# Quod Erat Demonstrandum

## 2008/09/01

### 利用對稱性解概率問題

Filed under: Additional / Applied Mathematics,HKALE — johnmayhk @ 11:17 下午

X = 紅色骰子的總點數
Y = 藍色骰子的總點數

「X = Y」，「X > Y」及「X < Y」。

$P(X > Y) = \frac{1 - P(X = Y)}{2}$

X 可取值由 2 至 12，又用所謂對稱性，易知

$P(X = 2) = P(X = 12) = \frac{1}{36}$
$P(X = 3) = P(X = 11) = \frac{2}{36}$
$P(X = 4) = P(X = 10) = \frac{3}{36}$
$P(X = 5) = P(X = 9) = \frac{4}{36}$
$P(X = 6) = P(X = 8) = \frac{5}{36}$

$P(X = 7) = \frac{6}{36}$

$P(X = Y = 2) = P(X = 2)P(Y = 2) = (\frac{1}{36})^2$

$P(X = Y) = 2\{(\frac{1}{36})^2 + (\frac{2}{36})^2 + \dots + (\frac{5}{36})^2\} + (\frac{6}{36})^2$

$n$ 個點把 $(a , b)$ 分成 $n + 1$ 部份，設各部份之長度依次為 $X_1, X_2, \dots, X_{n + 1}$

$X_1, X_2, \dots, X_{n + 1}$$n + 1$ 個隨機變量有同等的分佈，故

$E(X_1) = E(X_2) = \dots = E(X_{n + 1})$ – – – – – – (*)

$X_1 + X_2 + \dots + X_{n + 1} = b - a$

$E(X_1) + E(X_2) + \dots + E(X_{n + 1}) = b - a$

[Hint：仿傚例二]

## 4 則迴響 »

1. 我想問…

你寫的數學式是用什麼軟件編出來的-.-?

例子一很爽, 我也想不出來..也許我笨

迴響 由 Humdrum — 2008/09/08 @ 8:34 下午 | 回應

2. Humdrum,

you may read the following post

http://faq.wordpress.com/2007/02/18/can-i-put-math-or-equations-in-my-posts/

for details

迴響 由 johnmayhk — 2008/09/09 @ 7:56 上午 | 回應

3. I find it much harder to “survive" in form 3…
Form two I can get into the top ten doing my best, but I don’t think so now, I’ve gotta do more than I can, else eliminated.