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 2024年度 確率・統計

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 \(N\).

Kai

Let \(A\) denote the event "both balls are white", then we have

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

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

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

when \(k = 0\), \(N_1\) get the minimum value \(3\).

(2)

Let \(N_2 = 2k\). Then we have

\[ N_1^2 - (4k + 1)N_1 + 2k - 4k^2 = 0 \]

from which we get

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

when \(k = 3\), \(N_1\) get the minimum value \(15\).

(3)

From (i) and (ii) , we easily know that the larger \(N_1\), the larger \(N_2\) we have and \(8N_2^2 + 1\) must be a number of squares.

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 \(2N_2^2 = p(p+1)\). Easy to find that \(p = 1\), \(p = 8\) and \(p = 49\) are three solutions and the corresponding value of \(N\) is \(4\), \(21\) and \(120\).