# Quod Erat Demonstrandum

## 2015/02/01

### 黑白球

Filed under: mathematics,NSS — johnmayhk @ 10:33 上午
Tags: , ,

In a game, Evan has to draw balls from a bag containing 2 black balls and 3 white balls one by one without replacement. If he gets two consecutive black balls, he wins; otherwise he loses. Find the probability that he wins.

P(wins)
=P(BB)+P(WBB)+P(WWBB)+P(WWWBB)
=$\frac{2}{5}\frac{1}{4}+\frac{3}{5}\frac{3}{4}\frac{2}{3}+\frac{3}{5}\frac{2}{4}\frac{2}{3}\frac{1}{2}+\frac{3}{5}\frac{2}{4}\frac{1}{3}$
=$\frac{2}{5}$

$\frac{3}{7}\frac{2}{6}+\frac{4}{7}\frac{3}{6}\frac{2}{5}+\frac{4}{7}\frac{3}{6}\frac{3}{5}\frac{2}{4}+\frac{4}{7}\frac{3}{6}\frac{2}{5}\frac{3}{4}\frac{2}{3}+\frac{4}{7}\frac{3}{6}\frac{2}{5}\frac{1}{4}=\frac{3}{7}$

（一）設盒內共有 n 球（$n\ge 2$），黑球 2 個。求連取 2 黑球之機會。

$P(E)=\frac{2\times (n-1)!}{n!}=\frac{2}{n}$

（二）設盒內共有 n 球，黑球 m 個（$n\ge m$）。求連取 m 黑球之機會。

$P(E)=\frac{(n-m+1)!m!}{n!}=\frac{n-m+1}{C^n_m}$

（三）設盒內共有 n 球，黑球 m 個（$n\ge m\ge 2$）。求連取 2 黑球之機會。

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$P(E)=\frac{C^{n-1}_{m-1}}{C^n_m}=\frac{m}{n}$

（四）設盒內共有 n 球，黑球 m 個（$n\ge m\ge r$）。求連取 r 黑球之機會。

https://johnmayhk.wordpress.com/2012/02/13/core-math-problem-probability/

## 1 則迴響 »

1. Just suggest another solution to the original question:

Since there are 2 black (indistinguishable) balls in a box of 5 balls, the probability that a black ball is drawn at the 1st,2nd,3rd,4th,5th drawings is 2/5.

Under the condition that a black ball is drawn at the i-th drawing (i=1,2,3,4), the (conditional) probability that the other black ball is drawn at the (i+1)th drawing is 1/4.

Hence, the probability that two black balls are drawn at the ith and (i+1)th drawing is (2/5)(1/4)=1/10.

Since i=1,2,3 or 4;

the probability that two black balls are drawn consecutively is 4*(1/10)=2/5.

迴響 由 johnmayhk — 2015/02/02 @ 11:07 上午 | 回應