九州大学 システム情報科学府 情報理工学専攻・電気電子工学専攻 2023年8月実施 確率・統計

标签：

• Kyushu-University
• Combinatorial-Probability
• Hypergeometric-Distribution

Author

Casablanca, 祭音Myyura

Description

箱の中に $N_1$ 個の白いボールと $N_2$ 個の黒いボールがあり, その総数を $N = N_1 + N_2$ とする。この箱から $2$ つのボールをランダムに選び, 両方が白いボールである確率は $1/2$ であるとする。

(1) $N_2$ が奇数のとき $N_1$ の最小値を求めよ。

(2) $N_2$ が偶数のとき $N_1$ の最小値を求めよ。

(3) $N$ を値の小さい順に $3$ つ求めよ。

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A box contains $N_1$ white and $N_2$ black balls, and the total number of balls is $N = N_1 + N_2$. When two balls are randomly drawn from the box, the probability that both balls are white is $1/2$.

(1) Find the minimum value of $N_1$ when $N_2$ is an odd number.

(2) Find the minimum value of $N_1$ when $N_2$ is an even number.

(3) Find the three smallest values of $N$.

题目描述

盒中有 $N_1$ 个白球和 $N_2$ 个黑球，总数为 $N=N_1+N_2$。从盒中随机抽取两个球，已知两球均为白球的概率为 $1/2$。

1. 当 $N_2$ 为奇数时，求 $N_1$ 的最小值。
2. 当 $N_2$ 为偶数时，求 $N_1$ 的最小值。
3. 按从小到大的顺序，求 $N$ 最小的三个可能值。

Kai

Let $A$ denote the event "both balls are white", then we have

$$P(A) = \frac{\binom{N_1}{2}}{\binom{N_1+N_2}{2}} = \frac{N_1(N_1 - 1)}{(N_1 + N_2)(N_1 + N_2 - 1)}$$

Since $P(A) = \frac{1}{2}$, we have

$$N_1^2 - N_1 - 2N_1N_2 - N_2^2 + N_2 = 0$$

(1)

Let $N_2 = 2k + 1$. Then we have

$$N_1^2 - (4k + 3)N_1 + (2k + 1) - (2k + 1)^2 = 0$$

from which we have

$$N_1 = \frac{4k + 3 \pm \sqrt{8(2k + 1)^2 + 1}}{2} \tag{i}$$

Only the plus sign gives a positive $N_1$. For $k=0$ it gives $N_1=3$, while for $k\geq1$ it is greater than $(4k+3)/2\geq7/2$. Hence the minimum is $3$.

(2)

Let $N_2 = 2k$. Then we have

$$N_1^2 - (4k + 1)N_1 + 2k - 4k^2 = 0$$

from which we get

$$N_1 = \frac{4k + 1 \pm \sqrt{32k^2 + 1}}{2} \tag{ii}$$

For an admissible solution $k\geq1$, and only the plus sign gives a positive $N_1$. For $k=1,2$, the discriminants $33,129$ are not squares; for $k=3$, it is $17^2$ and $N_1=15$. If $k\geq4$, then $N_1>4k+1/2>15$. Hence the minimum is $15$.

(3)

By (i) and (ii), an integer solution requires $8N_2^2+1$ to be a perfect square.

Let $8N_2^2 + 1 = K^2$, we have $8N_2^2 = (K - 1)(K + 1)$, which implies that $K$ is odd.

Let $K = 2p + 1$. Then we have $2N_2^2 = p(p+1)$. The positive solutions of $K^2-8N_2^2=1$ are generated by

$$K+\sqrt{8}N_2=(3+\sqrt{8})^m \qquad (m\geq 1).$$

The first three are $(K,N_2)=(3,1),(17,6),(99,35)$. Since

$$N_1=\frac{2N_2+1+K}{2},$$

the corresponding values of $N=N_1+N_2$ are $4,21,120$. Both coordinates increase with $m$, so these are the three smallest values.