值 方差 条件概率

解题技巧和信息​

在处理类似的概率问题时，关键是分解事件的组成部分，并分别计算各个部分的概率。求和公式在计算期望和方差时非常有用。此外，掌握常见的离散概率分布和相关的求和公式可以简化计算过程。

重点词汇​

  • Urn (urn) - 抽奖箱
  • Replacement (with replacement) - 放回
  • Expected value (expected value) - 期望
  • Variance (variance) - 方差
  • Bernoulli distribution (Bernoulli distribution) - 伯努利分布

Reference​

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