Quod Erat Demonstrandum

2012/02/13

Core Math 某題:概率

Filed under: NSS,Teaching — johnmayhk @ 11:14 上午
Tags:

In a lucky draw of a car, only 1 key out of 10 can open the door of the car. Chris, Rachel and Mike take turns to draw a key at random without replacement. The person who can open the car door will get the car. Find the probabilities of the following events happening.

(a) Chris will get the car.
(b) Rachel will get the car.
(c) Mike will get the car.

在堂上計算 (a) 時,我寫

P(Chris wins)

=\frac{1}{10}+\frac{9}{10}\frac{8}{9}\frac{7}{8}\frac{1}{7}

+\frac{9}{10}\frac{8}{9}\frac{7}{8}\frac{6}{7}\frac{5}{6}\frac{4}{5}\frac{1}{4}

+\frac{9}{10}\frac{8}{9}\frac{7}{8}\frac{6}{7}\frac{5}{6}\frac{4}{5}\frac{3}{4}\frac{2}{3}\frac{1}{2}\frac{1}{1}

=\frac{4}{10}=\frac{2}{5}

在這之前,我列了三人(Chris, Rachel and Mike)取匙的順序:

C,R,M,C,R,M,C,R,M,C

徐同學問:為何不直接計 \frac{4}{10}?因為 10 個字母中有 4 個 C。

答案正確,但可以這樣計嗎?

我們再看看,用「算字母」方法,得

P(Rachel wins) =\frac{3}{10}
P(Mike wins) = \frac{3}{10}

答案也正確。

先觀察上面計算 P(Chris wins) 的算式,把「複雜」的分數乘法約簡,其實是

\frac{1}{10}+\frac{1}{10}+\frac{1}{10}+\frac{1}{10}

當中 4 項 \frac{1}{10} 代表著以下 4 個概率:

P(Chris selects lucky key at the 1st draw);
P(Chris selects lucky key at the 4th draw);
P(Chris selects lucky key at the 7th draw);
P(Chris selects lucky key at the 10th draw)

即是說,對於 Chris,無論哪一次,抽出中獎車匙的機會都是 \frac{1}{10}

其實對於 Rachel 和 Mike 都是一樣;無論是哪一次,抽出中獎車匙的機會都是 \frac{1}{10}

為何?我們可以如此理解:

三人得匙,可以自己去抽,又或有主持人把匙按以下順序

C,R,M,C,R,M,C,R,M,C

派給他們。中獎車匙只有一條,主持人先派完所有匙後,暫時不公佈哪條才是中獎車匙。

那麼,對於任何一條匙,它是中獎匙的概率就是 \frac{1}{10}

C,R,M,C,R,M,C,R,M,C

上述順序,顯示了 Chris 有 4 個位置(1st,4th,7th,10th)可以中獎,即對應著主持人派給他的 4 條匙。

而每條匙中獎的概率是 \frac{1}{10},那麼 Chris 中獎的概率就是 \frac{4}{10} 了。

又或可以想像,盒子裡有貼上了字母的 10 個球:

{C,R,M,C,R,M,C,R,M,C}

隨便一球被抽出的概率也一樣(都是 \frac{1}{10}),

故抽出 C 球的概率是 \frac{4}{10}

不過,要說明「抽匙」和「派匙」在計算概率時是「等價」的,似乎也不是太易,大家有沒有好方法?

4 則迴響 »

  1. Suppose that John is a lucky person. Namely, his chance of making a right guess is 2 times higher than an ordinary person, then your student’s argument will not work. It is just an coincidence that your student’s interpretation works.

    When I was a secondary school, I drew the tree diagram to solve this type of problem.

    迴響 由 Mt — 2012/02/14 @ 11:04 上午 | 回覆

  2. Can you explain that either distributing the key or picking it on your own are equivalent by the concept of information.
    Since the one who receive the key have no any extra information known before they receive all of the keys (i.e. knowing nothing at the beginning). They only know whether they win the car at the moment they try to use the keys.

    迴響 由 Justin — 2012/02/14 @ 4:57 下午 | 回覆

  3. Thank you Mt and Justin!

    Justin, it is good to think about information. But students may feel puzzled that when picking up keys one by one, people will try the key immediately, when the lucky guy open the door, no more keys will be picked by the followers. However, when distributing keys, everyone gets a key. So, apparently the situations are different. To persuade them, I just asked them to show the probability that the lucky guy appears at the nth draw is always 1/10.

    迴響 由 johnmayhk — 2012/02/14 @ 6:00 下午 | 回覆


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