# Probability

Tossing two fair coins once. Find the conditional probability that both

coins show tails given that

(a )the first coin shows a tail;

(b) at least one coin shows a tail

### 2 個解答

• 最愛解答

Let Ti denote the event that the i-th coin shows a tail.

(a)

Pr(both coins show tails | the first coin shows a tail)

= Pr(T1 and T2 | T1)

= Pr(T1 and T2 and T1) / Pr(T1)

= Pr(T1 and T2) / Pr(T1)

= [(1/2)(1/2)] / (1/2)

= 1/2

(b)

Pr(both coins show tails | at least one coin shows a tail)

= Pr(T1 and T2 | T1 or T2)

= Pr(T1 and T2 and "T1 or T2") / Pr(T1 or T2)

= Pr( (T1 and T2 and T1) or (T1 and T2 and T2) ) / Pr(T1 or T2)

= Pr( (T1 and T2) or (T1 and T2) ) / Pr(T1 or T2)

= Pr(T1 and T2) / Pr(T1 or T2)

= Pr(T1 and T2) / [ Pr(T1) + Pr(T2) - Pr(T1 and T2) ]

= [(1/2)(1/2)] / [1/2 + 1/2 - (1/2)(1/2)]

= (1/4) / (1 - 1/4)

= 1/(4 - 1)

= 1/3

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Alternative Method

There are only four possible situations, with equal chances:

(1st T, 2nd T)

(1st T, 2nd H)

(1st H, 2nd T)

(1st H, 2nd H)

(a)

Pr(both coins show tails | the first coin shows a tail)

= Pr( {(1st T, 2nd T)} | {(1st T, 2nd T), (1st T, 2nd H)} )

= Pr( {(1st T, 2nd T)} and {(1st T, 2nd T), (1st T, 2nd H)} ) /

Pr({(1st T, 2nd T), (1st T, 2nd H)})

= Pr( {(1st T, 2nd T)} ) / Pr({(1st T, 2nd T), (1st T, 2nd H)})

= #{(1st T, 2nd T)} / #{(1st T, 2nd T), (1st T, 2nd H)}

= 1/2

(b)

Pr(both coins show tails | at least one coin shows a tail)

= Pr( {(1st T, 2nd T)} | {(1st T, 2nd T), (1st T, 2nd H), (1st H, 2nd T)} )

= Pr( {(1st T, 2nd T)} and {(1st T, 2nd T), (1st T, 2nd H), (1st H, 2nd T)} ) / Pr( {(1st T, 2nd T), (1st T, 2nd H), (1st H, 2nd T)} )

= Pr( {(1st T, 2nd T)} ) / Pr( {(1st T, 2nd T), (1st T, 2nd H), (1st H, 2nd T)} )

= #{(1st T, 2nd T)} / #{(1st T, 2nd T), (1st T, 2nd H), (1st H, 2nd T)}

= 1/3

2015-02-01 13:00:35 補充：

呢個問題問得好好！

呢個係典型的 conditional probability 會造成的一些有趣的 paradox 的原因。

〔稍後你有時間可以看看 Monte Hall Problem，你會見到很有趣！〕

先答你的問題。

我們明顯知道有四個情況：

1. (T, T)

2. (T, H)

3. (H, T)

4. (H, H)

首先，已知 at least 1 tail，即是否決了 4.，但有可能是 1, 2, 3。

在這三個選項中，問 both tail （即是 1) 的機就，就是三個之中的一個。

而且機會均等，所以是 1/3。

2015-02-01 13:10:22 補充：

從數學看你詢問的那個分子：

P(both tails AND at least 1 tail)

= P( situation(T, T) AND situations{(T, H), (H, T), (T, T)} )

= P( situation(T, T) )

= (1/2)²

= 1/4

你想分子的時候，要用 AND 的概念去思考。

因為你已經把 given 的意思轉化成公式：

P( A | B ) = P( A and B ) / P(B)

那麼你思考分子 P( A and B ) 的一刻就要想 AND，不要再想 given。

2015-02-01 13:13:26 補充：

為了令你更明白本題的精髓，我再舉一個常見的例子。

假設男女機會均等。

(a)

在一個有兩個孩子的家庭中，已知年長的是男孩，問年幼的是男孩的機會。

答案： 1/2

(b)

在一個有兩個孩子的家庭中，已知其中一個是男孩，問另一個是男孩的機會。

答案： 1/3

知否為何兩題的答案不一？

這就是類似意見 002 的有趣之處。

2015-02-01 13:14:55 補充：

設符號標記為（年長，年幼）

可能性：

(男，男)

(男，女)

(女，男)

(女，女)

(a) 己知年長的是男孩，可能性是 { (男，男), (男，女) }

(b) 已知其中一個是男孩，可能性是 { (男，男), (男，女), (女，男) }

就是這個原因。

2015-02-01 13:23:36 補充：

以上我在 (a) 和 (b) 問的：

{年幼的是男孩} 和 {另一個是男孩} 的機會，

其實已經是在各題中相當於 {兩個皆是男孩} 的機會。

看回原題：

你詢問為什麼不是單單的乘 1/2 就是因為你以為情況是 (a) 的類型，但其實題目是問 (b) 的類型。

{at least one coin shows a tail} 類似於 {其中一個是男孩}

{both coins show tails} 類似於 {兩個皆是男孩}

2015-02-02 22:47:58 補充：

Cheers~

╭∧---∧╮

│ .✪‿✪ │

╰/) ⋈ (\\╯

• P(both tails | at least 1 tail) = (1/2)^2 / (1-1/4)

我想問點解 P(both tails AND at least 1 tail) 唔係 1/2 既?

已知有一個tail, 咁要另外果個都係tail唔係只乘 1/2 就得喇咩?

求指教, thanks!

2015-02-02 20:55:58 補充：

非常感謝你咁詳細既解釋!!!!!!!!!

我下次會列晒啲結果出黎, 咁就一目了然~ ^^

THX A MILLION!!