# 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.

迴響 由 Edmund — 2008/09/09 @ 8:44 下午 | 回應

4. Edmund, “doing your best" should be the best way to enjoy your schooling. Top ten or not, well, take it easy, at least, you are in the elite class.

Though I’d written this blog for 1 year, I really don’t know what is RSS (so poor) and some functions of this blog, like “Blogroll", may be someone may tell me what are they. XP

迴響 由 johnmayhk — 2008/09/10 @ 9:00 上午 | 回應